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Proposition 4.2. Suppose \( X \) is a nonempty connected space. Then every continuous map from \( X \) to a discrete space is constant.
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Proof. Let \( Y \) be a discrete space, and suppose \( f : X \rightarrow Y \) is continuous. Choose any \( x \in X \), and let \( c = f\left( x\right) \) . Because the singleton \( \{ c\} \) is both open and closed in \( Y \), its preimage \( {f}^{-1}\left( c\right) \) is both open and closed in \( X \) . Since it is not empty by hypothesis, it must be all of \( X \) . Thus \( f \) is constant.
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Yes
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Theorem 4.7 (Main Theorem on Connectedness). Let \( X, Y \) be topological spaces and let \( f : X \rightarrow Y \) be a continuous map. If \( X \) is connected, then \( f\left( X\right) \) is connected.
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Proof. By replacing \( Y \) with \( f\left( X\right) \), we may as well assume that \( f \) is surjective. We prove the contrapositive. If \( Y \) is disconnected, then it is the union of two nonempty, disjoint, open subsets \( U, V \) . It follows immediately that \( {f}^{-1}\left( U\right) \) and \( {f}^{-1}\left( V\right) \) disconnect \( X \), so \( X \) is also disconnected.
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Yes
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Theorem 4.7 (Main Theorem on Connectedness). Let \( X, Y \) be topological spaces and let \( f : X \rightarrow Y \) be a continuous map. If \( X \) is connected, then \( f\left( X\right) \) is connected.
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Proof. By replacing \( Y \) with \( f\left( X\right) \), we may as well assume that \( f \) is surjective. We prove the contrapositive. If \( Y \) is disconnected, then it is the union of two nonempty, disjoint, open subsets \( U, V \) . It follows immediately that \( {f}^{-1}\left( U\right) \) and \( {f}^{-1}\left( V\right) \) disconnect \( X \), so \( X \) is also disconnected.
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Yes
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Proposition 4.11. A nonempty subset of \( \mathbb{R} \) is connected if and only if it is a singleton or an interval.
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Proof. Singletons are obviously connected, so we may as well assume that \( J \subseteq \mathbb{R} \) contains at least two points. First assume that \( J \) is an interval. If it is not connected, there are open subsets \( U, V \subseteq \mathbb{R} \) such that \( U \cap J \) and \( V \cap J \) disconnect \( J \) . Choose \( a \in U \cap J, b \in V \cap J \), and assume (interchanging \( U \) and \( V \) if necessary) that \( a < b \) (Fig. 4.3). Then \( \left\lbrack {a, b}\right\rbrack \subseteq J \) because \( J \) is an interval. Since \( U \) and \( V \) are both open, there exists \( \varepsilon > 0 \) such that \( \lbrack a, a + \varepsilon ) \subseteq U \cap J \) and \( (b - \varepsilon, b\rbrack \subseteq V \cap J \) . Let \( c = \sup \left( {U \cap \left\lbrack {a, b}\right\rbrack }\right) \) . By our choice of \( \varepsilon \), we have \( a + \varepsilon \leq c \leq b - \varepsilon \) . In particular, \( c \) is between \( a \) and \( b \), so \( c \in J \subseteq U \cup V \) . But if \( c \) were in \( U \), it would have a neighborhood \( \left( {c - \delta, c + \delta }\right) \subseteq U \), which would contradict the definition of \( c \) . On the other hand, if \( c \) were in \( V \), it would have a neighborhood \( \left( {c - \delta, c + \delta }\right) \subseteq V \) , which is disjoint from \( U \), again contradicting the definition of \( c \) . Therefore, \( J \) is connected. Conversely, assume that \( J \) is not an interval. This means that there exist \( a < c < b \) with \( a, b \in J \) but \( c \notin J \) . Then the sets \( \left( {-\infty, c}\right) \cap J \) and \( \left( {c,\infty }\right) \cap J \) disconnect \( J \), so \( J \) is not connected.
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Yes
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Theorem 4.12 (Intermediate Value Theorem). Suppose \( X \) is a connected topological space, and \( f : X \rightarrow \mathbb{R} \) is continuous. If \( p, q \in X \), then \( f \) attains every value between \( f\left( p\right) \) and \( f\left( q\right) \) .
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Proof. By the main theorem on connectedness, \( f\left( X\right) \) is connected, so it must be a singleton or an interval.
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No
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Theorem 4.15. Path connectedness implies connectedness.
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Proof. Suppose \( X \) is path-connected, and fix \( p \in X \) . For each \( q \in X \), let \( {B}_{q} \) be the image of a path in \( X \) from \( p \) to \( q \) . By Proposition 4.11 and the main theorem on connectedness, each \( {B}_{q} \) is connected. Thus by Proposition 4.9(d), \( X = \mathop{\bigcup }\limits_{{q \in X}}{B}_{q} \) is connected.
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Yes
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Proposition 4.18. If \( X \) is any topological space, its components form a partition of \( X \) .
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Proof. We need to show that the components are disjoint and their union is \( X \) . To see that distinct components are disjoint, suppose \( U \) and \( V \) are components that are not disjoint. Then they have a point in common, and Proposition 4.9(d) implies that \( U \cup V \) is connected. By maximality, therefore, \( U \cup V = U = V \), so \( U \) and \( V \) are not distinct.\n\nTo see that the union of the components is \( X \), let \( x \in X \) be arbitrary. There is at least one connected set containing \( x \), namely \( \{ x\} \) . If \( U \) is the union of all connected sets containing \( x \), then \( U \) is connected by Proposition 4.9(d), and it certainly is maximal, so it is a component containing \( x \) .
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Yes
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## Proposition 4.20 (Properties of Components). Let \( X \) be a nonempty topological space.\n\n(a) Each component of \( X \) is closed in \( X \) .\n\n(b) Any nonempty connected subset of \( X \) is contained in a single component.
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Proof. If \( B \) is any component of \( X \), it follows from Proposition 4.9(c) that \( \bar{B} \) is a connected set containing \( B \) . Since components are maximal connected sets, \( \bar{B} = B \) , so \( B \) is closed.\n\nSuppose \( A \subseteq X \) is connected. Because the components cover \( X \), if \( A \) is nonempty, it has a point in common with some component \( B \) . By Proposition 4.9(d), \( A \cup B \) is connected, so by maximality of \( B, A \cup B \) must be equal to \( B \) . This means that \( A \subseteq B \) .
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Yes
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Proposition 4.25 (Properties of Locally Connected Spaces). Suppose \( X \) is a locally connected space.\n\n(a) Every open subset of \( X \) is locally connected.\n\n(b) Every component of \( X \) is open.
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Proof. If \( U \) is an open subset of \( X \) and \( \mathcal{B} \) is a basis for \( X \) consisting of connected open subsets, then the subset of \( \mathcal{B} \) consisting of sets contained in \( U \) is a basis for \( U \) . This proves (a).\n\nTo prove (b), let \( A \) be a component of \( X \) . If \( p \in A \), then \( p \) has a connected neighborhood \( U \) by local connectedness, and this neighborhood must lie entirely in \( A \) by Proposition 4.20(b). Thus every point of \( A \) has a neighborhood in \( A \), so \( A \) is open.
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Yes
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Proposition 4.26 (Properties of Locally Path-Connected Spaces). Suppose \( X \) is a locally path-connected space.\n\n(a) \( X \) is locally connected.\n\n(b) Every open subset of \( X \) is locally path-connected.\n\n(c) Every path component of \( X \) is open.\n\n(d) The path components of \( X \) are equal to its components.\n\n(e) \( X \) is connected if and only if it is path-connected.
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Proof. Part (a) follows immediately from Theorem 4.15; parts (b) and (c) are proved exactly as in the locally connected case. To prove (d), let \( p \in X \), and let \( A \) and \( B \) be the component and the path component containing \( p \), respectively. By Proposition 4.21(b), we know that \( B \subseteq A \) and \( A \) can be written as a disjoint union of path components, each of which is open in \( X \) and thus in \( A \) . If \( B \) is not the only path component in \( A \), then the sets \( B \) and \( A \smallsetminus B \) disconnect \( A \), which is a contradiction because \( A \) is connected. This proves that \( A = B \) . Finally, for (e), \( X \) is connected if and only if it has exactly one component, which by (d) is the same as having exactly one path component, which in turn is equivalent to being path-connected.
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Yes
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Proposition 4.31. Suppose \( X \) is a topological space, and \( \left( {x}_{i}\right) \) is a sequence of points in \( X \) converging to \( x \in X \) . Then the set \( A = \left\{ {{x}_{i} : i \in \mathbb{N}}\right\} \cup \{ x\} \) is compact in the subspace topology.
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Proof. Suppose \( \mathcal{U} \) is a cover of \( A \) by open subsets of \( X \) . There is some set \( U \in \mathcal{U} \) containing \( x \), and \( U \) must contain \( {x}_{i} \) for all but finitely many \( i \) . Choosing one set in \( \mathcal{U} \) for each of those finitely many elements, we obtain a finite subcover of \( A \) .
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Yes
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Theorem 4.32 (Main Theorem on Compactness). Let \( X \) and \( Y \) be topological spaces, and let \( f : X \rightarrow Y \) be a continuous map. If \( X \) is compact, then \( f\left( X\right) \) is compact.
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Proof. Let \( \mathcal{U} \) be a cover of \( f\left( X\right) \) by open subsets of \( Y \) . For each \( U \in \mathcal{U},{f}^{-1}\left( U\right) \) is an open subset of \( X \) . Since \( \mathcal{U} \) covers \( f\left( X\right) \), every point of \( X \) is in some set \( {f}^{-1}\left( U\right) \) , so the collection \( \left\{ {{f}^{-1}\left( U\right) : U \in \mathcal{U}}\right\} \) is an open cover of \( X \) . By compactness of \( X \) , some finite number of these, say \( \left\{ {{f}^{-1}\left( {U}_{1}\right) ,\ldots ,{f}^{-1}\left( {U}_{k}\right) }\right\} \), cover \( X \) . Then it follows that \( \left\{ {{U}_{1},\ldots ,{U}_{k}}\right\} \) cover \( f\left( X\right) \) .
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Yes
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Lemma 4.34 (Compact Subsets Can Be Separated by Open Subsets). If \( X \) is a Hausdorff space and \( A, B \subseteq X \) are disjoint compact subsets, there exist disjoint open subsets \( U, V \subseteq X \) such that \( A \subseteq U \) and \( B \subseteq V \) .
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Proof. First consider the case in which \( B = \{ q\} \) is a singleton (Fig. 4.5). For each \( p \in A \), there exist disjoint open subsets \( {U}_{p} \) containing \( p \) and \( {V}_{p} \) containing \( q \) by the Hausdorff property. The collection \( \left\{ {{U}_{p} : p \in A}\right\} \) is an open cover of \( A \), so it has a finite subcover: call it \( \left\{ {{U}_{{p}_{1}},\ldots ,{U}_{{p}_{k}}}\right\} \) . Let \( \mathbb{U} = {U}_{{p}_{1}} \cup \cdots \cup {U}_{{p}_{k}} \) and \( \mathbb{V} = \) \( {V}_{{p}_{1}} \cap \cdots \cap {V}_{{p}_{k}} \) . Then \( \mathbb{U} \) and \( \mathbb{V} \) are disjoint open subsets with \( A \subseteq \mathbb{U} \) and \( \{ q\} \subseteq \mathbb{V} \) , so this case is proved.\n\nNext consider the case of a general compact subset \( B \) . The argument above shows that for each \( q \in B \) there exist disjoint open subsets \( {\mathbb{U}}_{q},{\mathbb{V}}_{q} \subseteq X \) such that \( A \subseteq {\mathbb{U}}_{q} \) and \( q \in {\mathbb{V}}_{q} \) . By compactness of \( B \), finitely many of these, say \( \left\{ {{\mathbb{V}}_{{q}_{1}},\ldots ,{\mathbb{V}}_{{q}_{m}}}\right\} \), cover \( B \) . Then setting \( U = {\mathbb{U}}_{{q}_{1}} \cap \cdots \cap {\mathbb{U}}_{{q}_{m}} \) and \( V = {\mathbb{V}}_{{q}_{1}} \cup \cdots \cup {\mathbb{V}}_{{q}_{m}} \) proves the result.
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Yes
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Lemma 4.35 (Tube Lemma). Let \( X \) be any space and let \( Y \) be a compact space. If \( x \in X \) and \( U \subseteq X \times Y \) is an open subset containing \( \{ x\} \times Y \), then there is a neighborhood \( V \) of \( x \) in \( X \) such that \( V \times Y \subseteq U \) .
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Proof. Because product open subsets are a basis for the product topology, for each \( y \in Y \) there is a product open subset \( V \times W \subseteq X \times Y \) such that \( \left( {x, y}\right) \in V \times W \subseteq U \) . The \
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No
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Theorem 4.39. Every closed, bounded interval in \( \mathbb{R} \) is compact.
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Proof. Let \( \left\lbrack {a, b}\right\rbrack \subseteq \mathbb{R} \) be such an interval, and let \( \mathcal{U} \) be a cover of \( \left\lbrack {a, b}\right\rbrack \) by open subsets of \( \mathbb{R} \) . Define a subset \( X \subseteq (a, b\rbrack \) by\n\n\[ X = \{ x \in (a, b\rbrack : \left\lbrack {a, x}\right\rbrack \text{is covered by finitely many sets of}\mathcal{U}\} \text{.} \]\n\nOne of the sets \( {U}_{1} \in \mathcal{U} \) contains \( a \) . Because \( {U}_{1} \) is open, there must be some \( x > a \) such that \( \left\lbrack {a, x}\right\rbrack \subseteq {U}_{1} \), which implies that \( X \) is not empty. Let \( c = \sup X \) . Then there is some set \( {U}_{0} \in \mathcal{U} \) containing \( c \), and because \( {U}_{0} \) is open there is some \( \varepsilon > 0 \) such that \( (c - \varepsilon, c\rbrack \subseteq {U}_{0} \) . By our choice of \( c \), there exists \( x \in X \) such that \( c - \varepsilon < x < c \) . This means that \( \left\lbrack {a, x}\right\rbrack \) is covered by finitely many sets of \( \mathcal{U} \), say \( {U}_{1},\ldots ,{U}_{k} \), and thus \( \left\lbrack {a, c}\right\rbrack \subseteq {U}_{1} \cup \cdots \cup {U}_{k} \cup {U}_{0} \) . If \( c = b \), we are done. On the other hand, if \( c < b \) , then because \( {U}_{0} \) is open there is some \( x > c \) such that \( \left\lbrack {a, x}\right\rbrack \subseteq {U}_{1} \cup \cdots \cup {U}_{k} \cup {U}_{0} \) , which contradicts our choice of \( c \) .
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Yes
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Theorem 4.40 (Heine-Borel). The compact subsets of \( {\mathbb{R}}^{n} \) are exactly the closed and bounded ones.
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Proof. If \( K \subseteq {\mathbb{R}}^{n} \) is compact, it follows from Proposition 4.36 that it is closed and bounded. Conversely, suppose \( K \subseteq {\mathbb{R}}^{n} \) is closed and bounded. Then there is some \( R > 0 \) such that \( K \) is contained in the cube \( {\left\lbrack -R, R\right\rbrack }^{n} \) . Now, \( \left\lbrack {-R, R}\right\rbrack \) is compact by Theorem 4.39, and thus \( {\left\lbrack -R, R\right\rbrack }^{n} \) is compact by Theorem 4.36(d). Because \( K \) is a closed subset of a compact set, it is compact by Theorem 4.36(a).
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Yes
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Theorem 4.41 (Extreme Value Theorem). If \( X \) is a compact space and \( f : X \rightarrow \mathbb{R} \) is continuous, then \( f \) is bounded and attains its maximum and minimum values on \( X \) .
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Proof. By the main theorem on compactness, \( f\left( X\right) \) is a compact subset of \( \mathbb{R} \), so by parts (b) and (c) of Proposition 4.36 it is closed and bounded. In particular, it contains its supremum and infimum.
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Yes
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Lemma 4.42. Compactness implies limit point compactness.
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Proof. Suppose \( X \) is compact, and let \( S \subseteq X \) be an infinite subset. If \( S \) has no limit point in \( X \), then every point \( x \in X \) has a neighborhood \( U \) such that \( U \cap S \) is either empty or \( \{ x\} \) . Finitely many of these neighborhoods cover \( X \) . But since each such neighborhood contains at most one point of \( S \), this implies that \( S \) is finite, which is a contradiction.
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Yes
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Lemma 4.43. For first countable Hausdorff spaces, limit point compactness implies sequential compactness.
