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Proposition 3.42 (Other Properties of Disjoint Union Spaces). Let \( {\left( {X}_{\alpha }\right) }_{\alpha \in A} \) be an indexed family of topological spaces.\n\n(a) A subset of \( \mathop{\coprod }\limits_{{\alpha \in A}}{X}_{\alpha } \) is closed if and only if its intersection with each \( {X}_{\alpha } \) is clo... | Exercise 3.43. Prove Proposition 3.42. | No |
Define an equivalence relation on the square \( I \times I \) by setting \( \left( {x,0}\right) \sim \) \( \left( {x,1}\right) \) for all \( x \in I \), and \( \left( {0, y}\right) \sim \left( {1, y}\right) \) for all \( y \in I \) (Fig. 3.9). This can be visualized as the space obtained by pasting the top boundary seg... | Later we will prove that the resulting quotient space is homeomorphic to the torus (see Example 4.52). | No |
Define \( {\mathbb{P}}^{n} \), the real projective space of dimension \( \mathbf{n} \), to be the set of 1-dimensional linear subspaces (lines through the origin) in \( {\mathbb{R}}^{n + 1} \). There is a natural map \( q : {\mathbb{R}}^{n + 1} \smallsetminus \{ 0\} \rightarrow {\mathbb{P}}^{n} \) defined by sending a ... | Projective space can also be viewed in another way. If we define an equivalence relation on \( {\mathbb{R}}^{n + 1} \smallsetminus \{ 0\} \) by declaring two points \( x, y \) to be equivalent if \( x = {\lambda y} \) for some nonzero real number \( \lambda \), then there is an obvious identification between \( {\mathb... | Yes |
Proposition 3.56. Suppose \( P \) is a second countable space and \( M \) is a quotient space of \( P \) . If \( M \) is locally Euclidean, then it is second countable. Thus if \( M \) is locally Euclidean and Hausdorff, it is a manifold. | Proof. Let \( q : P \rightarrow M \) denote the quotient map, and let \( \mathcal{U} \) be a cover of \( M \) by coordinate balls. The collection \( \left\{ {{q}^{-1}\left( U\right) : U \in \mathcal{U}}\right\} \) is an open cover of \( P \), which has a countable subcover by Theorem 2.50 . If we let \( {\mathcal{U}}^{... | Yes |
Proposition 3.57. Suppose \( q : X \rightarrow Y \) is an open quotient map. Then \( Y \) is Hausdorff if and only if the set \( \mathcal{R} = \left\{ {\left( {{x}_{1},{x}_{2}}\right) : q\left( {x}_{1}\right) = q\left( {x}_{2}\right) }\right\} \) is closed in \( X \times X \) . | Proof. First assume \( Y \) is Hausdorff. If \( \left( {{x}_{1},{x}_{2}}\right) \notin \mathcal{R} \), then there are disjoint neighborhoods \( {V}_{1} \) of \( q\left( {x}_{1}\right) \) and \( {V}_{2} \) of \( q\left( {x}_{2}\right) \), and it follows that \( {q}^{-1}\left( {V}_{1}\right) \times {q}^{-1}\left( {V}_{2}... | Yes |
Consider the map \( \omega : I \rightarrow {\mathbb{S}}^{1} \) that wraps the interval once around the circle at constant speed, given (in complex notation) by \( \omega \left( s\right) = {e}^{2\pi is} \). This map is continuous and surjective. To show that it is a quotient map, let \( U \subseteq {\mathbb{S}}^{1} \) b... | If \( U \) is open, then \( {\omega }^{-1}\left( U\right) \) is open by continuity. Conversely, suppose \( {\omega }^{-1}\left( U\right) \) is open, and let \( z \) be a point in \( U \). If \( z \neq 1 \), then \( z = \omega \left( {s}_{0}\right) \) for a unique \( {s}_{0} \in \left( {0,1}\right) \), and there is some... | Yes |
Proposition 3.67. If \( q : X \rightarrow Y \) is a surjective continuous map that is also an open or closed map, then it is a quotient map. | Proof. If \( q \) is open, it takes saturated open subsets to open subsets (because it takes all open subsets to open subsets). If \( q \) is closed, it takes saturated closed subsets to closed subsets. In either case, it is a quotient map by Proposition 3.60. | Yes |
If \( {X}_{1},\ldots ,{X}_{k} \) are topological spaces, then each canonical projection \( {\pi }_{i} : {X}_{1} \times \cdots \times {X}_{k} \rightarrow {X}_{i} \) is a quotient map. | because it is continuous, surjective, and open. | No |
Theorem 3.70 (Characteristic Property of the Quotient Topology). Suppose \( X \) and \( Y \) are topological spaces and \( q : X \rightarrow Y \) is a quotient map. For any topological space \( Z \), a map \( f : Y \rightarrow Z \) is continuous if and only if the composite map \( f \circ q \) is continuous: | Proof. This result follows immediately from the fact that for any open subset \( U \subseteq \) \( Z,{f}^{-1}\left( U\right) \) is open in \( Y \) if and only if \( {q}^{-1}\left( {{f}^{-1}\left( U\right) }\right) = {\left( f \circ q\right) }^{-1}\left( U\right) \) is open in \( X \) . | Yes |
