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\[ {\pi }_{1}\left( {n\text{-dimensional torus}}\right) = {\pi }_{1}\left( {{\mathbb{R}}^{n}/{\mathbb{Z}}^{n}}\right) \; \cong \;{\mathbb{Z}}^{n} \] | Proof (*). By Proposition 14.14 we know that for any \( n \geq 2 \) the sphere \( {S}^{n} \) is simply connected. Furthermore we know by the discussion on page [468] that \( {\mathbb{R}}^{n}, n \geq 1 \), and that the strip \( \mathbb{R} \times \left\lbrack {-1,1}\right\rbrack \) are simply connected. We saw on page 50... | No |
Lemma 16.19. (*) Let \( {\varphi }_{1} : {G}_{1} \rightarrow \pi \) and \( {\varphi }_{2} : {G}_{2} \rightarrow \pi \) be two group homomorphisms. If for every \( {g}_{1} \in {G}_{1} \) and \( {g}_{2} \in {G}_{2} \) the images \( {\varphi }_{1}\left( {g}_{1}\right) \in \pi \) and \( {\varphi }_{2}\left( {g}_{2}\right) ... | Proof \( \left( *\right) \) . We write \( \Phi = {\varphi }_{1} \times {\varphi }_{2} \) . Let \( {g}_{1},{g}_{1}^{\prime } \in {G}_{1} \) and let \( {g}_{2},{g}_{2}^{\prime } \in {G}_{2} \) . We have\n\n\[ \n\Phi \left( {{g}_{1},{g}_{2}}\right) \cdot \Phi \left( {{g}_{1}^{\prime },{g}_{2}^{\prime }}\right) = {\varphi ... | Yes |
Proposition 16.20. Let \( A \) and \( B \) be two topological spaces and let \( {a}_{0} \in A \) and \( {b}_{0} \in B \) . We consider the inclusion maps\n\n\[ \n\begin{aligned} i : A & \rightarrow A \times B \\ a & \mapsto \left( {a,{b}_{0}}\right) \end{aligned}\;\text{ and }\;\begin{aligned} j : B & \rightarrow A \ti... | Proof of Proposition 16.20 (*). It is clear that\n\n\[ \n\Psi = \left( {{p}_{ * },{q}_{ * }}\right) : {\pi }_{1}\left( {A \times B,\left( {{a}_{0},{b}_{0}}\right) }\right) \rightarrow {\pi }_{1}\left( {A,{a}_{0}}\right) \times {\pi }_{1}\left( {B,{b}_{0}}\right) \n\]\n\nis a homomorphism. Thus it suffices to show that ... | No |
Theorem 16.21. (Fundamental Theorem of Algebra) Every nonconstant polynomial with coefficients in \( \mathbb{C} \) has a zero in \( \mathbb{C} \) . | Proof. Let \( q\left( z\right) \) be a polynomial with complex coefficients with \( q\left( z\right) \neq 0 \) for all \( z \in \mathbb{C} \) with \( \left| z\right| = r \) . We consider the path\n\n\[ \n{f}_{r}^{q\left( z\right) } : \left\lbrack {0,1}\right\rbrack \rightarrow {S}^{1} \n\] \n\n\[ \ns \mapsto \frac{q\le... | No |
Lemma 17.2. (*) Let \( p : \widetilde{X} \rightarrow X \) be a countable covering of a topological space \( X \) . If \( X \) is second-countable, then \( \widetilde{X} \) is also second-countable. | Proof (*). Let \( p : \widetilde{X} \rightarrow X \) be a countable covering of a topological space \( X \) . Recall that we say \( U \subset X \) is uniformly covered, if \( {p}^{-1}\left( U\right) \) is the union of disjoint open subsets \( {\left\{ {\widetilde{U}}_{i}\right\} }_{i \in I} \) with the property that th... | Yes |
Proposition 17.3. (1) To each smooth manifold \( M \) we can canonically associate an oriented smooth manifold \( \widetilde{M} \) together with a 2-fold covering \( p : \widetilde{M \rightarrow }M \) and to each local diffeomorphism \( f : M \rightarrow N \) we can associate an orientation-preserving local diffeomorph... | Proof. We start out with a general construction. Let \( W \) be an \( n \) -dimensional smooth manifold. We consider the set \[ \widetilde{W} = \left\{ {\left( {Q, O}\right) \mid Q \in W\text{ and }O\text{ is an orientation of }{\mathrm{T}}_{Q}W}\right\} \] together with the map \[ p = {p}_{W} : \widetilde{W} \rightarr... | No |