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Proof. Suppose \( X \) is first countable, Hausdorff, and limit point compact, and let \( {\left( {p}_{n}\right) }_{n \in \mathbb{N}} \) be any sequence of points in \( X \) . If the sequence takes on only finitely many values, then it has a constant subsequence, which is certainly convergent. So we may suppose it takes on infinitely many values.\n\nBy hypothesis the set of values \( \left\{ {p}_{n}\right\} \) has a limit point \( p \in X \) . If \( p \) is actually equal to \( {p}_{n} \) for infinitely many values of \( n \), again there is a constant subsequence and we are done; so by discarding finitely many terms at the beginning of the sequence if necessary we may assume \( {p}_{n} \neq p \) for all \( n \) . Because \( X \) is first countable, Lemma 2.47 shows that there is a nested neighborhood basis at \( p \), say \( {\left( {B}_{n}\right) }_{n \in \mathbb{N}} \) . For such a neighborhood basis, it is easy to see that any subsequence \( \left( {p}_{{n}_{i}}\right) \) such that \( {p}_{{n}_{i}} \in {B}_{i} \) converges to \( p \) .\n\nSince \( p \) is a limit point, we can choose \( {n}_{1} \) such that \( {p}_{{n}_{1}} \in {B}_{1} \) . Suppose by induction that we have chosen \( {n}_{1} < {n}_{2} < \cdots < {n}_{k} \) with \( {p}_{{n}_{i}} \in {B}_{i} \) . By Proposition 2.39, the sequence takes on infinitely many values in \( {B}_{k + 1} \), so we can choose some \( {n}_{k + 1} > {n}_{k} \) such that \( {p}_{{n}_{k + 1}} \in {B}_{k + 1} \) . This completes the induction, and proves that there is a subsequence \( \left( {p}_{{n}_{i}}\right) \) converging to \( p \) .
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Yes
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Lemma 4.44. For metric spaces and second countable topological spaces, sequential compactness implies compactness.
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Proof. Suppose first that \( X \) is second countable and sequentially compact, and let \( \mathcal{U} \) be an open cover of \( X \) . By Theorem 2.50(c), \( \mathcal{U} \) has a countable subcover \( {\left\{ {U}_{i}\right\} }_{i \in \mathbb{N}} \) . Assume no finite subcollection of \( {U}_{i} \) ’s covers \( X \) . Then for each \( i \) there exists \( {q}_{i} \in X \) such that \( {q}_{i} \notin {U}_{1} \cup \cdots \cup {U}_{i} \) . By hypothesis, the sequence \( \left( {q}_{i}\right) \) has a convergent subsequence \( {q}_{{i}_{k}} \rightarrow q \) . Now, \( q \in {U}_{m} \) for some \( m \) because the \( {U}_{i} \) ’s cover \( X \), and then convergence of the subsequence means that \( {q}_{{i}_{k}} \in {U}_{m} \) for all but finitely many values of \( k \) . But by construction, \( {q}_{{i}_{k}} \notin {U}_{m} \) as soon as \( {i}_{k} \geq m \), which is a contradiction. This proves that second countable sequentially compact spaces are compact.
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Yes
|
Lemma 4.50 (Closed Map Lemma). Suppose \( F \) is a continuous map from a compact space to a Hausdorff space.\n\n(a) \( F \) is a closed map.\n\n(b) If \( F \) is surjective, it is a quotient map.\n\n(c) If \( F \) is injective, it is a topological embedding.\n\n(d) If \( F \) is bijective, it is a homeomorphism.
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Proof. Let \( F : X \rightarrow Y \) be such a map. If \( A \subseteq X \) is closed, then it is compact, because every closed subset of a compact space is compact (Proposition 4.36(a)). Therefore, \( F\left( A\right) \) is compact by the main theorem on compactness, and closed in \( Y \) because compact subsets of Hausdorff spaces are closed (Proposition 4.36(b)). This shows that \( F \) is a closed map. The other three results follow from Proposition 3.69.
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No
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In Example 3.76, we showed that the circle is homeomorphic to a quotient of the unit interval. The only tedious part of the proof was the argument of Example 3.66 showing that the map \( \omega : I \rightarrow {\mathbb{S}}^{1} \) is a quotient map.
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Now we can simply say that \( \omega \) is a quotient map by the closed map lemma.
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Yes
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In Example 3.49, we constructed a quotient space of the square \( I \times I \) by pasting the side boundary segments together and the top and bottom boundary segments together, and we claimed that it was homeomorphic to the torus \( {\mathbb{T}}^{2} = \) \( {\mathbb{S}}^{1} \times {\mathbb{S}}^{1} \).
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Here is a proof. Construct another map \( q : I \times I \rightarrow {\mathbb{T}}^{2} \) by setting \( q\left( {u, v}\right) = \) \( \left( {{e}^{2\pi iu},{e}^{2\pi iv}}\right) \) . By the closed map lemma, this is a quotient map. Since it makes the same identifications as the quotient map we started with, the original quotient of \( I \times I \) must be homeomorphic to the torus by the uniqueness of quotient spaces.
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Yes
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In Proposition 3.36, we used a rather laborious explicit computation to show that the doughnut surface \( D \) is homeomorphic to the torus. Now that proposition can be proved much more simply, as follows.
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Consider the map \( F : {\mathbb{R}}^{2} \rightarrow D \) defined in Example 3.22. The restriction of this map to \( I \times I \) is a quotient map by the closed map lemma. Since it makes the same identifications as the map \( q \) in Example 4.52, the two quotient spaces \( D \) and \( {\mathbb{T}}^{2} \) are homeomorphic. (The homeomorphism is the map that sends \( q\left( {u, v}\right) \) to \( F\left( {u, v}\right) \).)
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Yes
|
In Example 4.54, we defined projective space \( {\mathbb{P}}^{n} \) as a quotient of \( {\mathbb{R}}^{n + 1} \smallsetminus \{ 0\} \) . It can also be represented as a quotient of the sphere with antipodal points identified. Let \( \sim \) denote the equivalence relation on \( {\mathbb{S}}^{n} \) generated by \( x \sim - x \) for each \( x \in {\mathbb{S}}^{n} \) . To see that \( {\mathbb{S}}^{n}/ \sim \) is homeomorphic to \( {\mathbb{P}}^{n} \)
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let \( p : {\mathbb{S}}^{n} \rightarrow {\mathbb{S}}^{n}/ \sim \) denote the quotient map. Consider also the composite map\n\n\[ \n{\mathbb{S}}^{n}\overset{\iota }{ \hookrightarrow }{\mathbb{R}}^{n + 1} \smallsetminus \{ 0\} \overset{q}{ \rightarrow }{\mathbb{P}}^{n} \n\] \n\nwhere \( \iota \) is inclusion and \( q \) is the quotient map defining \( {\mathbb{P}}^{n} \) . Note that \( q \circ \iota \) is a quotient map by the closed map lemma. It makes exactly the same identifications as \( p \) , so by uniqueness of quotient spaces \( {\mathbb{P}}^{n} \) is homeomorphic to \( {\mathbb{S}}^{n}/ \sim \) . This representation also yields an important fact about \( {\mathbb{P}}^{n} \) that might not have been evident from its definition: because it is a quotient of a compact space, it is compact.
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Yes
|
In Example 3.52, we described the space \( {\overline{\mathbb{B}}}^{n}/{\mathbb{S}}^{n - 1} \) obtained by collapsing the boundary of \( {\overline{\mathbb{B}}}^{n} \) to a point. To see that this space is homeomorphic to \( {\mathbb{S}}^{n} \), we just need to construct a surjective continuous map \( q : {\overline{\mathbb{B}}}^{n} \rightarrow {\mathbb{S}}^{n} \) that makes the same identifications; such a map is automatically a quotient map by the closed map lemma. One such map, suggested schematically in Fig. 4.7, is given by the formula
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\[ q\left( x\right) = \left( {2\sqrt{1 - {\left| x\right| }^{2}}x,2{\left| x\right| }^{2} - 1}\right) . \]
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Yes
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In Example 3.53, for any topological space \( X \), we defined the cone \( {CX} \) as the quotient space \( \left( {X \times I}\right) /\left( {X\times \{ 0\} }\right) \) . We can now show that \( C{\mathbb{S}}^{n} \) is homeomorphic to \( {\overline{\mathbb{B}}}^{n + 1} \) .
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The continuous surjective map \( F : {\mathbb{S}}^{n} \times I \rightarrow {\overline{\mathbb{B}}}^{n + 1} \) defined by \( F\left( {x, s}\right) = {sx} \) is a quotient map by the closed map lemma. It maps the set \( {\mathbb{S}}^{n} \times \{ 0\} \) to \( 0 \in {\overline{\mathbb{B}}}^{n + 1} \) and is injective elsewhere, so it makes exactly the same identifications as the quotient map \( {\mathbb{S}}^{n} \times I \rightarrow C{\mathbb{S}}^{n} \) . Thus \( C{\mathbb{S}}^{n} \approx {\overline{\mathbb{B}}}^{n + 1} \) by uniqueness of quotient spaces. It is easy to check that the homeomorphism restricts to the identity on \( {\mathbb{S}}^{n} \) (considered as a subspace of \( C{\mathbb{S}}^{n} \) as in Example 3.65).
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Yes
|
Lemma 4.59. Let \( M \) be an \( n \) -manifold. If \( {B}^{\prime } \subseteq M \) is any coordinate ball and \( \varphi : {B}^{\prime } \rightarrow {B}_{{r}^{\prime }}\left( x\right) \subseteq {\mathbb{R}}^{n} \) is a homeomorphism, then \( {\varphi }^{-1}\left( {{B}_{r}\left( x\right) }\right) \) is a regular coordinate ball whenever \( 0 < r < {r}^{\prime } \) .
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Proof. Suppose \( \varphi : {B}^{\prime } \rightarrow {B}_{{r}^{\prime }}\left( x\right) \) is such a homeomorphism, and \( 0 < r < {r}^{\prime } \) . It is clear that \( \varphi \) restricts to a homeomorphism of \( B = {\varphi }^{-1}\left( {{B}_{r}\left( x\right) }\right) \) onto \( {B}_{r}\left( x\right) \) . The only subtle point that needs to be checked is that \( \varphi \) maps \( \bar{B} \) (the closure of \( B \) in \( M \) ) onto \( {\bar{B}}_{r}\left( x\right) \), or equivalently that\n\n\[{\varphi }^{-1}\left( {{\bar{B}}_{r}\left( x\right) }\right) = \bar{B}\]\n\n(4.3)\n\nRegard (the restriction of) \( {\varphi }^{-1} \) as a map from \( {\bar{B}}_{r}\left( x\right) \) to \( M \) ; then the closed map lemma guarantees that it is a closed map, and therefore (4.3) follows from Proposition 2.30.
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Yes
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Proposition 4.60. Every manifold has a countable basis of regular coordinate balls.
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Proof. Let \( M \) be an \( n \) -manifold. Every point of \( M \) is contained in a Euclidean neighborhood, and since \( M \) is second countable, a countable collection \( \left\{ {{U}_{i} : i \in \mathbb{N}}\right\} \) of such neighborhoods covers \( M \) by Theorem 2.50 . For each of these open subsets \( {U}_{i} \), choose a homeomorphism \( {\varphi }_{i} \) from \( {U}_{i} \) to an open subset \( {\widehat{U}}_{i} \subseteq {\mathbb{R}}^{n} \) . For each \( x \in {\widehat{U}}_{i} \), the fact that \( {\widehat{U}}_{i} \) is open means that there is some positive number \( r\left( x\right) \) such that \( {B}_{r\left( x\right) }\left( x\right) \subseteq {\widehat{U}}_{i} \) .\n\nNow let \( \mathcal{B} \) be the collection of all open subsets of \( M \) of the form \( {\varphi }_{i}^{-1}\left( {{B}_{r}\left( x\right) }\right) \) , where \( x \in {\widehat{U}}_{i} \) is a point with rational coordinates and \( r \) is any positive rational number strictly less than \( r\left( x\right) \) . Then it follows from Lemma 4.59 that each such set is a regular coordinate ball. Since there are only countably many such balls for each \( {U}_{i} \), the collection \( \mathcal{B} \) is countable.\n\nIt remains only to check that the collection \( \mathcal{B} \) is a basis for \( M \), which we leave as an exercise.
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No
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Proposition 4.63. Let \( X \) be a Hausdorff space. The following are equivalent.\n\n(a) \( X \) is locally compact.\n\n(b) Each point of \( X \) has a precompact neighborhood.\n\n(c) \( X \) has a basis of precompact open subsets.
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Proof. Clearly,(c) \( \Rightarrow \) (b) \( \Rightarrow \) (a), so all we have to prove is (a) \( \Rightarrow \) (c). It suffices to show that if \( X \) is a locally compact Hausdorff space, then each point \( x \in X \) has a neighborhood basis of precompact open subsets. Let \( K \subseteq X \) be a compact set containing a neighborhood \( U \) of \( x \) . The collection \( \mathcal{V} \) of all neighborhoods of \( x \) contained in \( U \) is clearly a neighborhood basis at \( x \) .\n\nBecause \( X \) is Hausdorff, \( K \) is closed in \( X \) . If \( V \in \mathcal{V} \), then \( \bar{V} \subseteq K \) (because \( V \subseteq U \subseteq K \) and \( K \) is closed), and therefore \( \bar{V} \) is compact (because a closed subset of a compact set is compact). Thus \( \mathcal{V} \) is the required neighborhood basis.
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Yes
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Proposition 4.64. Every manifold with or without boundary is locally compact.
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Proof. Proposition 4.60 showed that every manifold has a basis of regular coordinate balls. Every regular coordinate ball is precompact, because its closure is homeomorphic to a compact set of the form \( {\bar{B}}_{r}\left( x\right) \subseteq {\mathbb{R}}^{n} \) . The statement for manifolds with boundary follows from Exercise 4.62.
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No
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Lemma 4.65. Let \( X \) be a locally compact Hausdorff space. If \( x \in X \) and \( U \) is any neighborhood of \( x \), there exists a precompact neighborhood \( V \) of \( x \) such that \( \bar{V} \subseteq \) \( U \) .
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Proof. Suppose \( U \) is a neighborhood of \( x \) . If \( W \) is any precompact neighborhood of \( x \), then \( \bar{W} \smallsetminus U \) is closed in \( \bar{W} \) and therefore compact. Because compact subsets can be separated by open subsets in a Hausdorff space, there are disjoint open subsets \( Y \) containing \( x \) and \( {Y}^{\prime } \) containing \( \bar{W} \smallsetminus U \) (Fig. 4.8). Let \( V = Y \cap W \) . Because \( \bar{V} \subseteq \bar{W} \) , \( \bar{V} \) is compact.\n\nBecause \( V \subseteq Y \subseteq X \smallsetminus {Y}^{\prime } \), we have \( \bar{V} \subseteq \bar{Y} \subseteq \overline{X \smallsetminus {Y}^{\prime }} = X \smallsetminus {Y}^{\prime } \), and thus \( \bar{V} \subseteq \) \( \bar{W} \smallsetminus {Y}^{\prime } \) . Now the fact that \( \bar{W} \smallsetminus U \subseteq {Y}^{\prime } \) means that \( \bar{W} \smallsetminus {Y}^{\prime } \subseteq U \), so \( \bar{V} \subseteq U \) .
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Yes
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Proposition 4.66. Any open or closed subset of a locally compact Hausdorff space is a locally compact Hausdorff space.
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Proof. Let \( X \) be a locally compact Hausdorff space. Note that every subspace of \( X \) is Hausdorff, so only local compactness needs to be checked. If \( Y \subseteq X \) is open, Lemma 4.65 says that any point in \( Y \) has a neighborhood whose closure is compact and contained in \( Y \), so \( Y \) is locally compact. Suppose \( Z \subseteq X \) is closed. Any \( x \in Z \) has a precompact neighborhood \( U \) in \( X \) . Since \( \overline{U \cap Z} \) is a closed subset of the compact set \( \bar{U} \), it is compact. Since \( \overline{U \cap Z} \subseteq \bar{Z} = Z \) it follows that \( U \cap Z \) is a precompact neighborhood of \( x \) in \( Z \) . Thus \( Z \) is locally compact.
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Yes
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Theorem 4.68 (Baire Category Theorem). In a locally compact Hausdorff space or a complete metric space, every countable collection of dense open subsets has a dense intersection.