Theorem 3.73 (Passing to the Quotient). Suppose \( q : X \rightarrow Y \) is a quotient map, \( Z \) is a topological space, and \( f : X \rightarrow Z \) is any continuous map that is constant on the fibers of \( q \) (i.e., if \( q\left( x\right) = q\left( {x}^{\prime }\right) \), then \( \left. {f\left( x\right) = f... | Proof. The existence and uniqueness of \( \widetilde{f} \) follow from elementary set theory: given \( y \in Y \), there is some \( x \in X \) such that \( q\left( x\right) = y \), and we can set \( \widetilde{f}\left( y\right) = f\left( x\right) \) for any such \( x \) . The hypothesis on \( f \) guarantees that \( \w... | Yes |
Theorem 3.75 (Uniqueness of Quotient Spaces). Suppose \( {q}_{1} : X \rightarrow {Y}_{1} \) and \( {q}_{2} : X \rightarrow {Y}_{2} \) are quotient maps that make the same identifications (i.e., \( {q}_{1}\left( x\right) = \left. {{q}_{1}\left( {x}^{\prime }\right) \text{if and only if}{q}_{2}\left( x\right) = {q}_{2}\l... | Proof. By Theorem 3.73, both \( {q}_{1} \) and \( {q}_{2} \) pass uniquely to the quotient as in the following diagrams:\n\n\n\n\nSince both of these diagrams commute, it follows that\n\n\[{\widetilde{q}}_{1} \circ \le... | Yes |
To show that the quotient space \( I/ \sim \) of Example 3.47 is homeomorphic to the circle, all we need to do is exhibit a quotient map \( \omega : I \rightarrow {\mathbb{S}}^{1} \) that makes the same identifications as \( \sim \) . | The map described in Example 3.66 is such a map. | No |
Proposition 3.77 (Properties of Adjunction Spaces). Let \( X{ \cup }_{f}Y \) be an adjunction space, and let \( q : X \coprod Y \rightarrow X{ \cup }_{f}Y \) be the associated quotient map.\n\n(a) The restriction of \( q \) to \( X \) is a topological embedding, whose image set \( q\left( X\right) \) is a closed subspa... | Proof. We begin by showing that \( {\left. q\right| }_{X} \) is a closed map. Suppose that \( B \) is a closed subset of \( X \) . To show that \( q\left( B\right) \) is closed in the quotient space, we need to show that \( {q}^{-1}\left( {q\left( B\right) }\right) \) is closed in \( X \coprod Y \), which is equivalent... | Yes |
Example 3.86 (More Topological Groups). In view of Proposition 3.84, each of the following is a topological group, with the product topology or subspace topology as appropriate:\n\n- Euclidean space \( {\mathbb{R}}^{n} = \mathbb{R} \times \cdots \times \mathbb{R} \) as a group under vector addition\n\n- the group \( {\... | If \( G \) is a topological group and \( g \in G \), left translation by \( g \) is the map \( {L}_{g} : G \rightarrow \) \( G \) defined by \( {L}_{g}\left( {g}^{\prime }\right) = g{g}^{\prime } \) . It is continuous, because it is equal to the composition\n\n\[ G\overset{ig}{ \rightarrow }G \times G\overset{m}{ \righ... | Yes |
Proposition 3.87. Suppose \( G \) is a topological group acting on a topological space \( X \) .\n\n(a) If the action is continuous, then it is an action by homeomorphisms.\n\n(b) If \( G \) has the discrete topology, then the action is continuous if and only if it is an action by homeomorphisms. | Proof. First suppose the action is continuous. This means, in particular, that for each \( g \in G \) the map \( x \mapsto g \cdot x \) is continuous from \( X \) to itself, because it is the composition \( x \mapsto \left( {g, x}\right) \mapsto g \cdot x \) . Each such map is a homeomorphism, because the definition of... | Yes |
Example 3.89. As we mentioned above, the action of \( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) on \( {\mathbb{R}}^{n} \) by matrix multiplication has two orbits, so the quotient space has exactly two points: \( a = \) \( q\left( {{\mathbb{R}}^{n}\smallsetminus \{ 0\} }\right) \) and \( b = q\left( {\{ 0\} }\right) ... | This quotient space is not Hausdorff. | No |
As an application, let us consider the coset space \( \mathbb{R}/\mathbb{Z} \) . Because \( \mathbb{Z} \) is a subgroup of the topological group \( \mathbb{R} \), there is a natural free continuous action of \( \mathbb{Z} \) on \( \mathbb{R} \) by translation: \( n \cdot x = n + x \) . (Because \( \mathbb{R} \) is abel... | Consider also the map \( \varepsilon : \mathbb{R} \rightarrow {\mathbb{S}}^{1} \) defined (in complex notation) by\n\n\[ \varepsilon \left( r\right) = {e}^{2\pi ir}. \]\n\nIt is straightforward to check that this is a local homeomorphism and thus an open map, so it is a quotient map. Because it makes the same identific... | Yes |
Proposition 4.1. A topological space \( X \) is connected if and only if the only subsets of \( X \) that are both open and closed in \( X \) are \( \varnothing \) and \( X \) itself. | Proof. Assume first that \( X \) is connected, and suppose that \( U \subseteq X \) is open and closed. Then \( V = X \smallsetminus U \) is also open and closed. If both \( U \) and \( V \) were nonempty, then they would disconnect \( X \) ; therefore, either \( V \) is empty, which means that \( U = X \), or \( U \) ... | Yes |