(1) If \( M \) is a connected non-orientable smooth manifold, then there exists an epimorphism \( {\pi }_{1}\left( M\right) \rightarrow {\mathbb{Z}}_{2} \) . | Proof. The second statement of the corollary is an immediate consequence of the first statement. Thus it suffices to prove the first statement.\n\nLet \( M \) be a connected non-orientable smooth manifold. By Proposition 17.3 there exists a connected 2-fold covering \( p : \widetilde{M} \rightarrow M \) such that \( \w... | No |
Lemma 17.5. Let \( M \) be a connected non-orientable smooth manifold and let \( {x}_{0} \in M \) be a point. The map\n\n\[ \Phi : {\pi }_{1}\left( {M,{x}_{0}}\right) \rightarrow \{ \pm 1\} \]\n\n\[ \left\lbrack {\gamma : \left\lbrack {0,1}\right\rbrack \rightarrow M}\right\rbrack \; \mapsto \;\left\{ \begin{matrix} 1,... | Proof. We denote by \( p : \widetilde{M} \rightarrow M \) the orientation covering. The following three easy observations imply the lemma.\n\n(1) It follows immediately from Corollary 16.13 that \( \Phi \) is well-defined.\n\n(2) Using Lemma 16.4 one easily verifies that \( \Phi \) is a homomorphism.\n\n(3) We denote b... | Yes |
Lemma 18.1. Let \( X \) be a topological space and let \( Y \subset {\mathbb{R}}^{n} \) be a subset. Furthermore let \( {f}_{0},{f}_{1} : X \rightarrow Y \) be two maps. If \( Y \) is convex, then\n\n\[ F : X \times \left\lbrack {0,1}\right\rbrack \rightarrow Y \]\n\n\[ \left( {x, t}\right) \mapsto \underset{ \in Y\tex... | Proof (*). Basically by definition we have \( {F}_{0} = {f}_{0} \) and \( {F}_{1} = {f}_{1} \) . But we are not done yet, the conscientious reader will not have failed to notice that we need to show that the map \( F \) is actually continuous. The map \( F \) can be written as the composition of the following two maps:... | Yes |
Lemma 18.2. (*) Let \( A \) be a bounded closed convex subset of \( {\mathbb{R}}^{n} \) with non-empty interior. Let \( Q \) be a point in the interior of \( A \). There exists a canonical isotopy \( F : A \times \left\lbrack {0,1}\right\rbrack \rightarrow {\mathbb{R}}^{n} \) with the following three properties:\n\n(1)... | Proof (*). First consider the case that \( Q \) is the origin. In the following we use the notation that we introduced in the proof of Proposition 2.53 (2). Similar to the discussion on page 128 we consider the map\n\n\[ F : A \times \left\lbrack {0,1}\right\rbrack \rightarrow {\bar{B}}^{n} \]\n\n\[ \left( {x, t}\right... | No |
Lemma 18.3. (*) Let \( X \) and \( Y \) be topological spaces and let \( A \subset X \) be a (possibly empty) subset of \( X \) . Let \( G, H : X \times \left\lbrack {0,1}\right\rbrack \rightarrow Y \) be two homotopies rel \( A \) . If \( {G}_{1} = {H}_{0} \), then the map\n\n\[ \nG * H : X \times \left\lbrack {0,1}\r... | Proof \( \left( *\right) \) . We only need to verify that \( G * H \) is indeed continuous, but that is a consequence of Lemma 14.3 together with our hypothesis that \( {G}_{1} = {H}_{0} \) . | Yes |
Lemma 18.5. Let \( n \in \mathbb{N} \) .\n\n(1) Any reflection in a hyperplane of \( {\mathbb{R}}^{n} \) has the following properties:\n\n(a) it is a diffeomorphism,\n\n(b) it is an isometry (i.e. it preserves the Euclidean distance) and preserves the origin,\n\n(c) it restricts to diffeomorphisms of \( {\bar{B}}^{n} \... | Proof. Let \( n \in \mathbb{N} \) .\n\n(1) Let \( \rho \) be a reflection in a hyperplane of \( {\mathbb{R}}^{n} \) . Clearly \( \rho \) is smooth and it satisfies \( \rho \circ \rho = {\operatorname{id}}_{{\mathbb{R}}^{n}} \) . This implies immediately that \( \rho \) is a diffeomorphism. A short amusing calculation s... | Yes |