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Proof. Let \( X \) be a space satisfying either of the hypotheses, and suppose \( {\left\{ {V}_{n}\right\} }_{n \in \mathbb{N}} \) is a countable collection of dense open subsets of \( X \) . By Exercise 2.11, to show that \( \mathop{\bigcap }\limits_{n}{V}_{n} \) is dense, it suffices to show that every nonempty open subset of \( X \) contains at least one point of the intersection. Let \( U \subseteq X \) be an arbitrary nonempty open subset.\n\nFirst consider the case in which \( X \) is a locally compact Hausdorff space. Since \( {V}_{1} \) is dense, \( U \cap {V}_{1} \) is nonempty, so by Lemma 4.65 there is a nonempty precompact open subset \( {W}_{1} \) such that \( {\bar{W}}_{1} \subseteq U \cap {V}_{1} \) . Similarly, there is a nonempty precompact open subset \( {W}_{2} \) such that \( {\bar{W}}_{2} \subseteq {W}_{1} \cap {V}_{2} \subseteq U \cap {V}_{1} \cap {V}_{2} \) . Continuing by induction, we obtain a sequence of nested nonempty compact sets \( {\bar{W}}_{1} \supseteq {\bar{W}}_{2} \supseteq \cdots \supseteq {\bar{W}}_{n} \supseteq \cdots \) such that \( {\bar{W}}_{n} \subseteq U \cap {V}_{1} \cap \cdots \cap {V}_{n} \) . By Exercise 4.38, there is a point \( x \in \mathop{\bigcap }\limits_{n}{\bar{W}}_{n} \) , which is clearly in \( U \) and also in \( \mathop{\bigcap }\limits_{n}{V}_{n} \) .\n\nIn the case that \( X \) is a complete metric space, we modify the above proof as follows. At the inductive step, since \( {W}_{n - 1} \cap {V}_{n} \) is open and nonempty, there is some point \( {x}_{n} \) and positive number \( {\varepsilon }_{n} \) such that \( {B}_{{\varepsilon }_{n}}\left( {x}_{n}\right) \subseteq {W}_{n - 1} \cap {V}_{n} \) . Choosing \( {r}_{n} < \) \( \min \left( {{\varepsilon }_{n},1/n}\right) \), we obtain a sequence of nested closed balls such that \( {\bar{B}}_{{r}_{n}}\left( {x}_{n}\right) \subseteq U \cap \) \( {V}_{1} \cap \cdots \cap {V}_{n} \) . Because \( {r}_{n} \rightarrow 0 \), the centers \( \left( {x}_{n}\right) \) form a Cauchy sequence, which converges to a point \( x \in U \cap \left( {\mathop{\bigcap }\limits_{n}{V}_{n}}\right) \) .
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No
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Lemma 4.74. Let \( X \) be a topological space and \( \mathcal{A} \) be a collection of subsets of \( X \) . Then \( \mathcal{A} \) is locally finite if and only \( \overline{\mathcal{A}} \) is locally finite.
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Proof. If \( \overline{\mathcal{A}} \) is locally finite, then it follows immediately that \( \mathcal{A} \) is locally finite. Conversely, suppose \( \mathcal{A} \) is locally finite. Given \( x \in X \), let \( W \) be a neighborhood of \( x \) that intersects only finitely many of the sets in \( \mathcal{A} \), say \( {A}_{1},\ldots ,{A}_{n} \) . If \( W \) contains a point of \( \bar{A} \) for some \( A \in \mathcal{A} \), then Proposition 2.8(d) shows that \( W \) also contains a point of \( A \), so \( A \) must be one of the sets \( {A}_{1},\ldots ,{A}_{n} \) . Thus the same neighborhood \( W \) intersects \( \bar{A} \) for only finitely many \( \bar{A} \in \overline{\mathcal{A}} \) .
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Yes
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Lemma 4.75. If \( \mathcal{A} \) is a locally finite collection of subsets of \( X \), then\n\n\[ \overline{\mathop{\bigcup }\limits_{{A \in \mathcal{A}}}A} = \mathop{\bigcup }\limits_{{A \in \mathcal{A}}}\bar{A} \]\n\n(4.4)
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Proof. Problem 2-4 shows that the right-hand side of (4.4) is contained in the left-hand side even without the assumption of local finiteness, so we need only prove the reverse containment. We will prove the contrapositive: assuming \( x \in X \) is not an element of \( \mathop{\bigcup }\limits_{{A \in \mathcal{A}}}\bar{A} \), we show it is not an element of \( \overline{\mathop{\bigcup }\limits_{{A \in \mathcal{A}}}A} \) either. By Lemma 4.74, \( x \) has a neighborhood \( U \) that intersects only finitely many sets in \( \overline{\mathcal{A}} \), say \( {\bar{A}}_{1},\ldots ,{\bar{A}}_{k} \) . Then \( U \smallsetminus \left( {{\bar{A}}_{1} \cup \cdots \cup {\bar{A}}_{k}}\right) \) is a neighborhood of \( x \) that intersects none of the sets in \( \mathcal{A} \), so \( x \notin \overline{\mathop{\bigcup }\limits_{{A \in \mathcal{A}}}A} \) .
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Yes
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Proposition 4.76. A second countable, locally compact Hausdorff space admits an exhaustion by compact sets.
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Proof. If \( X \) is a locally compact Hausdorff space, it has a basis of precompact open subsets; if in addition \( X \) is second countable, it is covered by countably many such sets. Let \( {\left\{ {U}_{i}\right\} }_{i = 1}^{\infty } \) be such a countable cover.\n\nTo prove the theorem, it suffices to construct a sequence \( {\left( {K}_{j}\right) }_{j = 1}^{\infty } \) of compact sets satisfying \( {U}_{j} \subseteq {K}_{j} \) and \( {K}_{j} \subseteq \operatorname{Int}{K}_{j + 1} \) for each \( j \) . We construct such a sequence by induction.\n\nBegin by setting \( {K}_{1} = {\bar{U}}_{1} \) . Now assume by induction that we have compact sets \( {K}_{1},\ldots ,{K}_{n} \) satisfying the required conditions. Because \( {K}_{n} \) is compact, there is some integer \( {k}_{n} \) such that \( {K}_{n} \subseteq {U}_{1} \cup \cdots \cup {U}_{{k}_{n}} \) . If we define \( {K}_{n + 1} \) to be \( {\bar{U}}_{1} \cup \cdots \cup {\bar{U}}_{{k}_{n}} \) , then \( {K}_{n + 1} \) is a compact set whose interior contains \( {K}_{n} \) . If in addition we choose \( {k}_{n} > n + 1 \), then we also have \( {U}_{n + 1} \subseteq {K}_{n + 1} \) . This completes the induction.
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Yes
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Theorem 4.77 (Paracompactness Theorem). Every second countable, locally compact Hausdorff space (and in particular, every topological manifold with or without boundary) is paracompact.
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Proof. Suppose \( X \) is a second countable, locally compact Hausdorff space, and \( \mathcal{U} \) is an open cover of \( X \) . Let \( {\left( {K}_{j}\right) }_{j = 1}^{\infty } \) be an exhaustion of \( X \) by compact sets. For each \( j \), let \( {A}_{j} = {K}_{j + 1} \smallsetminus \operatorname{Int}{K}_{j} \) and \( {W}_{j} = \operatorname{Int}{K}_{j + 2} \smallsetminus {K}_{j - 1} \) (where we interpret \( {K}_{j} \) as the empty set if \( j < 1 \) ). Then \( {A}_{j} \) is a compact subset contained in the open subset \( {W}_{j} \) (see Fig. 4.10). For each \( x \in {A}_{j} \), choose \( {U}_{x} \in \mathcal{U} \) containing \( x \), and let \( {V}_{x} = {U}_{x} \cap {W}_{j} \) . The collection of all such sets \( {V}_{x} \) as \( x \) ranges over \( {A}_{j} \) is an open cover of \( {A}_{j} \), and thus has a finite subcover. The union of all such finite subcovers as \( j \) ranges over \( \mathbb{N} \) forms an open cover of \( M \) that refines \( \mathcal{U} \) . Because \( {W}_{j} \) intersects \( {W}_{{j}^{\prime }} \) only for \( j - 2 \leq {j}^{\prime } \leq j + 2 \), the resulting cover is locally finite.
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Yes
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Lemma 4.80. Let \( X \) be a Hausdorff space. Then \( X \) is normal if and only if it satisfies the following condition: whenever \( A \) is a closed subset of \( X \) and \( U \) is a neighborhood of \( A \), there exists a neighborhood \( V \) of \( A \) such that \( \bar{V} \subseteq U \) .
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Proof. This is easily seen to be equivalent to the definition of normality by taking \( B = X \smallsetminus U \) .
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No
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Theorem 4.81. Every paracompact Hausdorff space is normal.
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Proof. Suppose \( X \) is a paracompact Hausdorff space, and let \( A \) and \( B \) be disjoint closed subsets of \( X \) . Just as in the proof of Lemma 4.34, we begin with the special case in which \( B = \{ q\} \) is a singleton; in other words, we prove that \( X \) is regular. For each \( p \in A \), because \( X \) is Hausdorff there exist disjoint open subsets \( {U}_{p} \) and \( {V}_{p} \) such that \( p \in {U}_{p} \) and \( q \in {V}_{p} \) . The collection \( \left\{ {{U}_{p} : p \in A}\right\} \cup \{ X \smallsetminus A\} \) is an open cover of \( X \) . By paracompactness, it has a locally finite open refinement \( \mathcal{W} \) . Each of the open subsets in \( W \) is contained either in \( {U}_{p} \) for some \( p \), or in \( X \smallsetminus A \) . Let \( \mathcal{U} \) be the collection of all the sets in \( \mathcal{W} \) that are contained in some \( {U}_{p} \) . Then \( \mathcal{U} \) is still locally finite, and it is an open cover of \( A \) . Moreover, if \( U \in \mathcal{U} \), then there is a neighborhood \( {V}_{p} \) of \( q \) disjoint from \( U \), so \( \bar{U} \) does not contain \( q \) .\n\nLet \( \mathbb{U} = \mathop{\bigcup }\limits_{{U \in \mathcal{U}}}U \), and \( \mathbb{V} = X \smallsetminus \overline{\mathbb{U}} \) . Because \( \mathcal{U} \) is locally finite, \( \overline{\mathbb{U}} = \mathop{\bigcup }\limits_{{U \in \mathcal{U}}}\bar{U} \) by Lemma 4.75. Thus \( q \notin \overline{\mathbb{U}} \), so \( \mathbb{U} \) and \( \mathbb{V} \) are disjoint open subsets of \( X \) containing \( A \) and \( q \), respectively. This completes the proof that \( X \) is regular.\n\nNext consider arbitrary disjoint closed subsets \( A \) and \( B \) . Exactly the same argument works in this case, using regularity in place of the Hausdorff condition.
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Yes
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Corollary 4.83 (Existence of Bump Functions). Let \( X \) be a normal space. If \( A \) is a closed subset of \( X \) and \( U \) is a neighborhood of \( A \), there exists a bump function for A supported in \( U \) .
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Proof. Just apply Urysohn’s Lemma with \( B = X \smallsetminus U \) .
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Yes
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Lemma 4.84. Suppose \( X \) is a paracompact Hausdorff space. If \( \mathcal{U} = {\left( {U}_{\alpha }\right) }_{\alpha \in A} \) is an indexed open cover of \( X \), then \( \mathcal{U} \) admits a locally finite open refinement \( \mathcal{V} = \) \( {\left( {V}_{\alpha }\right) }_{\alpha \in A} \) indexed by the same set, such that \( {\bar{V}}_{\alpha } \subseteq {U}_{\alpha } \) for each \( \alpha \) .
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Proof. By Lemma 4.80, each \( x \in X \) has a neighborhood \( {Y}_{x} \) such that \( {\bar{Y}}_{x} \subseteq {U}_{\alpha } \) for some \( \alpha \in A \) . The open cover \( \left\{ {{Y}_{x} : x \in X}\right\} \) has a locally finite open refinement. Let us index this refinement by some set \( B \), and denote it by \( Z = {\left( {Z}_{\beta }\right) }_{\beta \in B} \) . For each \( \beta \) , there is some \( x \in X \) such that \( {Z}_{\beta } \subseteq {Y}_{x} \), and therefore there is some \( \alpha \in A \) such that \( {\bar{Z}}_{\beta } \subseteq {\bar{Y}}_{x} \subseteq {U}_{\alpha } \) . Define a function \( a : B \rightarrow A \) by choosing some such index \( \alpha \in A \) for each \( \beta \in B \), and setting \( a\left( \beta \right) = \alpha \) .\n\nFor each \( \alpha \in A \), define an open subset \( {V}_{\alpha } \subseteq X \) by\n\n\[ \n{V}_{\alpha } = \mathop{\bigcup }\limits_{{\beta : a\left( \beta \right) = \alpha }}{Z}_{\beta }\n\]\n\n(If there are no indices \( \beta \) such that \( a\left( \beta \right) = \alpha \), then \( {V}_{\alpha } = \varnothing \) .) Because the family \( Z \) is locally finite, the closure of \( {V}_{\alpha } \) is equal to \( \mathop{\bigcup }\limits_{{\beta : a\left( \beta \right) = \alpha }}{\bar{Z}}_{\beta } \) (Lemma 4.75), which is contained in \( {U}_{\alpha } \) as required.
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Yes
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Theorem 4.85 (Existence of Partitions of Unity). Let \( X \) be a paracompact Haus-dorffspace. If \( \mathcal{U} \) is any indexed open cover of \( X \), then there is a partition of unity subordinate to \( \mathcal{U} \) .
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Proof. Let \( \mathcal{U} = {\left( {U}_{\alpha }\right) }_{\alpha \in A} \) be an indexed open cover of \( X \) . Applying Lemma 4.84 twice, we obtain locally finite open covers \( \mathcal{V} = {\left( {V}_{\alpha }\right) }_{\alpha \in A} \) and \( \mathcal{W} = {\left( {W}_{\alpha }\right) }_{\alpha \in A} \) such that \( {\bar{W}}_{\alpha } \subseteq {V}_{\alpha } \) and \( {\bar{V}}_{\alpha } \subseteq {U}_{\alpha } \) . Now, for each \( \alpha \in A \), let \( {f}_{\alpha } : X \rightarrow \left\lbrack {0,1}\right\rbrack \) be a bump function for \( {\bar{W}}_{\alpha } \) supported in \( {V}_{\alpha } \) . Define \( f : X \rightarrow \mathbb{R} \) by \( f\left( p\right) = \mathop{\sum }\limits_{\alpha }{f}_{\alpha }\left( p\right) \) . Because supp \( {f}_{\alpha } \subseteq {V}_{\alpha } \), the family of supports (supp \( {f}_{\alpha }{)}_{\alpha \in A} \) is locally finite; thus each point of \( X \) has a neighborhood on which only finitely many terms of this sum are nonzero, so \( f \) is continuous. Because the sets \( \left\{ {W}_{\alpha }\right\} \) cover \( X \), for each \( p \in X \) there is at least one \( \alpha \) such that \( p \in {W}_{\alpha } \) and thus \( {f}_{\alpha }\left( p\right) = 1 \), so \( f \) is everywhere positive. Therefore, we can define \( {\psi }_{\alpha }\left( p\right) = \) \( {f}_{\alpha }\left( p\right) /f\left( p\right) \), and we see that \( {\psi }_{\alpha } \) is continuous, \( 0 \leq {\psi }_{\alpha }\left( p\right) \leq 1 \), and \( \mathop{\sum }\limits_{\alpha }{\psi }_{\alpha }\left( p\right) = 1 \) everywhere on \( X \) . Thus \( {\left( {\psi }_{\alpha }\right) }_{\alpha \in A} \) is the desired partition of unity.
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Yes
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Theorem 4.86 (Embeddability of Compact Manifolds). Every compact manifold is homeomorphic to a subset of some Euclidean space.
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Proof. Suppose \( M \) is a compact \( n \) -manifold. By compactness we can obtain a cover of \( M \) by finitely many open subsets \( {U}_{1},\ldots ,{U}_{k} \), each of which is homeomorphic to \( {\mathbb{R}}^{n} \) . For each \( i \), let \( {\varphi }_{i} : {U}_{i} \rightarrow {\mathbb{R}}^{n} \) be a homeomorphism. Let \( \left( {\psi }_{i}\right) \) be a partition of unity subordinate to this cover, and define functions \( {F}_{i} : M \rightarrow {\mathbb{R}}^{n} \) by\n\n\[ \n{F}_{i}\left( x\right) = \left\{ \begin{array}{ll} {\psi }_{i}\left( x\right) {\varphi }_{i}\left( x\right) , & x \in {U}_{i} \\ 0, & x \in M \smallsetminus \operatorname{supp}{\psi }_{i}. \end{array}\right.\n\]\n\nEach \( {F}_{i} \) is continuous by the gluing lemma.\n\nNow define \( F : M \rightarrow {\mathbb{R}}^{{nk} + k} \) by\n\n\[ \nF\left( x\right) = \left( {{F}_{1}\left( x\right) ,\ldots ,{F}_{k}\left( x\right) ,{\psi }_{1}\left( x\right) ,\ldots ,{\psi }_{k}\left( x\right) }\right) .\n\]\n\nThen \( F \) is continuous, and if we can show it is injective, it is a topological embedding by the closed map lemma.\n\nSuppose \( F\left( x\right) = F\left( y\right) \) for some points \( x, y \in M \) . Since \( \mathop{\sum }\limits_{i}{\psi }_{i}\left( x\right) \equiv 1 \), there is some \( i \) such that \( {\psi }_{i}\left( x\right) > 0 \) and therefore \( x \in {U}_{i} \) . Because \( F\left( x\right) = F\left( y\right) \) implies \( {\psi }_{i}\left( y\right) = {\psi }_{i}\left( x\right) > 0 \), it follows that \( y \in {U}_{i} \) as well. Then we see that \( {F}_{i}\left( x\right) = {F}_{i}\left( y\right) \) implies \( {\varphi }_{i}\left( x\right) = {\varphi }_{i}\left( y\right) \), which in turn implies that \( x = y \) since \( {\varphi }_{i} \) is injective on \( {U}_{i} \) .
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Yes
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Theorem 4.88 (Zero Sets of Continuous Functions). Suppose \( M \) is a topological manifold, and \( B \subseteq M \) is any closed subset. Then there exists a continuous function \( f : M \rightarrow \lbrack 0,\infty ) \) whose zero set is exactly \( B \) .