Proposition 4.2. Suppose \( X \) is a nonempty connected space. Then every continuous map from \( X \) to a discrete space is constant. | 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 n... | Yes |
Each of the following topological spaces is disconnected. | (a) \( \mathbb{R} \smallsetminus \{ 0\} \) (Fig. 4.1) is disconnected by the two open subsets \( \{ x : x > 0\} \) and \( \{ x : x < 0\} \) . | Yes |
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. | 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\... | Yes |
Proposition 4.11. A nonempty subset of \( \mathbb{R} \) is connected if and only if it is a singleton or an interval. | 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 \( J \) . Ch... | Yes |
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 \( f\\left( q\\right) \) . | Proof. By the main theorem on connectedness, \( f\\left( X\\right) \) is connected, so it must be a singleton or an interval. | No |
Theorem 4.15. Path connectedness implies connectedness. | 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_... | Yes |
Proposition 4.18. If \( X \) is any topological space, its components form a partition of \( X \) . | 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... | Yes |
## 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. | 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 \... | Yes |
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. | 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 ... | Yes |
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 co... | 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 \( B \subseteq A \) an... | Yes |
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. | 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 ... | Yes |
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. | 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 colle... | Yes |
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 \( B \subseteq V \) . | 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... | Yes |
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 \( V \times Y \subseteq U \) . | 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 \ | No |
Theorem 4.39. Every closed, bounded interval in \( \mathbb{R} \) is compact. | 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\rb... | Yes |
Theorem 4.40 (Heine-Borel). The compact subsets of \( {\mathbb{R}}^{n} \) are exactly the closed and bounded ones. | 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 \( {\left\lbrack -R, R\right\rbrack }^{n} \) . ... | Yes |
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 \( X \) . | 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. | Yes |
Lemma 4.42. Compactness implies limit point compactness. | 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 ne... | Yes |
Lemma 4.43. For first countable Hausdorff spaces, limit point compactness implies sequential compactness. | 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 take... | Yes |
Lemma 4.44. For metric spaces and second countable topological spaces, sequential compactness implies compactness. | 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 \( X \) .... | 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. | 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 Hau... | No |
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. | Now we can simply say that \( \omega \) is a quotient map by the closed map lemma. | Yes |
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 \( {\mathbb{T}}^{2} = \) \( {\mathbb{S}}^{1} \times {\mathbb{S}}^{1} \). | 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... | Yes |
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. | 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 homeomo... | Yes |
In Example 3.51, 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... | \[ {\mathbb{S}}^{n}\overset{\iota }{ \hookrightarrow }{\mathbb{R}}^{n + 1} \smallsetminus \{ 0\} \overset{q}{ \rightarrow }{\mathbb{P}}^{n} \] where \( \iota \) 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 make... | 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\n\n\[ q\left( x\rig... | Yes |
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 \( {\overline{\mathbb{B}}}^{n + 1} \). | 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 else... | 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 coordi... | 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 o... | Yes |
Proposition 4.60. Every manifold has a countable basis of regular coordinate balls. | 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} \... | No |