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Proof. First, consider the special case in which \( M = {\mathbb{R}}^{n} \) and \( B \subseteq {\mathbb{R}}^{n} \) is a closed subset. It is straightforward to check that\n\n\[ u\left( x\right) = \inf \{ \left| {x - y}\right| : y \in B\} \]\n\ndoes the trick. (This function \( u \) is called the distance to \( \mathbf{B} \) .)\n\nNow, let \( M \) be an arbitrary \( n \) -manifold and let \( B \) be a closed subset of \( M \) . Let \( \mathcal{U} = {\left( {U}_{\alpha }\right) }_{\alpha \in A} \) be a cover of \( M \) by open subsets homeomorphic to \( {\mathbb{R}}^{n} \), and let \( {\left( {\psi }_{\alpha }\right) }_{\alpha \in A} \) be a subordinate partition of unity. For each \( \alpha \), the construction in the preceding paragraph yields a continuous function \( {u}_{\alpha } : {U}_{\alpha } \rightarrow \lbrack 0,\infty ) \) such that \( {u}_{\alpha }^{-1}\left( 0\right) = B \cap {U}_{\alpha } \) . Define \( f : M \rightarrow \mathbb{R} \) by\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{\alpha \in A}}{\psi }_{\alpha }\left( x\right) {u}_{\alpha }\left( x\right) \]\n\nwhere each summand is to be interpreted as zero outside the support of \( {\psi }_{\alpha } \) . Each term in this sum is continuous by the gluing lemma, and only finitely many terms are nonzero in a neighborhood of each point, so this defines a continuous function on \( M \) . It is easy to check that it is zero exactly on \( B \) .
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Yes
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Corollary 4.89. Suppose \( M \) is a topological manifold, and \( A, B \) are disjoint closed subsets of \( M \). Then there exists a continuous function \( f : M \rightarrow \left\lbrack {0,1}\right\rbrack \) such that \( {f}^{-1}\left( 1\right) = A \) and \( {f}^{-1}\left( 0\right) = B \).
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Proof. Using the previous theorem, we can find \( u, v : M \rightarrow \lbrack 0,\infty ) \) such that \( u \) vanishes exactly on \( A \) and \( v \) vanishes exactly on \( B \), and then the function \( f\left( x\right) = \) \( v\left( x\right) /\left( {u\left( x\right) + v\left( x\right) }\right) \) satisfies the conclusion of the corollary.
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Yes
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Theorem 4.90 (Existence of Exhaustion Functions). Every manifold admits a positive exhaustion function.
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Proof. Let \( M \) be a manifold, let \( \left\{ {U}_{i}\right\} \) be a countable open cover of \( M \) by precom-pact open subsets, and let \( \left\{ {\psi }_{i}\right\} \) be a partition of unity subordinate to this cover. Define \( f : M \rightarrow \mathbb{R} \) by\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{k = 1}}^{\infty }k{\psi }_{k}\left( x\right) \]\n\nThen \( f \) is continuous because only finitely many terms are nonzero in a neighborhood of each point, and positive because \( f\left( x\right) \geq \mathop{\sum }\limits_{k}{\psi }_{k}\left( x\right) = 1 \) . For any positive integer \( m \), if \( x \notin \mathop{\bigcup }\limits_{{k = 1}}^{m}{\bar{U}}_{k} \), then \( {\psi }_{k}\left( x\right) = 0 \) for \( 1 \leq k \leq m \), so\n\n\[ f\left( x\right) = \mathop{\sum }\limits_{{k = m + 1}}^{\infty }k{\psi }_{k}\left( x\right) > \mathop{\sum }\limits_{{k = m + 1}}^{\infty }m{\psi }_{k}\left( x\right) = m\mathop{\sum }\limits_{{k = 1}}^{\infty }{\psi }_{k}\left( x\right) = m. \]\n\nThe contrapositive of this last statement is that \( f\left( x\right) \leq m \) implies \( x \in \mathop{\bigcup }\limits_{{k = 1}}^{m}{\bar{U}}_{k} \) . Let \( c \in \mathbb{R} \) be arbitrary, and let \( m \) be any positive integer greater than \( c \) . It follows that \( {f}^{-1}(\left( {-\infty, c\rbrack }\right) \) is a closed subset of the compact set \( \mathop{\bigcup }\limits_{{k = 1}}^{m}{\bar{U}}_{k} \), and is therefore compact.
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Yes
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Lemma 4.91. Suppose \( X \) is a first countable Hausdorff space. A sequence in \( X \) diverges to infinity if and only if it has no convergent subsequence.
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Proof. Assume first that \( \left( {x}_{i}\right) \) is a sequence in \( X \) that diverges to infinity. If there is a subsequence \( \left( {x}_{{i}_{j}}\right) \) that converges to \( x \in X \), then the set \( K = \left\{ {{x}_{{i}_{j}} : j \in \mathbb{N}}\right\} \cup \{ x\} \) is compact (see Proposition 4.31) and contains infinitely many terms of the sequence, which is a contradiction. (This implication does not require the hypothesis that \( X \) is first countable and Hausdorff.)\n\nConversely, assume that \( \left( {x}_{i}\right) \) has no convergent subsequence. If \( K \subseteq X \) is any compact set that contains \( {x}_{i} \) for infinitely many \( i \), then there is a subsequence \( \left( {x}_{ij}\right) \) such that \( {x}_{ij} \in K \) for all \( j \) . Because a compact, first countable Hausdorff space is sequentially compact, this subsequence in turn has a convergent subsequence, which is a contradiction.
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Yes
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Proposition 4.92. Suppose \( X \) and \( Y \) are topological spaces and \( F : X \rightarrow Y \) is a proper map. Then \( F \) takes every sequence diverging to infinity in \( X \) to a sequence diverging to infinity in \( Y \) .
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Proof. Suppose \( \left( {x}_{i}\right) \) is a sequence in \( X \) that diverges to infinity. If \( \left( {F\left( {x}_{i}\right) }\right) \) does not diverge to infinity, then there is a compact subset \( K \subseteq Y \) that contains \( F\left( {x}_{i}\right) \) for infinitely many values of \( i \) . It follows that \( {x}_{i} \) lies in the compact set \( {F}^{-1}\left( K\right) \) for these values of \( i \), which contradicts the assumption that \( \left( {x}_{i}\right) \) diverges to infinity.
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Yes
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Proposition 4.93 (Sufficient Conditions for Properness). Suppose \( X \) and \( Y \) are topological spaces, and \( F : X \rightarrow Y \) is a continuous map.\n\n(a) If \( X \) is compact and \( Y \) is Hausdorff, then \( F \) is proper.
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Proof. We begin with (a). Suppose \( X \) is compact and \( Y \) is Hausdorff. If \( K \subseteq Y \) is compact, then it is closed in \( Y \) because \( Y \) is Hausdorff. By continuity, \( {F}^{-1}\left( K\right) \) is closed in \( X \) and therefore compact.
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Yes
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Lemma 4.94. First countable spaces and locally compact spaces are compactly generated.
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Proof. Let \( X \) be a space satisfying either of the two hypotheses, and let \( A \subseteq X \) be a subset whose intersection with each compact set \( K \subseteq X \) is closed in \( K \) . Suppose \( x \in \bar{A} \) ; we need to show that \( x \in A \) .\n\nFirst assume that \( X \) is first countable. By the sequence lemma, there is a sequence \( \left( {a}_{i}\right) \) of points in \( A \) converging to \( x \) . The set \( K = \left\{ {{a}_{i} : i \in \mathbb{N}}\right\} \cup \{ x\} \) is compact by Proposition 4.31, so \( A \cap K \) is closed in \( K \) by hypothesis. Since \( x \) is the limit of a sequence of points in \( A \cap K \), it must also be in \( A \cap K \subseteq A \) .\n\nNow assume \( X \) is locally compact. Let \( K \) be a compact subset of \( X \) containing a neighborhood \( U \) of \( x \) . If \( V \) is any neighborhood of \( x \), then the fact that \( x \in \bar{A} \) implies that \( V \cap U \) contains a point of \( A \), so \( V \) contains a point of \( A \cap K \) . Thus \( x \in \overline{A \cap K} \) . Since \( A \) is closed in \( K \) and \( K \) is closed in \( X \) (because \( X \) is Hausdorff), it follows that \( A \cap K \) is closed in \( X \), so \( x \in A \cap K \subseteq A \) .
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Yes
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Theorem 4.95 (Proper Continuous Maps are Closed). Suppose \( X \) is any topological space, \( Y \) is a compactly generated Hausdorff space (e.g., any subset of a manifold with or without boundary), and \( F : X \rightarrow Y \) is a proper continuous map. Then \( F \) is a closed map.
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Proof. Let \( A \subseteq X \) be a closed subset. We show that \( F\left( A\right) \) is closed in \( Y \) by showing that its intersection with each compact subset is closed. If \( K \subseteq Y \) is compact, then \( {F}^{-1}\left( K\right) \) is compact, and so is \( A \cap {F}^{-1}\left( K\right) \) because it is a closed subset of a compact set. By the main theorem on compactness, \( F\left( {A \cap {F}^{-1}\left( K\right) }\right) \) is compact as well, and by Exercise A.4(k), this set is equal to \( F\left( A\right) \cap K \) . Because \( K \) is Hausdorff, \( F\left( A\right) \cap K \) is closed in \( K \) .
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No
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Corollary 4.96. If \( X \) is a topological space and \( Y \) is a compactly generated Haus-dorffspace, an embedding \( F : X \rightarrow Y \) is proper if and only if it has closed image.
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Proof. This follows from Theorem 4.95 and Proposition 4.93(d).
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No
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Corollary 4.97. Suppose \( F \) is a proper continuous map from a topological space to a compactly generated Hausdorff space.\n\n(a) If \( F \) is surjective, it is a quotient map.\n\n(b) If \( F \) is injective, it is a topological embedding.\n\n(c) If \( F \) is bijective, it is a homeomorphism.
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Proof. Theorem 4.95 and Proposition 3.69.
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No
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Proposition 5.4. Let \( X \) be a Hausdorff space, and let \( \mathcal{E} \) be a cell decomposition of \( X \) . If \( \mathcal{E} \) is locally finite, then it is a \( {CW} \) decomposition.
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Proof. To prove condition (C), observe that for each \( e \in \mathcal{E} \), every point of \( \bar{e} \) has a neighborhood that intersects only finitely many cells of \( \mathcal{E} \) . Because \( \bar{e} \) is compact, it is covered by finitely many such neighborhoods.\n\nTo prove (W), suppose \( A \subseteq X \) is a subset whose intersection with \( \bar{e} \) is closed in \( \bar{e} \) for each \( e \in \mathcal{E} \) . Given \( x \in X \smallsetminus A \), let \( W \) be a neighborhood of \( x \) that intersects the closures of only finitely many cells, say \( {\bar{e}}_{1},\ldots ,{\bar{e}}_{m} \) . Since \( A \cap {\bar{e}}_{i} \) is closed in \( {\bar{e}}_{i} \) and thus in \( X \), it follows that\n\n\[ W \smallsetminus A = W \smallsetminus \left( {\left( {A \cap {\bar{e}}_{1}}\right) \cup \cdots \cup \left( {A \cap {\bar{e}}_{m}}\right) }\right) \]\n\nis a neighborhood of \( x \) contained in \( X \smallsetminus A \) . Thus \( X \smallsetminus A \) is open, so \( A \) is closed.
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Yes
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Proposition 5.5. Suppose \( X \) is an \( n \) -dimensional \( {CW} \) complex. Then every \( n \) -cell of \( X \) is an open subset of \( X \) .
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Proof. Suppose \( {e}_{0} \) is an \( n \) -cell of \( X \) . If \( \Phi : D \rightarrow X \) is a characteristic map for \( {e}_{0} \) , then \( \Phi \), considered as a map onto \( {\bar{e}}_{0} \), is a quotient map by the closed map lemma. Since \( {\Phi }^{-1}\left( {e}_{0}\right) = \operatorname{Int}D \) is open in \( D \), it follows that \( {e}_{0} \) is open in \( {\bar{e}}_{0} \) . On the other hand, if \( e \) is any other cell of \( X \), then \( {e}_{0} \cap e = \varnothing \), so \( {e}_{0} \cap \bar{e} \) is contained in \( \bar{e} \smallsetminus e \) , which in turn is contained in a union of finitely many cells of dimension less than \( n \) . Since \( {e}_{0} \) has dimension \( n \), it follows that \( {e}_{0} \cap \bar{e} = \varnothing \) . Thus the intersection of \( {e}_{0} \) with the closure of every cell is open, so \( {e}_{0} \) is open in \( X \) by condition (W).
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Yes
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Theorem 5.6. Suppose \( X \) is a \( {CW} \) complex and \( Y \) is a subcomplex of \( X \). Then \( Y \) is closed in \( X \), and with the subspace topology and the cell decomposition that it inherits from \( X \), it is a \( {CW} \) complex.
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Proof. Obviously \( Y \) is Hausdorff, and by definition it is the disjoint union of its cells. Let \( e \subseteq Y \) denote such a cell. Since \( \bar{e} \subseteq Y \), the finitely many cells of \( X \) that have nontrivial intersections with \( \bar{e} \) must also be cells of \( Y \), so condition (C) is automatically satisfied by \( Y \). In addition, any characteristic map \( \Phi : D \rightarrow X \) for \( e \) in \( X \) also serves as a characteristic map for \( e \) in \( Y \).\n\nTo prove that \( Y \) satisfies condition (W), suppose \( S \subseteq Y \) is a subset such that \( S \cap \bar{e} \) is closed in \( \bar{e} \) for every cell \( e \) contained in \( Y \). Let \( e \) be a cell of \( X \) that is not contained in \( Y \). We know that \( \bar{e} \smallsetminus e \) is contained in the union of finitely many cells of \( X \) ; some of these, say \( {e}_{1},\ldots ,{e}_{k} \), might be contained in \( Y \). Then \( {\bar{e}}_{1} \cup \cdots \cup {\bar{e}}_{k} \subseteq Y \),\n\nand\n\[ S \cap \bar{e} = S \cap \left( {{\bar{e}}_{1} \cup \cdots \cup {\bar{e}}_{k}}\right) \cap \bar{e} = \left( {\left( {S \cap {\bar{e}}_{1}}\right) \cup \cdots \cup \left( {S \cap {\bar{e}}_{k}}\right) }\right) \cap \bar{e}, \]\n\nwhich is closed in \( \bar{e} \). It follows that \( S \) is closed in \( X \) and therefore in \( Y \).\n\nFinally, to show that \( Y \) is closed in \( X \), just apply the argument in the preceding paragraph with \( S = Y \).
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Yes
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Proposition 5.7. If \( X \) is any \( {CW} \) complex, the topology of \( X \) is coherent with the collection of subspaces \( \left\{ {{X}_{n} : n \geq 0}\right\} \) .
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Proof. Problem 5.7.
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No
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Theorem 5.11. For a CW complex \( X \), the following are equivalent.\n\n(a) \( X \) is path-connected.\n\n(b) \( X \) is connected.\n\n(c) The 1-skeleton of \( X \) is connected.\n\n(d) Some n-skeleton of \( X \) is connected.
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Proof. Obviously, \( \left( \mathrm{a}\right) \Rightarrow \left( \mathrm{b}\right) \) and \( \left( \mathrm{c}\right) \Rightarrow \left( \mathrm{d}\right) \), so it suffices to show that \( \left( \mathrm{b}\right) \Rightarrow \left( \mathrm{c}\right) \) and (d) \( \Rightarrow \) (a).\n\nTo prove (b) \( \Rightarrow \) (c), we prove the contrapositive. Suppose that \( {X}_{1} = {X}_{1}^{\prime } \cup {X}_{1}^{\prime \prime } \) is a disconnection of the 1 -skeleton of \( X \) . We show by induction on \( n \) that for each \( n > 1 \), the \( n \) -skeleton has a disconnection \( {X}_{n} = {X}_{n}^{\prime } \cup {X}_{n}^{\prime \prime } \) such that \( {X}_{n - 1}^{\prime } \subseteq {X}_{n}^{\prime } \) and \( {X}_{n - 1}^{\prime \prime } \subseteq {X}_{n}^{\prime \prime } \) for each \( n \) . Suppose \( {X}_{n - 1} = {X}_{n - 1}^{\prime } \cup {X}_{n - 1}^{\prime \prime } \) is a disconnection of \( {X}_{n - 1} \) for some \( n > 1 \) . For each \( n \) -cell \( e \) of \( X \), let \( \Phi : D \rightarrow {X}_{n} \) be a characteristic map. Its restriction to \( \partial D \) is a continuous map into \( {X}_{n - 1} \) ; since \( \partial D \approx {\mathbb{S}}^{n - 1} \) is connected, its image must lie entirely in one of the sets \( {X}_{n - 1}^{\prime },{X}_{n - 1}^{\prime \prime } \) . Thus \( \bar{e} = \Phi \left( D\right) \) has a nontrivial intersection with either \( {X}_{n - 1}^{\prime } \) or \( {X}_{n - 1}^{\prime \prime } \), but not both. Divide the \( n \) -cells into two disjoint collections \( {\mathcal{E}}^{\prime } \) and \( {\mathcal{E}}^{\prime \prime } \), according to whether their closures intersect \( {X}_{n - 1}^{\prime } \) or \( {X}_{n - 1}^{\prime \prime } \), respectively, and let\n\n\[{X}_{n}^{\prime } = {X}_{n
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Yes
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Theorem 5.11. For a CW complex \( X \), the following are equivalent.\n\n(a) \( X \) is path-connected.\n\n(b) \( X \) is connected.\n\n(c) The 1-skeleton of \( X \) is connected.\n\n(d) Some n-skeleton of \( X \) is connected.
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Proof. Obviously, \( \left( \mathrm{a}\right) \Rightarrow \left( \mathrm{b}\right) \) and \( \left( \mathrm{c}\right) \Rightarrow \left( \mathrm{d}\right) \), so it suffices to show that \( \left( \mathrm{b}\right) \Rightarrow \left( \mathrm{c}\right) \) and (d) \( \Rightarrow \) (a).\n\nTo prove (b) \( \Rightarrow \) (c), we prove the contrapositive. Suppose that \( {X}_{1} = {X}_{1}^{\prime } \cup {X}_{1}^{\prime \prime } \) is a disconnection of the 1 -skeleton of \( X \) . We show by induction on \( n \) that for each \( n > 1 \), the \( n \) -skeleton has a disconnection \( {X}_{n} = {X}_{n}^{\prime } \cup {X}_{n}^{\prime \prime } \) such that \( {X}_{n - 1}^{\prime } \subseteq {X}_{n}^{\prime } \) and \( {X}_{n - 1}^{\prime \prime } \subseteq {X}_{n}^{\prime \prime } \) for each \( n \) . Suppose \( {X}_{n - 1} = {X}_{n - 1}^{\prime } \cup {X}_{n - 1}^{\prime \prime } \) is a disconnection of \( {X}_{n - 1} \) for some \( n > 1 \) . For each \( n \) -cell \( e \) of \( X \), let \( \Phi : D \rightarrow {X}_{n} \) be a characteristic map. Its restriction to \( \partial D \) is a continuous map into \( {X}_{n - 1} \) ; since \( \partial D \approx {\mathbb{S}}^{n - 1} \) is connected, its image must lie entirely in one of the sets \( {X}_{n - 1}^{\prime },{X}_{n - 1}^{\prime \prime } \) . Thus \( \bar{e} = \Phi \left( D\right) \) has a nontrivial intersection with either \( {X}_{n - 1}^{\prime } \) or \( {X}_{n - 1}^{\prime \prime } \), but not both. Divide the \( n \) -cells into two disjoint collections \( {\mathcal{E}}^{\prime } \) and \( {\mathcal{E}}^{\prime \prime } \), according to whether their closures intersect \( {X}_{n - 1}^{\prime } \) or \( {X}_{n - 1}^{\prime \prime } \), respectively, and let\n\n\[{X}_{n}^{\prime } = {X}_{n - 1}^{\prime } \cup \left( {\mathop{\bigcup }\limits_{{e \in {\mathcal{E}}^{\prime }}}e}\right) ,\;{X}_{n}^{\prime \prime } = {X}_{n - 1}^{\prime \prime } \cup \left( {\mathop{\bigcup }\limits_{{e \in {\mathcal{E}}^{\prime \prime }}}e}\right) .\n\]\n\nClearly \( {X}_{n} \) is the disjoint union of \( {X}_{n}^{\prime } \) and \( {X}_{n}^{\prime \prime } \), and both sets are nonempty because \( {X}_{n - 1}^{\prime } \) and \( {X}_{n - 1}^{\prime \prime } \) are nonempty by the inductive hypothesis. If \( e \) is any cell of \( {X}_{n} \) , its closure is entirely contained in one of these two sets, so \( {X}_{n}^{\prime } \cap \bar{e} \) is either \( \bar{e} \) or \( \varnothing \) , as is \( {X}_{n}^{\prime \prime } \cap \bar{e} \) . It follows from condition (W) that both \( {X}_{n - 1}^{\prime } \) and
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Yes
|
Lemma 5.12. In any CW complex, the closure of each cell is contained in a finite subcomplex.
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Proof. Let \( X \) be a CW complex, and let \( e \subseteq X \) be an \( n \) -cell; we prove the lemma by induction on \( n \) . If \( n = 0 \), then \( \bar{e} = e \) is itself a finite subcomplex, so assume the lemma is true for every cell of dimension less than \( n \) . Then by condition (C), \( \bar{e} \smallsetminus e \) is contained in the union of finitely many cells of lower dimension, each of which is contained in a finite subcomplex by the inductive hypothesis. The union of these finite subcomplexes together with \( e \) is a finite subcomplex containing \( \bar{e} \) .
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Yes
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Lemma 5.13. Let \( X \) be a CW complex. A subset of \( X \) is discrete if and only if its intersection with each cell is finite.
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Proof. Suppose \( S \subseteq X \) is discrete. For each cell \( e \) of \( X \), the intersection \( S \cap \bar{e} \) is a discrete subset of the compact set \( \bar{e} \), so it is finite, and thus so also is \( S \cap e \) .\n\nConversely, suppose \( S \) is a subset whose intersection with each cell is finite. Because the closure of each cell is contained in a finite subcomplex, the hypothesis implies that \( S \cap \bar{e} \) is finite for each \( e \) . This means that \( S \cap \bar{e} \) is closed in \( \bar{e} \), and thus by condition (W), \( S \) is closed in \( X \) . But the same argument applies to every subset of \( S \) ; thus every subset of \( S \) is closed in \( X \), which implies that the subspace topology on \( S \) is discrete.
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Yes
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Theorem 5.14. Let \( X \) be a CW complex. A subset of \( X \) is compact if and only if it is closed in \( X \) and contained in a finite subcomplex.
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Proof. Every finite subcomplex \( Y \subseteq X \) is compact, because it is the union of finitely many compact sets of the form \( \bar{e} \) . Thus if \( K \subseteq X \) is closed and contained in a finite subcomplex, it is also compact.\n\nConversely, suppose \( K \subseteq X \) is compact. If \( K \) intersects infinitely many cells, by choosing one point of \( K \) in each such cell we obtain an infinite discrete subset of \( K \) , which is impossible. Therefore, \( K \) is contained in the union of finitely many cells, and thus in a finite subcomplex by Lemma 5.12.
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Yes
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Proposition 5.16. A CW complex is locally compact if and only if it is locally finite.
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Proof. Problem 5-11.
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No
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Lemma 5.17. Suppose \( X \) is a CW complex, \( {\left\{ {e}_{\alpha }\right\} }_{\alpha \in A} \) is the collection of cells of \( X \) , and for each \( \alpha \in A,{\Phi }_{\alpha } : {D}_{\alpha } \rightarrow X \) is a characteristic map for the cell \( {e}_{\alpha } \) . Then the map \( \Phi : \mathop{\coprod }\limits_{\alpha }{D}_{\alpha } \rightarrow X \) whose restriction to each \( {D}_{\alpha } \) is \( {\Phi }_{\alpha } \) is a quotient map.
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Proof. The map \( \Phi \) can be expressed as the composition of two maps: the map \( {\Phi }_{1} : \mathop{\coprod }\limits_{\alpha }{D}_{\alpha } \rightarrow \mathop{\coprod }\limits_{\alpha }{\bar{e}}_{\alpha } \) whose restriction to each \( {D}_{\alpha } \) is \( {\Phi }_{\alpha } : {D}_{\alpha } \rightarrow {\bar{e}}_{\alpha } \), followed by the map \( {\Phi }_{2} : \mathop{\coprod }\limits_{\alpha }{\bar{e}}_{\alpha } \rightarrow X \) induced by inclusion of each set \( {\bar{e}}_{\alpha } \) . The first is a quotient map by the closed map lemma and Proposition 3.62(e), and the second by Proposition 5.2(b).
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Yes
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Proposition 5.18. Let \( X \) be a CW complex. Each skeleton \( {X}_{n} \) is obtained from \( {X}_{n - 1} \) by attaching a collection of \( n \) -cells.
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Proof. Let \( \left\{ {e}_{\alpha }^{n}\right\} \) be the collection of \( n \) -cells of \( X \), and for each \( n \) -cell \( {e}_{\alpha }^{n} \), let \( {\Phi }_{\alpha }^{n} : {D}_{\alpha }^{n} \rightarrow X \) be a characteristic map. Define \( \varphi : \mathop{\coprod }\limits_{\alpha }\partial {D}_{\alpha }^{n} \rightarrow X \) to be the map whose restriction to each \( \partial {D}_{\alpha }^{n} \) is equal to the restriction of \( {\Phi }_{\alpha }^{n} \) . By definition of a cell complex, \( \varphi \) takes its values in \( {X}_{n - 1} \), so we can form the adjunction space \( {X}_{n - 1}{ \cup }_{\varphi }\left( {\mathop{\coprod }\limits_{\alpha }{D}_{\alpha }^{n}}\right) \) . The map \( \Phi : {X}_{n - 1} \coprod \left( {\mathop{\coprod }\limits_{\alpha }{D}_{\alpha }^{n}}\right) \rightarrow {X}_{n} \) that is equal to inclusion on \( {X}_{n - 1} \) and to \( {\Phi }_{\alpha }^{n} \) on each \( {D}_{\alpha }^{n} \) makes the same identifications as the quotient map defining the adjunction space, so if we can show that \( \Phi \) is a quotient map, then uniqueness of quotient spaces shows that \( {X}_{n} \) is homeomorphic to the adjunction space described in the preceding paragraph. Suppose therefore that \( A \) is a saturated closed subset of the disjoint union, and let \( B = \Phi \left( A\right) \), so that \( A = {\Phi }^{-1}\left( B\right) \) . The hypothesis means that \( A \cap {X}_{n - 1} \) is closed in \( {X}_{n - 1} \) and \( A \cap {D}_{\alpha }^{n} \) is closed in \( {D}_{\alpha }^{n} \) for each \( \alpha \) . The first assertion implies that \( B \cap \bar{e} \) is closed in \( \bar{e} \) for every cell \( e \) of dimension less than \( n \) ; and the second implies that \( B \cap \overline{{e}_{\alpha }^{n}} \) is closed in \( \overline{{e}_{\alpha }^{n}} \) for each \( n \) -cell because \( {\Phi }_{\alpha }^{n} : {D}_{\alpha }^{n} \rightarrow \overline{{e}_{\alpha }^{n}} \) is a closed map by the closed map lemma. Thus \( B \) is closed in \( {X}_{n} \) . It follows from Proposition 3.60 that \( \Phi \) is a quotient map.
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Yes
|
Theorem 5.20 (CW Construction Theorem). Suppose \( {X}_{0} \subseteq {X}_{1} \subseteq \cdots \subseteq {X}_{n - 1} \subseteq \) \( {X}_{n} \subseteq \cdots \) is a sequence of topological spaces satisfying the following conditions:\n\n(i) \( {X}_{0} \) is a nonempty discrete space.\n\n(ii) For each \( n \geq 1,{X}_{n} \) is obtained from \( {X}_{n - 1} \) by attaching a (possibly empty) collection of \( n \) -cells.\n\nThen \( X = \mathop{\bigcup }\limits_{n}{X}_{n} \) has a unique topology coherent with the family \( \left\{ {X}_{n}\right\} \), and a unique cell decomposition making it into a CW complex whose n-skeleton is \( {X}_{n} \) for each \( n \) .
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Proof. Give \( X \) a topology by declaring a subset \( B \subseteq X \) to be closed if and only if \( B \cap {X}_{n} \) is closed in \( {X}_{n} \) for each \( n \) . It is immediate that this is a topology, and it is obviously the unique topology coherent with \( \left\{ {X}_{n}\right\} \) . With this topology, each \( {X}_{n} \) is a subspace of \( X \) : if \( B \) is closed in \( X \), then \( B \cap {X}_{n} \) is closed in \( {X}_{n} \) by definition of the topology on \( X \) ; conversely, if \( B \) is closed in \( {X}_{n} \), then by virtue of the fact that each \( {X}_{m - 1} \) is closed in \( {X}_{m} \) by Proposition 3.77, it follows that \( B \cap {X}_{m} \) is closed in \( {X}_{m} \) for each \( m \) and thus \( B \) is also closed in \( X \) .\n\nNext we define the cell decomposition of \( X \) . The 0 -cells are just the points of the discrete space \( {X}_{0} \) . For each \( n \geq 1 \), let\n\n\[ \n{q}_{n} : {X}_{n - 1} \coprod \left( {\mathop{\coprod }\limits_{{\alpha \in {A}_{n}}}{D}_{\alpha }^{n}}\right) \rightarrow {X}_{n} \n\]\n\nbe a quotient map realizing \( {X}_{n} \) as an adjunction space. Proposition 3.77 shows that \( {X}_{n} \smallsetminus {X}_{n - 1} \) is an open subset of \( {X}_{n} \) homeomorphic to \( \mathop{\coprod }\limits_{\alpha }\operatorname{Int}{D}_{\alpha }^{n} \), which is a disjoint union of open \( n \) -cells, so we can define the \( n \) -cells of \( \bar{X} \) to be the components \( \left\{ {e}_{\alpha }^{n}\right\} \) of \( {X}_{n} \smallsetminus {X}_{n - 1} \) . These are subspaces of \( {X}_{n} \) and hence of \( X \), and \( X \) is the disjoint union of all of its cells.\n\nFor each \( n \) -cell \( {e}_{\beta }^{n} \), define a characteristic map \( {\Phi }_{\beta }^{n} : {D}_{\beta }^{n} \rightarrow X \) as the composition\n\n\[ \n{D}_{\beta }^{n} \hookrightarrow {X}_{n - 1} \coprod \left( {\mathop{\coprod }\limits_{{\alpha \in {A}_{n}}}{D}_{\alpha }^{n}}\right) \overset{{q}_{n}}{ \rightarrow }{X}_{n} \hookrightarrow X. \n\]\n\nClearly \( {\Phi }_{\beta }^{n} \) maps \( \partial {D}_{\beta }^{n} \) into \( {X}_{n - 1} \), and its restriction to Int \( {D}_{\beta }^{n} \) is a bijective continuous map onto \( {e}_{\beta }^{n} \), so we need only show that this restriction is a homeomorphism onto its image. This follows because \( {\left. {\Phi }_{\beta }^{n}\right| }_{\operatorname{Int}{D}_{\beta }^{n}} \) is equal to the inclusion of Int \( {D}_{\beta }^{n} \) into the disjoint union, followed by the restriction of \( {q}_{n} \) to the saturated open subset Int \( {D}_{\beta }^{n} \), which is a bijective quotient map onto \( {e}_{\beta }^{n} \) . This proves that \( X \) has a cell decomposition for which \( {X}_{n} \) is the \( n \) -skeleton for each \( n \) . Because the \( n \) -cells of any such decomposition are the components of \( {X}_{n} \smallsetminus {X}_{n - 1} \) this is the unique such cell decomposition.
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Yes
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Proposition 5.23. Suppose \( X \) is a CW complex with countably many cells. If \( X \) is locally Euclidean, then it is a manifold.
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Proof. Every CW complex is Hausdorff by definition. Lemma 5.17 shows that \( X \) is a quotient of a disjoint union of countably many closed cells of various dimensions. Such a disjoint union is easily seen to be second countable, and then Proposition 3.56 implies that \( X \) is also second countable.
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Yes
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Proposition 5.24. If \( M \) is a nonempty \( n \) -manifold and a CW complex, then the dimension of \( M \) as a \( {CW} \) complex is also \( n \) .
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Proof. For this proof, we assume the theorem on invariance of dimension. Let \( M \) be an \( n \) -manifold with a given CW decomposition. Because every manifold is locally compact, the CW decomposition is locally finite by Proposition 5.16. Let \( x \in M \) be arbitrary. Then \( x \) has a neighborhood \( W \) that intersects the closures of only finitely many cells. Suppose \( k \) is the maximum dimension of such cells, and let \( {e}_{0} \) be an open \( k \) -cell of maximum dimension \( k \) such that \( {\bar{e}}_{0} \) has a nontrivial intersection with \( W \) . Because \( W \) is open, it must contain a point of \( {e}_{0} \) as well. Let \( U = W \cap {e}_{0} \) . Then \( U \) is open in \( {e}_{0} \), so it is a \( k \) -manifold. We will show below, using an adaptation of the argument of Proposition 5.5, that \( U \) is open in \( M \), from which it follows that it is also an \( n \) -manifold, so \( k = n \) . Since this shows that every point has a neighborhood that intersects no cell of dimension larger than \( n \), this completes the proof.\n\nTo show that \( U \) is open, we show that \( U \cap \bar{e} \) is open in \( \bar{e} \) for every cell \( e \) . It suffices to consider only cells whose closures have nontrivial intersections with \( W \) . Note that \( U \cap {e}_{0} = W \cap {e}_{0} \) is open in \( {e}_{0} \) and hence in \( {\bar{e}}_{0} \) (since \( {e}_{0} \) is open in \( {\bar{e}}_{0} \) ). On the other hand, if \( e \) is any other open cell whose closure intersects \( W \), then \( U \) is disjoint from \( e \) (because the open cells are disjoint), and \( \bar{e} \smallsetminus e \) is contained in a union of cells of dimension less than \( k \), so \( U \cap \bar{e} = \varnothing \) . It follows that \( U \) is open in \( M \), which completes the proof.