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. | 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 cont... | Yes |
Proposition 4.64. Every manifold with or without boundary is locally compact. | 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 ... | No |
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 \( \bar{V} \subseteq \) \( U \) . | 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 \( Y \) contain... | Yes |
Proposition 4.66. Any open or closed subset of a locally compact Hausdorff space is a locally compact Hausdorff space. | 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 ... | Yes |
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. | 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 ... | No |
It is easy to show that the solution set of any polynomial equation in two variables is nowhere dense in \( {\mathbb{R}}^{2} \) . | Since there are only countably many polynomials with rational coefficients, the Baire category theorem implies that there are points in the plane (a dense set of them, in fact) that satisfy no rational polynomial equation. | No |
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. | 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 ... | Yes |
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) | 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}}}\ba... | Yes |
Proposition 4.76. A second countable, locally compact Hausdorff space admits an exhaustion by compact sets. | 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 ... | Yes |
Theorem 4.77 (Paracompactness Theorem). Every second countable, locally compact Hausdorff space (and in particular, every topological manifold with or without boundary) is paracompact. | 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 \( {A}_{j} = {K}_{j + 1} \smallsetminus \operatorname{Int}{K}_{j} \) ... | Yes |
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 \( \bar{V} \subseteq U \) . | Proof. This is easily seen to be equivalent to the definition of normality by taking \( B = X \smallsetminus U \) . | No |
Theorem 4.81. Every paracompact Hausdorff space is normal. | 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... | Yes |
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 \( U \) . | Proof. Just apply Urysohn’s Lemma with \( B = X \smallsetminus U \) . | No |
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 sa... | 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 ... | Yes |
Theorem 4.85 (Existence of Partitions of Unity). Let \( X \) be a paracompact Hausdorff space. If \( \mathcal{U} \) is any indexed open cover of \( X \), then there is a partition of unity subordinate to \( \mathcal{U} \). | 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 t... | Yes |
Theorem 4.86 (Embeddability of Compact Manifolds). Every compact manifold is homeomorphic to a subset of some Euclidean space. | 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 homeomorphi... | Yes |
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 \( B \) . | 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{... | Yes |
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 \( {f}^{-1}\left( 0\right) = B \) . | 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 co... | Yes |
Theorem 4.90 (Existence of Exhaustion Functions). Every manifold admits a positive exhaustion function. | 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_{... | Yes |
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. | 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) an... | Yes |
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 \( Y \) . | 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 t... | Yes |
Proposition 4.93 (Sufficient Conditions for Properness). Suppose \( X \) and \( Y \) are topological spaces, and \( F : X \rightarrow Y \) is a continuous map. (a) If \( X \) is compact and \( Y \) is Hausdorff, then \( F \) is proper. | 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. | Yes |
Lemma 4.94. First countable spaces and locally compact spaces are compactly generated. | 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 se... | Yes |
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. | 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 ... | Yes |
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. | Proof. This follows from Theorem 4.95 and Proposition 4.93(d). | No |
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. | Proof. Theorem 4.95 and Proposition 3.69. | No |
Proposition 5.2. Suppose \( X \) is a topological space whose topology is coherent with a family \( \mathcal{B} \) of subspaces.\n\n(a) If \( Y \) is another topological space, then a map \( f : X \rightarrow Y \) is continuous if and only if \( {\left. f\right| }_{B} \) is continuous for every \( B \in \mathcal{B} \) ... | Exercise 5.3. Prove the preceding proposition. | No |