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Yes
|
Lemma 5.26. Suppose \( M \) is a 1-manifold endowed with a regular CW decomposition. Then the boundary of every \( 1 \) -cell of \( M \) consists of exactly two 0 -cells, and every 0 -cell of \( M \) is a boundary point of exactly two 1 -cells.
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Proof. By Proposition 5.24, the dimension of \( M \) as a CW complex is 1 . (Although the proof of that proposition depended on the theorem of invariance of dimension, for this proof we need only the 1-dimensional case, which is taken care of by Problem 4-2.)\n\n\n\nFig. 5.6: Classifying 1-manifolds.\n\nIf \( e \) is any 1 -cell of \( M \), then \( e \) is an open subset of \( M \) by Proposition 5.5. By definition of a regular CW decomposition, there is a homeomorphism from \( \left\lbrack {0,1}\right\rbrack \) to \( \bar{e} \) taking \( \left( {0,1}\right) \) to \( e \), so \( \partial e = \bar{e} \smallsetminus e \) consists of two points contained in the 0 -skeleton. This proves the first claim.\n\nTo prove the second, suppose \( v \) is a 0 -cell of \( M \), and let \( {e}_{1},\ldots ,{e}_{n} \) be the (finitely many) 1-cells that have \( v \) as a boundary point. Define\n\n\[ \n{Y}_{v} = \{ v\} \cup {e}_{1} \cup \cdots \cup {e}_{n} \n\]\n\n(5.1)\n\nWe show that \( {Y}_{v} \) is a neighborhood of \( v \) by showing that its intersection with the closure of each cell is open. The intersection of \( {Y}_{v} \) with \( \bar{v} = v \) is \( v \) itself, and the intersection of \( {Y}_{v} \) with each \( {\bar{e}}_{i} \) is \( {\bar{e}}_{i} \) minus a boundary point, hence open in \( {\bar{e}}_{i} \) . For any other cell \( e,{Y}_{v} \cap \bar{e} = \varnothing \) . It follows that \( {Y}_{v} \) is open in \( M \) . This implies in particular that \( {Y}_{v} \) is itself a 1-manifold.\n\nIf \( v \) is not a boundary point of any 1 -cell, then \( {Y}_{v} = \{ v\} \), so \( v \) is an isolated point of \( M \), contradicting the fact that \( M \) is a 1-manifold. If \( v \) is a boundary point of only one 1-cell, then \( {Y}_{v} \) is homeomorphic to \( \lbrack 0,1) \), which is a contradiction because \( \lbrack 0,1) \) is not a 1-manifold (see Problem 4-3). On the other hand, suppose \( v \) is a boundary point of the 1-cells \( {e}_{1},\ldots ,{e}_{k} \) for \( k \geq 3 \) . Because \( {Y}_{v} \) is a 1-manifold, \( v \) has a neighborhood \( W \subseteq {Y}_{v} \) that is homeomorphic to \( \mathbb{R} \), and therefore \( W \smallsetminus \{ v\} \) has exactly two components. But \( W \smallsetminus \{ v\} \) is the union of the disjoint open subsets \( W \cap {e}_{1},\ldots, W \cap {e}_{k} \) . Each of these is nonempty, because \( W \cap {\bar{e}}_{i} \) is nonempty and open in \( {\bar{e}}_{i} \) . Therefore, \( W \smallsetminus \{ v\} \) has at least \( k \) components, which is a contradiction. The only other possibility is that \( v \) is a boundary point of exactly two 1-cells.
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Yes
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Corollary 5.28 (Classification of 1-Manifolds with Boundary). A connected 1- manifold with nonempty boundary is homeomorphic to \( \left\lbrack {0,1}\right\rbrack \) if it is compact, and to \( \lbrack 0,\infty ) \) if not.
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Proof. Let \( M \) be such a manifold with boundary, and let \( D\left( M\right) \) be the double of \( M \) (see Example 3.80). Then \( D\left( M\right) \) is a connected 1-manifold without boundary (Exercise 4.10), so it is homeomorphic to either \( {\mathbb{S}}^{1} \) or \( \mathbb{R} \), and \( M \) is homeomorphic to a proper connected subspace of \( D\left( M\right) \) . If \( D\left( M\right) \approx {\mathbb{S}}^{1} \), we can choose a point \( p \in D\left( M\right) \smallsetminus M \) and obtain an embedding \( M \hookrightarrow D\left( M\right) \smallsetminus \{ p\} \approx \mathbb{R} \), so in either case, \( M \) is homeomorphic to a connected subset of \( \mathbb{R} \) containing more than one point, which is therefore an interval. Since \( M \) has a nonempty boundary, the interval must have at least one endpoint (here we are using the result of Problem 4-3 on the invariance of the boundary). If it is a closed bounded interval, it is a closed 1-cell and thus homeomorphic to \( \left\lbrack {0,1}\right\rbrack \) ; otherwise it is one of the types \( \lbrack a, b),\lbrack a,\infty ),\left( {a, b}\right\rbrack \), or \( \left( {-\infty, b\rbrack \text{, all of which are homeomorphic to}\lbrack 0,\infty }\right) \) .
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Yes
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Proposition 5.29. For any \( k + 1 \) distinct points \( {v}_{0},\ldots ,{v}_{k} \in {\mathbb{R}}^{n} \), the following are equivalent:\n\n(a) The set \( \left\{ {{v}_{0},\ldots ,{v}_{k}}\right\} \) is affinely independent.\n\n(b) The set \( \left\{ {{v}_{1} - {v}_{0},\ldots ,{v}_{k} - {v}_{0}}\right\} \) is linearly independent.\n\n(c) If \( {c}_{0},\ldots ,{c}_{k} \) are real numbers such that\n\n\n\n\[ \mathop{\sum }\limits_{{i = 0}}^{k}{c}_{i}{v}_{i} = 0\;\text{ and }\;\mathop{\sum }\limits_{{i = 0}}^{k}{c}_{i} = 0 \]\n\n(5.2)\n\n\[ \text{then}{c}_{0} = \cdots = {c}_{k} = 0\text{.} \]
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Proof. First we prove that (a) \( \Leftrightarrow \) (b). Let \( S \subseteq {\mathbb{R}}^{n} \) denote the linear span of \( \left\{ {{v}_{1} - {v}_{0},\ldots ,{v}_{k} - {v}_{0}}\right\} \) . First, if \( \left\{ {{v}_{1} - {v}_{0},\ldots ,{v}_{k} - {v}_{0}}\right\} \) is a linearly dependent set, then \( {v}_{0} + S \) is an affine subspace of dimension less than \( k \) containing \( \left\{ {{v}_{0},\ldots ,{v}_{k}}\right\} \) . Conversely, if \( A \) is some affine subspace of dimension less than \( k \) containing \( \left\{ {{v}_{0},\ldots ,{v}_{k}}\right\} \), then \( \left( {-{v}_{0}}\right) + A \) is a linear subspace of the same dimension containing \( S \), so the set \( \left\{ {{v}_{1} - {v}_{0},\ldots ,{v}_{k} - {v}_{0}}\right\} \) is linearly dependent.\n\nTo show that (b) \( \Leftrightarrow \) (c), suppose that equations (5.2) hold for some \( {c}_{0},\ldots ,{c}_{k} \) not all zero. Then since \( {c}_{0} = - \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i} \), we have\n\n\[ \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}\left( {{v}_{i} - {v}_{0}}\right) = \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{v}_{i} - \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{v}_{0} = \mathop{\sum }\limits_{{i = 1}}^{k}{c}_{i}{v}_{i} + {c}_{0}{v}_{0} = 0, \]\n\nwhich implies that the set \( \left\{ {{v}_{1} - {v}_{0},\ldots {v}_{k} - {v}_{0}}\right\} \) is linearly dependent. The converse is similar.
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Yes
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Proposition 5.32. Every k-simplex is a closed k-cell.
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Proof. Consider first the standard \( k \) -simplex \( {\Delta }_{k} = \left\lbrack {{e}_{0},\ldots ,{e}_{k}}\right\rbrack \subseteq {\mathbb{R}}^{k} \), where \( {e}_{0} = 0 \) and for \( i = 1,\ldots, k,{e}_{i} = \left( {0,\ldots ,1,\ldots ,0}\right) \) has a 1 in the \( i \) th place and zeros elsewhere. This simplex is just the set of points \( \left( {{t}_{1},\ldots ,{t}_{k}}\right) \in {\mathbb{R}}^{\widehat{k}} \) such that \( {t}_{i} \geq 0 \) for \( i = 1,\ldots, k \) and \( \mathop{\sum }\limits_{i}{t}_{i} \leq 1 \) . Any point for which all these inequalities are strict is an interior point, so \( {\Delta }_{k} \) is a closed \( k \) -cell by Proposition 5.1.\n\nNow suppose \( \sigma = \left\lbrack {{v}_{0},\ldots ,{v}_{k}}\right\rbrack \subseteq {\mathbb{R}}^{n} \) is an arbitrary \( k \) -simplex. Define a map \( F : {\Delta }_{k} \rightarrow \sigma \) by \( F\left( {{t}_{1},\ldots ,{t}_{k}}\right) = {t}_{0}{v}_{0} + {t}_{1}{v}_{1} + \cdots + {t}_{k}{v}_{k} \), where \( {t}_{0} = 1 - \mathop{\sum }\limits_{{i = 1}}^{k}{t}_{i} \) . This is a continuous bijection, and therefore a homeomorphism by the closed map lemma.
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Yes
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Theorem 5.36 (Triangulation Theorem for 2-Manifolds). Every 2-manifold is homeomorphic to the polyhedron of a 2-dimensional simplicial complex, in which every 1-simplex is a face of exactly two 2-simplices.
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The proof is highly technical and beyond our scope, so we can only describe some of the main ideas here. The basic approach is analogous to the proof of Theorem 5.25: cover the manifold with regular coordinate disks, and inductively show that each successive disk can be triangulated in a way that is compatible with the triangulations that have already been defined. In the case of surfaces, however, finding a triangulation of each disk that is compatible with the previous ones is much more difficult, primarily because the boundary of the new disk might intersect the boundaries of the already defined simplices infinitely many times. Even if there are only finitely many intersections, showing that the regions defined by the intersecting curves are homeomorphic to closed disks, and therefore triangulable, requires a delicate topological result known as the Schönflies theorem, which asserts that any topological embedding of the circle into \( {\mathbb{R}}^{2} \) extends to an embedding of the closed disk. The details of the proof are long and intricate and would take us too far from our main goals, so we leave it to the reader to look it up. Readable proofs can be found in [Moi77, Tho92].
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No
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Proposition 5.38. Let \( \sigma = \left\lbrack {{v}_{0},\ldots ,{v}_{k}}\right\rbrack \) be a \( k \) -simplex in \( {\mathbb{R}}^{n} \) . Given any \( k + 1 \) points \( {w}_{0},\ldots ,{w}_{k} \in {\mathbb{R}}^{m} \), there is a unique map \( f : \sigma \rightarrow {\mathbb{R}}^{m} \) that is the restriction of an affine map and takes \( {v}_{i} \) to \( {w}_{i} \) for each \( i \) .
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Proof. By applying the translations \( x \mapsto x - {v}_{0} \) and \( y \mapsto y - {w}_{0} \) (which are invertible affine maps), we may assume that \( {v}_{0} = 0 \) and \( {w}_{0} = 0 \) . Under this assumption, the set \( \left\{ {{v}_{1},\ldots ,{v}_{k}}\right\} \) is linearly independent, so we can let \( f : \sigma \rightarrow {\mathbb{R}}^{m} \) be the restriction of any linear map such that \( f\left( {v}_{i}\right) = {w}_{i} \) for \( i = 1,\ldots, k \) .\n\nA straightforward computation shows that if \( f : \sigma \rightarrow {\mathbb{R}}^{m} \) is the restriction of any affine map, then it satisfies\n\n\[ f\left( {\mathop{\sum }\limits_{{i = 0}}^{k}{t}_{i}{v}_{i}}\right) = \mathop{\sum }\limits_{{i = 0}}^{k}{t}_{i}f\left( {v}_{i}\right) \]\n\n(5.4)\n\nwhen applied to points in \( \sigma \) . This shows that \( f \) is uniquely determined by where it sends the vertices of \( \sigma \) .
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Yes
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Proposition 6.1. The sphere \( {\mathbb{S}}^{2} \) is homeomorphic to the following quotient spaces.\n\n(a) The closed disk \( {\overline{\mathbb{B}}}^{2} \subseteq {\mathbb{R}}^{2} \) modulo the equivalence relation generated by \( \left( {x, y}\right) \sim \) \( \left( {-x, y}\right) \) for \( \left( {x, y}\right) \in \partial {\overline{\mathbb{B}}}^{2} \) (Fig. 6.1).
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Proof. To see that each of these spaces is homeomorphic to the sphere, all we need to do is exhibit a quotient map from the given space to the sphere that makes the same identifications, and then appeal to uniqueness of quotient spaces (Theorem 3.75).\n\nFor (a), define a map from the disk to the sphere by wrapping each horizontal line segment around a \
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No
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Proposition 6.2. The projective plane \( {\mathbb{P}}^{2} \) is homeomorphic to each of the following quotient spaces (Fig. 6.3).\n\n(a) The closed disk \( {\overline{\mathbb{B}}}^{2} \) modulo the equivalence relation generated by \( \left( {x, y}\right) \sim \) \( \left( {-x, - y}\right) \) for each \( \left( {x, y}\right) \in \partial {\overline{\mathbb{B}}}^{2} \) .
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Proof. Let \( p : {\mathbb{S}}^{2} \rightarrow {\mathbb{P}}^{2} \) be the quotient map representing \( {\mathbb{P}}^{2} \) as a quotient of the sphere, as defined in Example 4.54. If \( F : {\overline{\mathbb{B}}}^{2} \rightarrow {\mathbb{S}}^{2} \) is the map sending the disk onto the upper hemisphere by \( F\left( {x, y}\right) = \left( {x, y,\sqrt{1 - {x}^{2} - {y}^{2}}}\right) \), then \( p \circ F : {\overline{\mathbb{B}}}^{2} \rightarrow {\mathbb{S}}^{2}/ \sim \) is easily seen to be surjective, and is thus a quotient map by the closed map lemma. It identifies only \( \left( {x, y}\right) \in \partial {\overline{\mathbb{B}}}^{2} \) with \( \left( {-x, - y}\right) \in \partial {\overline{\mathbb{B}}}^{2} \), so \( {\mathbb{P}}^{2} \) is homeomorphic to the resulting quotient space.
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Yes
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Proposition 6.4. Let \( {P}_{1},\ldots ,{P}_{k} \) be polygonal regions in the plane, let \( P = {P}_{1} \coprod \) \( \cdots \coprod {P}_{k} \), and suppose we are given an equivalence relation on \( P \) that identifies some of the edges of the polygons with others by means of affine homeomorphisms.\n\n(a) The resulting quotient space is a finite 2-dimensional CW complex whose 0- skeleton is the image under the quotient map of the set of vertices of \( P \), and whose 1-skeleton is the image of the union of the boundaries of the polygonal regions.\n\n(b) If the equivalence relation identifies each edge of each \( {P}_{i} \) with exactly one other edge in some \( {P}_{j} \) (which might or might not be equal to \( {P}_{i} \) ), then the resulting quotient space is a compact 2-manifold.
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Proof. Let \( M \) be the quotient space, let \( \pi : P \rightarrow M \) denote the quotient map, and let \( {M}_{0},{M}_{1} \), and \( {M}_{2} = M \) denote the images under \( \pi \) of the vertices, boundaries, and polygonal regions, respectively. It follows easily from the definition that \( {M}_{0} \) is discrete, and for \( k = 1,2,{M}_{k} \) is obtained from \( {M}_{k - 1} \) by attaching finitely many \( k \) -cells. Thus (a) follows from Theorem 5.20.\n\nNow assume the hypothesis of (b). By Proposition 5.23, to prove that \( M \) is a manifold, it suffices to show that it is locally Euclidean.\n\nBecause the 2-cells are open in \( M \), they are Euclidean neighborhoods of each of their points. Thus it suffices to show that each point in a 1-cell or a 0-cell has a Euclidean neighborhood.\n\nA point \( q \) in a 1-cell has exactly two preimages \( {q}_{1} \) and \( {q}_{2} \), each in the interior of a different edge. Since each \( {P}_{i} \) is a 2-manifold with boundary, and \( {q}_{1},{q}_{2} \) are boundary points, each \( {q}_{i} \) has a neighborhood \( {U}_{i} \) that is a regular coordinate half-ball (see Chapter 4). By shrinking the neighborhoods if necessary, we may assume that the equivalence relation identifies the boundary segment of \( {U}_{1} \) exactly with that of \( {U}_{2} \) . Then the same argument as in Theorem 3.79 shows that \( q \) has a Euclidean neighborhood.\n\nThe preimage of a 0 -cell \( v \) is a finite set of vertices \( \left\{ {{v}_{1},\ldots ,{v}_{k}}\right\} \subseteq P \) . For each of these vertices, we can choose \( \varepsilon \) small enough that the disk \( {B}_{\varepsilon }\left( {v}_{i}\right) \) contains no vertices other than \( {v}_{i} \), and intersects no edges other than the two that have \( {v}_{i} \) as endpoints. Because the interior of the polygonal region \( {P}_{j} \) of which \( {v}_{i} \) is a vertex is a regular coordinate ball, it lies on one side of its boundary, so \( {B}_{\varepsilon }\left( {v}_{i}\right) \cap {P}_{j} \) is equal to a \
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Yes
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The Klein bottle is the 2-manifold \( K \) obtained by identifying the edges of the square \( I \times I \) according to \( \left( {0, t}\right) \sim \left( {1, t}\right) \) and \( \left( {t,0}\right) \sim \left( {1 - t,1}\right) \) for \( 0 \leq t \leq 1 \).
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To visualize \( K \), think of attaching the left and right edges together to form a cylinder, and then passing the upper end of the cylinder through the cylinder wall near the lower end, in order to attach the upper circle to the lower one \
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No
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If \( M \) is any \( n \) -manifold, a connected sum \( M\# {\mathbb{S}}^{n} \) is homeomorphic to \( M \), at least if we make our choices carefully (Fig. 6.6).
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Let \( {B}_{2} \subseteq {\mathbb{S}}^{n} \) be the open lower hemisphere, so \( {\left( {\mathbb{S}}^{n}\right) }^{\prime } = {\mathbb{S}}^{n} \smallsetminus {B}_{2} \) is the closed upper hemisphere, which is homeomorphic to a closed ball. Then \( M\# {\mathbb{S}}^{n} \) is obtained from \( M \) by cutting out the open ball \( {B}_{1} \) and pasting back a closed ball along the boundary sphere, so we have not changed anything.
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Yes
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Proposition 6.12. Let \( {M}_{1} \) and \( {M}_{2} \) be surfaces that admit presentations \( \left\langle {{S}_{1} \mid {W}_{1}}\right\rangle \) and \( \left\langle {{S}_{2} \mid {W}_{2}}\right\rangle \), respectively, in which \( {S}_{1} \) and \( {S}_{2} \) are disjoint sets and each presentation has a single face. Then \( \left\langle {{S}_{1},{S}_{2} \mid {W}_{1}{W}_{2}}\right\rangle \) is a presentation of a connected sum \( {M}_{1}\# {M}_{2} \) . (Here \( {W}_{1}{W}_{2} \) denotes the word formed by concatenating \( {W}_{1} \) and \( {W}_{2} \) .)
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Proof. Consider the presentation \( \left\langle {{S}_{1}, a, b, c \mid {W}_{1}{c}^{-1}{b}^{-1}{a}^{-1},{abc}}\right\rangle \) (pictured in the left half of Fig. 6.16). Pasting along \( a \) and folding twice, we see that this presentation is equivalent to \( \left\langle {{S}_{1} \mid {W}_{1}}\right\rangle \) and therefore is a presentation of \( {M}_{1} \) . Let \( {B}_{1} \) denote the image in \( {M}_{1} \) of the interior of the polygonal region bounded by triangle \( {abc} \) . We will show below that \( {B}_{1} \) is a regular coordinate disk in \( {M}_{1} \) . Assuming this, it follows immediately that the geometric realization of \( \left\langle {{S}_{1}, a, b, c \mid {W}_{1}{c}^{-1}{b}^{-1}{a}^{-1}}\right\rangle \) is homeomorphic to \( {M}_{1} \smallsetminus {B}_{1} \) (which we denote by \( {M}_{1}^{\prime } \) ), and \( \partial {B}_{1} \) is the image of the edges \( {c}^{-1}{b}^{-1}{a}^{-1} \) . A similar argument shows that \( \left\langle {{S}_{2}, a, b, c \mid {abc}{W}_{2}}\right\rangle \) is a presentation of \( {M}_{2} \) with a coordinate disk removed (denoted by \( {M}_{2}^{\prime } \) ). Therefore, \( \left\langle {{S}_{1},{S}_{2}, a, b, c \mid {W}_{1}{c}^{-1}{b}^{-1}{a}^{-1},{abc}{W}_{2}}\right\rangle \) is a presentation of \( {M}_{1}^{\prime } \coprod {M}_{2}^{\prime } \) with the boundaries of the respective disks identified, which is \( {M}_{1}\# {M}_{2} \) . Pasting along \( a \) and folding twice, we arrive at the presentation \( \left\langle {{S}_{1},{S}_{2} \mid {W}_{1}{W}_{2}}\right\rangle \) .
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No
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Every compact surface admits a polygonal presentation.
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Proof. Let \( M \) be a compact surface. It follows from Theorem 5.36 that \( M \) is homeomorphic to the polyhedron of a 2-dimensional simplicial complex \( K \), in which each 1-simplex is a face of exactly two 2-simplices.\n\nFrom this complex, we can construct a surface presentation \( \mathcal{P} \) with one word of length 3 for each 2-simplex, and with edges having the same label if and only if they correspond to the same 1-simplex. We wish to show that the geometric realization of \( \mathcal{P} \) is homeomorphic to that of \( K \) . If \( P = {P}_{1} \coprod \cdots \coprod {P}_{k} \) denotes the disjoint union of the 2-simplices of \( K \), then we have two quotient maps \( {\pi }_{K} : P \rightarrow \left| K\right| \) and \( {\pi }_{\mathcal{P}} : P \rightarrow \left| \mathcal{P}\right| \), so it suffices to show that they make the same identifications. Both quotient maps are injective in the interiors of the 2-simplices, both make the same identifications of edges, and both identify vertices only with other vertices.\n\nTo complete the proof, we need to show that \( {\pi }_{K} \), like \( {\pi }_{\mathcal{P}} \), identifies vertices only when forced to do so by the relation generated by edge identifications. To prove this, suppose \( v \in K \) is any vertex. It must be the case that \( v \) belongs to some 1 -simplex, because otherwise it would be an isolated point of \( \left| K\right| \), contradicting the fact that \( \left| K\right| \) is a 2-manifold. Theorem 5.36 guarantees that this 1-simplex is a face of exactly two 2-simplices. Let us say that two 2-simplices \( \sigma ,{\sigma }^{\prime } \) containing \( v \) are edge-connected at \( v \) if there is a sequence \( \sigma = {\sigma }_{1},\ldots ,{\sigma }_{k} = {\sigma }^{\prime } \) of 2 -simplices containing \( v \) such that \( {\sigma }_{i} \) shares an edge with \( {\sigma }_{i + 1} \) for each \( i = 1,\ldots, k - 1 \) . Clearly edge-connectedness is an equivalence relation on the set of 2-simplices containing \( v \), so to prove the claim it suffices to show that there is only one equivalence class. If this is not the case, we can group the 2-simplices containing \( v \) into two disjoint sets \( \left\{ {{\sigma }_{1},\ldots ,{\sigma }_{k}}\right\} \) and \( \left\{ {{\tau }_{1},\ldots ,{\tau }_{m}}\right\} \), such that any \( {\sigma }_{i} \) and \( {\sigma }_{j} \) are edge-connected to each other, but no \( {\tau }_{i} \) is edge-connected to any \( {\sigma }_{j} \) . Let \( \varepsilon \) be chosen small enough that \( {B}_{\varepsilon }\left( v\right) \) intersects only those simplices that contain \( v \) . Then \( {B}_{\varepsilon }\left( v\right) \cap \left| K\right| \) is an open subset of \( \left| K\right| \) and thus a 2-manifold, so \( v \) has a neighborhood \( W \)
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No
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Lemma 6.16. The Klein bottle is homeomorphic to \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) .
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Proof. By a sequence of elementary transformations, we find that the Klein bottle has the following presentations (see Fig. 6.18):\n\n\[ \left\langle {a, b \mid {aba}{b}^{-1}}\right\rangle \]\n\n\[ \approx \left\langle {a, b, c \mid {abc},{c}^{-1}a{b}^{-1}}\right\rangle \;\text{ (cut along }c\text{ ) } \]\n\n\[ \approx \left\langle {a, b, c \mid {bca},{a}^{-1}{cb}}\right\rangle \;\text{(rotate and reflect)} \]\n\n\[ \approx \langle b, c \mid {bbcc}\rangle \;\text{(paste along}a\text{and rotate).} \]\n\nThe presentation in the last line is our standard presentation of a connected sum of two projective planes.
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Yes
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Lemma 6.17. The connected sum \( {\mathbb{T}}^{2}\# {\mathbb{P}}^{2} \) is homeomorphic to \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) .
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Proof. Start with \( \left\langle {a, b, c \mid {aba}{b}^{-1}{cc}}\right\rangle \) (Fig. 6.19), which is a presentation of \( K\# {\mathbb{P}}^{2} \) , and therefore by the preceding lemma is a presentation of \( {\mathbb{P}}^{2}\# {\mathbb{P}}^{2}\# {\mathbb{P}}^{2} \) . Following Fig. 6.19, we cut along \( d \), paste along \( c \), cut along \( e \), and paste along \( b \), rotating and reflecting as necessary, to obtain \( \left\langle {a, d, e \mid {a}^{-1}{d}^{-1}{adee}}\right\rangle \), which is a presentation of \( {\mathbb{T}}^{2}\# {\mathbb{P}}^{2} \) .
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Yes
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Proposition 6.18. The Euler characteristic of a polygonal presentation is unchanged by elementary transformations.
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Proof. It is immediate that relabeling, rotating, and reflecting do not change the Euler characteristic of a presentation, because they leave the numbers of 0-cells, 1-cells, and 2-cells individually unchanged. For the other transformations, we need only check that the changes to these three numbers cancel out. Subdividing increases both the number of 1-cells and the number of 0-cells by one, leaving the number of 2-cells unchanged. Cutting increases both the number of 1-cells and the number of 2-cells by one, and leaves the number of 0-cells unchanged. Unfolding increases the number of 1-cells and the number of 0-cells by one, and leaves the number of 2-cells unchanged. Finally, consolidating, pasting, and folding leave the Euler characteristic unchanged, since they are the inverses of subdividing, cutting, and unfolding, respectively.
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Yes
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Proposition 6.19 (Euler Characteristics of Compact Surfaces). The Euler characteristic of a standard surface presentation is equal to\n\n(a) 2 for the sphere,\n\n(b) \( 2 - {2n} \) for the connected sum of \( n \) tori,\n\n(c) \( 2 - n \) for the connected sum of \( n \) projective planes.
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Proof. Just compute.
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No
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Proposition 6.20. A compact surface is orientable if and only if it is homeomorphic to the sphere or a connected sum of one or more tori.
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Proof. The standard presentations of the sphere and the connected sums of tori are oriented, so these surfaces are certainly orientable. To show that an orientable surface is homeomorphic to one of these, let \( M \) be any surface that admits at least one orientable presentation. Starting with that presentation, follow the algorithm described in the proof of the classification theorem to transform it to one of the standard presentations. The only elementary transformation that can introduce a twisted pair into an oriented presentation is reflection. The only steps in which reflections are used are Steps 1, 3, 4, and 7, and you can check that none of those steps require any reflections if there were no twisted pairs to begin with. Thus the classification theorem tells us that the presentation can be reduced to one of the standard ones with no twisted pairs, which means that \( M \) is homeomorphic to a sphere or a connected sum of tori.
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Yes
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Proposition 7.1. For any topological spaces \( X \) and \( Y \), homotopy is an equivalence relation on the set of all continuous maps from \( X \) to \( Y \) .
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Proof. Any map \( f \) is homotopic to itself via the trivial homotopy \( H\left( {x, t}\right) = f\left( x\right) \) , so homotopy is reflexive. Similarly, if \( H : f \simeq g \), then a homotopy from \( g \) to \( f \) is given by \( \widetilde{H}\left( {x, t}\right) = H\left( {x,1 - t}\right) \), so homotopy is symmetric. Finally, if \( F : f \simeq g \) and \( G : g \simeq h \), define \( H : X \times I \rightarrow Y \) by following \( F \) at double speed for \( 0 \leq t \leq \frac{1}{2} \), and then following \( G \) at double speed for the remainder of the unit interval. Formally,\n\n\[ H\left( {x, t}\right) = \left\{ \begin{array}{ll} F\left( {x,{2t}}\right) , & 0 \leq t \leq \frac{1}{2} \\ G\left( {x,{2t} - 1}\right) , & \frac{1}{2} \leq t \leq 1 \end{array}\right. \]\n\nSince \( F\left( {x,1}\right) = g\left( x\right) = G\left( {x,0}\right) \), the two definitions of \( H \) agree at \( t = \frac{1}{2} \), where they overlap. Thus \( H \) is continuous by the gluing lemma, and is therefore a homotopy between \( f \) and \( h \) . This shows that homotopy is transitive.
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Yes
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Proposition 7.2. The homotopy relation is preserved by composition: if\n\n\\[ \n{f}_{0},{f}_{1} : X \rightarrow Y\\;\\text{ and }\\;{g}_{0},{g}_{1} : Y \rightarrow Z\n\\]\n\nare continuous maps with \\( {f}_{0} \simeq {f}_{1} \\) and \\( {g}_{0} \simeq {g}_{1} \\), then \\( {g}_{0} \circ {f}_{0} \simeq {g}_{1} \circ {f}_{1} \\) .
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Proof. Suppose \\( F : {f}_{0} \simeq {f}_{1} \\) and \\( G : {g}_{0} \simeq {g}_{1} \\) are homotopies. Define \\( H : X \times I \rightarrow \\) \\( Z \\) by \\( H\\left( {x, t}\\right) = G\\left( {F\\left( {x, t}\\right), t}\\right) \\) . At \\( t = 0, H\\left( {x,0}\\right) = G\\left( {{f}_{0}\\left( x\\right) ,0}\\right) = {g}_{0}\\left( {{f}_{0}\\left( x\\right) }\\right) \\), and at \\( t = 1, H\\left( {x,1}\\right) = G\\left( {{f}_{1}\\left( x\\right) ,1}\\right) = {g}_{1}\\left( {{f}_{1}\\left( x\\right) }\\right) \\) . Thus \\( H \\) is a homotopy from \\( {g}_{0} \circ {f}_{0} \\) to \\( {g}_{1} \circ {f}_{1} \\) .
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Yes
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Let \( B \subseteq {\mathbb{R}}^{n} \) and let \( X \) be any topological space. Suppose \( f, g : X \rightarrow \) \( B \) are any two continuous maps with the property that for all \( x \in X \), the line segment from \( f\left( x\right) \) to \( g\left( x\right) \) lies in \( B \) . This is the case, for example, if \( B \) is convex. Define a homotopy \( H : f \simeq g \) by letting \( H\left( {x, t}\right) \) trace out the line segment from \( f\left( x\right) \) to \( g\left( x\right) \) at constant speed as \( t \) goes from 0 to 1 (Fig. 7.2):
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\[ H\left( {x, t}\right) = f\left( x\right) + t\left( {g\left( x\right) - f\left( x\right) }\right) . \]
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Yes
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Example 7.5. Consider the path \( f : I \rightarrow \mathbb{C} \smallsetminus \{ 0\} \) defined (in complex notation) by\n\n\[ f\left( s\right) = {e}^{2\pi is} \]\n\nand the map \( H : I \times I \rightarrow \mathbb{C} \smallsetminus \{ 0\} \) by\n\n\[ H\left( {s, t}\right) = {e}^{2\pi ist}. \]\n\nAt each time \( t,{H}_{t} \) is a path that follows the circle only as far as angle \( {2\pi t} \), so \( {H}_{0} \) is the constant path \( {c}_{1}\left( s\right) \equiv 1 \) and \( {H}_{1} = f \) . Thus \( H \) is a homotopy from the constant path to \( f \) .
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This last example shows that the circular path around the origin is homotopic in \( {\mathbb{R}}^{2} \smallsetminus \{ 0\} \) to a constant path, so that simply asking whether a closed path is homotopic to a constant is not sufficient to detect holes. To remedy this, we need to consider homotopies of paths throughout which the endpoints stay fixed. More generally, it is useful to consider homotopies that fix an arbitrary subset of the domain.
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No
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Lemma 7.9. Any reparametrization of a path \( f \) is path-homotopic to \( f \) .
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Proof. Suppose \( f \circ \varphi \) is a reparametrization of \( f \), and let \( H : I \times I \rightarrow I \) denote the straight-line homotopy from the identity map to \( \varphi \) . Then \( f \circ H \) is a path homotopy from \( f \) to \( f \circ \varphi \) .