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. | 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 \( A \subseteq X \) ... | Yes |
Proposition 5.5. Suppose \( X \) is an \( n \) -dimensional \( {CW} \) complex. Then every \( n \) -cell of \( X \) is an open subset of \( X \) . | 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 ... | Yes |
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. | 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 automat... | Yes |
Proposition 5.7. If \( X \) is any \( {CW} \) complex, the topology of \( X \) is coherent with the collection of subspaces \( \left\{ {{X}_{n} : n \geq 0}\right\} \) . | Proof. Problem 5.7. | No |
Let \( X \subseteq {\mathbb{R}}^{2} \) be the union of the closed line segments from the origin to \( \left( {1,0}\right) \) and to the points \( \left( {1,1/n}\right) \) for \( n \in \mathbb{N} \), with the subspace topology (Fig. 5.3). Define a cell decomposition of \( X \) by declaring the 0 -cells to be \( \left( {... | This is easily seen to be a cell decomposition that satisfies condition (C). However, it does not satisfy condition \( \left( \mathrm{W}\right) \), because the set \( \left\{ {\left( {1/n,1/{n}^{2}}\right) : n \in \mathbb{N}}\right\} \) has a closed intersection with the closure of each cell, but is not closed in \( X ... | Yes |
Lemma 5.12. In any CW complex, the closure of each cell is contained in a finite subcomplex. | 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} \smallsetm... | Yes |
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. | 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. Bec... | Yes |
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. | 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 in... | Yes |
Proposition 5.16. A CW complex is locally compact if and only if it is locally finite. | Proof. Problem 5-11. | No |
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 }\l... | 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... | Yes |
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. | 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 }^{... | 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}... | 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 ... | Yes |
Proposition 5.23. Suppose \( X \) is a CW complex with countably many cells. If \( X \) is locally Euclidean, then it is a manifold. | 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. | Yes |
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 \( n \) . | 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 ... | 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. | 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\nIf \( e \) is any 1 -cell of \( M \), then \( e \) is an ... | Yes |
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. | 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... | Yes |
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 i... | 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 ... | Yes |
Proposition 5.32. Every \( k \) -simplex is a closed \( k \) -cell. | 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 simple... | Yes |
Proposition 5.33. If \( K \) is a Euclidean simplicial complex, then the collection consisting of the interiors of the simplices of \( K \) is a regular \( {CW} \) decomposition of \( \left| K\right| \) . | Exercise 5.34. Prove the preceding proposition. | No |
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. | 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... | No |
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 af... | 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 : \s... | Yes |
Proposition 5.41. Every finite abstract simplicial complex has a geometric realization. | Proof. Let \( {v}_{1},\ldots ,{v}_{m} \) be the vertices of \( \mathcal{K} \) in some order, and let \( K \subseteq {\mathbb{R}}^{m} \) be the complex whose vertices are the points \( \left\{ {{e}_{1},\ldots ,{e}_{m}}\right\} \), where \( {e}_{i} = \left( {0,\ldots ,1,\ldots ,0}\right) \) , with simplices \( \left\lbra... | Yes |
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) \i... | 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 eac... | No |
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}\... | 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... | No |
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 result... | 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... | Yes |
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 \( 0 \leq t \leq 1 \). | 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 \ | No |
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