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Yes
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Proposition 7.10 (Homotopy Invariance of Path Multiplication). The operation of path multiplication is well defined on path classes. More precisely, if \( {f}_{0} \sim {f}_{1} \) and \( {g}_{0} \sim {g}_{1} \), and if \( {f}_{0} \) and \( {g}_{0} \) are composable, then \( {f}_{1} \) and \( {g}_{1} \) are composable and \( {f}_{0} \cdot {g}_{0} \sim {f}_{1} \cdot {g}_{1} \)
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Proof. Let \( F : {f}_{0} \sim {f}_{1} \) and \( G : {g}_{0} \sim {g}_{1} \) be path homotopies (Fig. 7.3). The required homotopy \( H : {f}_{0} \cdot {g}_{0} \sim {f}_{1} \cdot {g}_{1} \) is given by\n\n\[ H\left( {s, t}\right) = \left\{ \begin{array}{ll} F\left( {{2s}, t}\right) ; & 0 \leq s \leq \frac{1}{2},0 \leq t \leq 1; \\ G\left( {{2s} - 1, t}\right) ; & \frac{1}{2} \leq s \leq 1,0 \leq t \leq 1. \end{array}\right. \]\n\n\n\nAgain, this is continuous by the gluing lemma.
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Yes
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Theorem 7.11 (Properties of Path Class Products). Let \( f \) be any path from p to \( q \) in a space \( X \), and let \( g \) and \( h \) be any paths in \( X \). Path multiplication satisfies the following properties:\n\n(a) \( \left\lbrack {c}_{p}\right\rbrack \cdot \left\lbrack f\right\rbrack = \left\lbrack f\right\rbrack \cdot \left\lbrack {c}_{q}\right\rbrack = \left\lbrack f\right\rbrack \).\n\n(b) \( \left\lbrack f\right\rbrack \cdot \left\lbrack \bar{f}\right\rbrack = \left\lbrack {c}_{p}\right\rbrack ;\left\lbrack \bar{f}\right\rbrack \cdot \left\lbrack f\right\rbrack = \left\lbrack {c}_{q}\right\rbrack \).\n\n(c) \( \left\lbrack f\right\rbrack \cdot \left( {\left\lbrack g\right\rbrack \cdot \left\lbrack h\right\rbrack }\right) = \left( {\left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack }\right) \cdot \left\lbrack h\right\rbrack \) whenever either side is defined.
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Proof. For (a), let us show that \( {c}_{p} \cdot f \sim f \) ; the product the other way works similarly. Define \( H : I \times I \rightarrow X \) (Fig. 7.4) by\n\n\[ H\left( {s, t}\right) = \left\{ \begin{array}{ll} p, & t \geq {2s} \\ f\left( \frac{{2s} - t}{2 - t}\right) , & t \leq {2s} \end{array}\right. \]\n\nGeometrically, this maps the portion of the square on the left of the line \( t = {2s} \) to the point \( p \), and it maps the portion on the right along the path \( f \) at increasing speeds as \( t \) goes from 0 to 1 . (The slanted lines in the picture are the level sets of \( H \), i.e., the lines along which \( H \) takes the same value.) This map is continuous by the gluing lemma, and you can check that \( H\left( {s,0}\right) = f\left( s\right) \) and \( H\left( {s,1}\right) = {c}_{p} \cdot f\left( s\right) \). Thus \( H : f \sim {c}_{p} \cdot f \).\n\nFor (b), we just show that \( f \cdot \bar{f} \sim {c}_{p} \). Since the reverse path of \( \bar{f} \) is \( f \), the other relation follows by interchanging the roles of \( f \) and \( \bar{f} \). Define a homotopy \( H : {c}_{p} \sim f \cdot \bar{f} \) by the following recipe (Fig. 7.5): at any time \( t \), the path \( {H}_{t} \) follows \( f \) as far as \( f\left( t\right) \) at double speed while the parameter \( s \) is in the interval \( \left\lbrack {0, t/2}\right\rbrack \) ; then for \( s \in \left\lbrack {t/2,1 - t/2}\right\rbrack \) it stays at \( f\left( t\right) \) ; then it retraces \( f \) at double speed back to \( p \). Formally,\n\n\[ H\left( {s, t}\right) = \left\{ \begin{array}{ll} f\left( {2s}\right) , & 0 \leq s \leq t/2 \\ f\left( t\right) , & t/2 \leq s \leq 1 - t/2 \\ f\left( {2 - {2s}}\right) & 1 - t/2 \leq s \leq 1 \end{array}\right. \]\n\nIt is easy to check that \( H \) is a homotopy from \( {c}_{p} \) to \( f \cdot \bar{f} \).\n\nFinally, to prove associativity, we need to show that \( \left( {f \cdot g}\right) \cdot h \sim f \cdot \left( {g \cdot h}\right) \). The first path follows \( f \) and then \( g \) at quadruple speed for \( s \in \left\lbrack {0,\frac{1}{2}}\right\rbrack \), and then follows \( h \) at double speed for \( s \in \left\lbrack {\frac{1}{2},1}\right\rbrack \), while the second follows \( f \) at double speed and then \( g \) and \( h \) at quadruple speed. The two paths are therefore reparametrizations of each other and thus homotopic.
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Yes
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Theorem 7.13 (Change of Base Point). Suppose \( X \) is path-connected, \( p, q \in X \) , and \( g \) is any path from \( p \) to \( q \) . The map \( {\Phi }_{g} : {\pi }_{1}\left( {X, p}\right) \rightarrow {\pi }_{1}\left( {X, q}\right) \) defined by\n\n\[ \n{\Phi }_{g}\left\lbrack f\right\rbrack = \left\lbrack \bar{g}\right\rbrack \cdot \left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack \n\]\n\nis an isomorphism, whose inverse is \( {\Phi }_{\bar{g}} \) .
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Proof. Before we begin, we should verify that \( {\Phi }_{g} \) makes sense (Fig. 7.6): since \( g \) goes from \( p \) to \( q \) and \( f \) goes from \( p \) to \( p \), paths in the class \( \left\lbrack \bar{g}\right\rbrack \cdot \left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack \) go from \( q \) to \( p \) (by \( \bar{g} \) ), then from \( p \) to \( p \) (by \( f \) ), and then from \( p \) back to \( q \) (by \( g \) ), so \( {\Phi }_{g}\left( f\right) \) does indeed define an element of \( {\pi }_{1}\left( {X, q}\right) \) .\n\nTo check that \( {\Phi }_{g} \) is a group homomorphism, use Theorem 7.11:\n\n\n\nFig. 7.6: Change of base point.\n\n\[ \n{\Phi }_{g}\left\lbrack {f}_{1}\right\rbrack \cdot {\Phi }_{g}\left\lbrack {f}_{2}\right\rbrack = \left\lbrack \bar{g}\right\rbrack \cdot \left\lbrack {f}_{1}\right\rbrack \cdot \left\lbrack g\right\rbrack \cdot \left\lbrack \bar{g}\right\rbrack \cdot \left\lbrack {f}_{2}\right\rbrack \cdot \left\lbrack g\right\rbrack \n\]\n\n\[ \n= \left\lbrack \bar{g}\right\rbrack \cdot \left\lbrack {f}_{1}\right\rbrack \cdot \left\lbrack {c}_{p}\right\rbrack \cdot \left\lbrack {f}_{2}\right\rbrack \cdot \left\lbrack g\right\rbrack \n\]\n\n\[ \n= \left\lbrack \bar{g}\right\rbrack \cdot \left\lbrack {f}_{1}\right\rbrack \cdot \left\lbrack {f}_{2}\right\rbrack \cdot \left\lbrack g\right\rbrack \n\]\n\n\[ \n= {\Phi }_{g}\left( {\left\lbrack {f}_{1}\right\rbrack \cdot \left\lbrack {f}_{2}\right\rbrack }\right) \n\]\n\n(This is one reason why we needed to prove the properties of Theorem 7.11 for paths that start and end at different points.)\n\nFinally, the fact that \( {\Phi }_{g} \) is an isomorphism follows easily from the fact that it has an inverse, given by \( {\Phi }_{\bar{g}} : {\pi }_{1}\left( {X, q}\right) \rightarrow {\pi }_{1}\left( {X, p}\right) \) .
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Yes
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Lemma 7.17 (Square Lemma). Let \( F : I \times I \rightarrow X \) be a continuous map, and let \( f, g, h \), and \( k \) be the paths in \( X \) defined by\n\n\[ f\left( s\right) = F\left( {s,0}\right) \]\n\n\[ g\left( s\right) = F\left( {1, s}\right) \]\n\n\[ h\left( s\right) = F\left( {0, s}\right) \]\n\n\[ k\left( s\right) = F\left( {s,1}\right) \text{.} \]\n\n(See Fig. 7.8.) Then \( f \cdot g \sim h \cdot k \) .
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Proof. See Problem 7-4.
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No
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Lemma 7.18 (Lebesgue Number Lemma). Every open cover of a compact metric space has a Lebesgue number.
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Proof. Let \( \mathcal{U} \) be an open cover of the compact metric space \( M \) . Each point \( x \in M \) is in some set \( U \in \mathcal{U} \) . Since \( U \) is open, there is some \( r\left( x\right) > 0 \) such that \( {B}_{{2r}\left( x\right) }\left( x\right) \subseteq U \) . The balls \( \left\{ {{B}_{r\left( x\right) }\left( x\right) : x \in M}\right\} \) form an open cover of \( M \), so finitely many of them, say \( {B}_{r\left( {x}_{1}\right) }\left( {x}_{1}\right) ,\ldots ,{B}_{r\left( {x}_{n}\right) }\left( {x}_{n}\right) \), cover \( M \) .\n\nWe will show that \( \delta = \min \left\{ {r\left( {x}_{1}\right) ,\ldots, r\left( {x}_{n}\right) }\right\} \) is a Lebesgue number for \( \mathcal{U} \) . To see why, suppose \( S \subseteq M \) is a nonempty set whose diameter is less than \( \delta \) . Let \( {y}_{0} \) be any point of \( S \) ; then there is some \( {x}_{i} \) such that \( {y}_{0} \in {B}_{r\left( {x}_{i}\right) }\left( {x}_{i}\right) \) (Fig. 7.9). It suffices to show that \( S \subseteq {B}_{{2r}\left( {x}_{i}\right) }\left( {x}_{i}\right) \), since the latter set is by construction contained in some \( U \in \mathcal{U} \) . If \( z \in S \), the triangle inequality gives\n\n\[ d\left( {z,{x}_{i}}\right) \leq d\left( {z,{y}_{0}}\right) + d\left( {{y}_{0},{x}_{i}}\right) < \delta + r\left( {x}_{i}\right) < {2r}\left( {x}_{i}\right) ,\]\n\nwhich proves the claim.
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Yes
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Lemma 7.19. Suppose \( M \) is a manifold of dimension \( n \geq 2 \). If \( f \) is a path in \( M \) from \( {p}_{1} \) to \( {p}_{2} \) and \( q \) is any point in \( M \) other than \( {p}_{1} \) or \( {p}_{2} \), then \( f \) is path-homotopic to a path that does not pass through \( q \).
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Proof. Consider the open cover \( \{ U, V\} \) of \( M \), where \( U \) is a coordinate ball centered at \( q \) and \( V = M \smallsetminus \{ q\} \). If \( f : I \rightarrow M \) is any path from \( {p}_{1} \) to \( {p}_{2} \), then \( \left\{ {{f}^{-1}\left( U\right) ,{f}^{-1}\left( V\right) }\right\} \) is an open cover of \( I \). Let \( \delta \) be a Lebesgue number for this cover, and let \( m \) be a positive integer such that \( 1/m < \delta \). It follows that on each subinterval \( \left\lbrack {k/m,\left( {k + 1}\right) /m}\right\rbrack, f \) takes its values either in \( U \) or in \( V \). If \( f\left( {k/m}\right) = q \) for some \( k \), then the two subintervals \( \left\lbrack {\left( {k - 1}\right) /m, k/m}\right\rbrack \) and \( \left\lbrack {k/m,\left( {k + 1}\right) /m}\right\rbrack \) must both be mapped into \( U \). Thus, letting \( 0 = {a}_{0} < \cdots < {a}_{l} = 1 \) be the points of the form \( k/m \) for which \( f\left( {a}_{i}\right) \neq q \), we obtain a sequence of curve segments \( {\left. f\right| }_{\left\lbrack {a}_{i - 1},{a}_{i}\right\rbrack } \) whose images lie either in \( U \) or in \( V \), and for which \( f\left( {a}_{i}\right) \neq q \).\n\nNow, \( U \smallsetminus \{ q\} \) is homeomorphic to \( {\mathbb{B}}^{n} \smallsetminus \{ 0\} \), which is path-connected. (Here is where the dimensional restriction comes in: when \( n = 1,{\mathbb{B}}^{n} \smallsetminus \{ 0\} \) is disconnected.) Thus, for each such segment that lies in \( U \), there is another path in \( U \smallsetminus \{ q\} \) with the same endpoints; since \( U \) is simply connected, these two paths are path-homotopic in \( U \) and thus in \( M \). Of course, each segment that lies in \( V \) already misses \( q \).
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Yes
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Theorem 7.21. The fundamental group of a manifold is countable.
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Proof. Let \( M \) be a manifold, and let \( \mathcal{U} \) be a countable cover of \( M \) by coordinate balls. For each \( U,{U}^{\prime } \in \mathcal{U} \) the intersection \( U \cap {U}^{\prime } \) has at most countably many components; choose a point in each such component and let \( X \) denote the (countable) set consisting of all the chosen points as \( U,{U}^{\prime } \) range over all the sets in \( \mathcal{U} \) . For each \( U \in \mathcal{U} \) and \( x,{x}^{\prime } \in \mathcal{X} \) such that \( x,{x}^{\prime } \in U \), choose a definite path \( {h}_{x,{x}^{\prime }}^{U} \) from \( x \) to \( {x}^{\prime } \) in \( U \) .\n\nNow choose any point \( p \in X \) as base point. Let us say that a loop based at \( p \) is special if it is a finite product of paths of the form \( {h}_{x,{x}^{\prime }}^{U} \) . Because both \( \mathcal{U} \) and \( x \) are countable sets, there are only countably many special loops. Each special loop determines an element of \( {\pi }_{1}\left( {M, p}\right) \) . If we can show that every element of \( {\pi }_{1}\left( {M, p}\right) \) is obtained in this way, we are done, because we will have exhibited a surjective map from a countable set onto \( {\pi }_{1}\left( {M, p}\right) \) .\n\nSo suppose \( f \) is any loop based at \( p \) . By the Lebesgue number lemma there is an integer \( n \) such that \( f \) maps each subinterval \( \left\lbrack {\left( {k - 1}\right) /n, k/n}\right\rbrack \) into one of the balls in \( \mathcal{U} \) ; call this ball \( {U}_{k} \) . Let \( {f}_{k} = {\left. f\right| }_{\left\lbrack \left( k - 1\right) /n, k/n\right\rbrack } \) reparametrized on the unit interval, so that \( \left\lbrack f\right\rbrack = \left\lbrack {f}_{1}\right\rbrack \cdot \cdots \cdot \left\lbrack {f}_{n}\right\rbrack \) (Fig. 7.10).\n\nFor each \( k = 1,\ldots, n - 1 \), the point \( f\left( {k/n}\right) \) lies in \( {U}_{k} \cap {U}_{k + 1} \) . Therefore, there is some \( {x}_{k} \in X \) that lies in the same component of \( {U}_{k} \cap {U}_{k + 1} \) as \( f\left( {k/n}\right) \) . Choose a path \( {g}_{k} \) in \( {U}_{k} \cap {U}_{k + 1} \) from \( {x}_{k} \) to \( f\left( {k/n}\right) \), and set \( {\widetilde{f}}_{k} = {g}_{k - 1} \cdot {f}_{k} \cdot {\bar{g}}_{k} \) (taking \( {x}_{k} = p \) and \( {g}_{k} \) to be the constant path \( {c}_{p} \) when \( k = 0 \) or \( n \) ). It is immediate that \( \left\lbrack f\right\rbrack = \left\lbrack {\widetilde{f}}_{1}\right\rbrack \cdot \cdots \cdot \left\lbrack {\widetilde{f}}_{n}\right\rbrack \), because all the \( {g}_{k} \) ’s cancel out. But for each \( k,{\widetilde{f}}_{k} \) is a path in \( {U}_{k} \) from \( {x}_{k - 1} \) to \( {x}_{k} \), and since \( {U}_{k} \) is simply connected, \( {\widetilde{f}}_{k} \) is path-homotopic to \( {h}_{{x}_{k - 1}{x}_{k}}^{{U}_{k}} \) . This shows that \( f \) is path-homotopic to a special loop and completes the proof.
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Yes
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Proposition 7.24. For any continuous map \( \varphi ,{\varphi }_{ * } \) is a group homomorphism.
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Proof. Just note that\n\n\[ \n{\varphi }_{ * }\left( {\left\lbrack f\right\rbrack \cdot \left\lbrack g\right\rbrack }\right) = {\varphi }_{ * }\left\lbrack {f \cdot g}\right\rbrack = \left\lbrack {\varphi \circ \left( {f \cdot g}\right) }\right\rbrack .\n\] \n\nThus it suffices to show that \( \varphi \circ \left( {f \cdot g}\right) = \left( {\varphi \circ f}\right) \cdot \left( {\varphi \circ g}\right) \) . This is immediate, because expanding both sides using the definition of path multiplication results in identical formulas.
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Yes
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