Lemma 18.6. Let \( X, Y \) and \( Z \) be topological spaces. Let \( f,{f}^{\prime } : X \rightarrow Y \) and \( g,{g}^{\prime } : Y \rightarrow Z \) be maps.\n\n(1) If \( f \) and \( {f}^{\prime } \) are homotopic and if \( g \) and \( {g}^{\prime } \) are homotopic, then the map \( g \circ f \) is homotopic to \( {g}... | Proof.\n\n(1) Let \( F : X \times \left\lbrack {0,1}\right\rbrack \rightarrow Y \) be a homotopy between \( f \) and \( {f}^{\prime } \) and let \( G : Y \times \left\lbrack {0,1}\right\rbrack \rightarrow Z \) be a homotopy between \( g \) and \( {g}^{\prime } \) . We consider the map\n\n\[ \begin{aligned} H : X \times... | Yes |
(1) Let \( f : X \rightarrow Y \) be a map between topological spaces. The element \( \left\lbrack f\right\rbrack \in \operatorname{Mor}\left( {X, Y}\right) \) is an isomorphism \( {}^{297} \) in the category HomTop if and only if \( f \) is a homotopy equivalence. | Proof.\n\n(1) This statement is a tautology, it follows immediately from the definitions. | No |
Let \( M \) be a smooth manifold and let \( C \subset M \) be a closed oriented curve in \( M \) . Furthermore let \( {x}_{0} \in C \) . For any two orientation-preserving diffeomorphisms \( \alpha ,\beta : {S}^{1} \rightarrow C \) with \( \alpha \left( 1\right) = \beta \left( 1\right) = {x}_{0} \) we have \( \left\lbr... | Proof. It is fairly straightforward to prove the corollary using Lemmas 18.9 (1) and 14.4 . We leave the task of filling in the details to the reader. | No |
(1) Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) be maps between topological spaces. If two out of the three maps \( f, g \) and \( g \circ f \) are homotopy equivalences, then so is the third. | Proof \( \left( *\right) \) .\n\n(1) Let \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) be maps between topological spaces. There are three slightly different statements that need to be proved.\n\n(a) First suppose that the maps \( f : X \rightarrow Y \) and \( g : Y \rightarrow Z \) admit homotopy inverses \(... | Yes |
Lemma 18.12. Let \( X \) be a topological space.\n\n(1) \( X \) is contractible \( \Leftrightarrow \) (2) there exists an \( {x}_{0} \in X \) such that the inclusion \( \left\{ {x}_{0}\right\} \rightarrow X \) is a homotopy equivalence\n\n\( \Leftrightarrow \) there exists an \( {x}_{0} \in X \) such that the identity ... | Proof (*).\n\n\( \left( 1\right) \Rightarrow \left( 2\right) \) First suppose that \( X \) is contractible. Let \( Y = \{ * \} \) be the topological space consisting of a single element. Since \( X \) is contractible we know that there exists a map \( g : \{ * \} \rightarrow X \) that is a homotopy equivalence. We set ... | Yes |
(1) If a topological space \( X \) admits a deformation retraction to a point \( {x}_{0} \in X \), then \( X \) is homotopy equivalent to \( \left\{ {x}_{0}\right\} \), in particular \( X \) is contractible. | (1) This statement follows immediately from Lemma 18.14. | No |
Proposition 18.16. (1) Let \( f, g : X \rightarrow Y \) be two maps between topological spaces and let \( {x}_{0} \in X \) . If \( f \) and \( g \) are homotopic, then there exists a path \( \alpha : \left\lbrack {0,1}\right\rbrack \rightarrow Y \) from \( f\left( {x}_{0}\right) \) to \( g\left( {x}_{0}\right) \) such ... | (1) Let \( f, g : X \rightarrow Y \) be two maps between topological spaces and let \( {x}_{0} \in X \) . Let \( H : X \times \left\lbrack {0,1}\right\rbrack \rightarrow Y \) be a homotopy between \( f \) and \( g \) . We denote by\n\n\[ \alpha : \left\lbrack {0,1}\right\rbrack \rightarrow Y \]\n\n\[ t \mapsto H\left( ... | Yes |
Lemma 18.17. The topological spaces \( {\mathbb{R}}^{2} \) and \( {\mathbb{R}}^{3} \) are not homeomorphic. | Proof. Let us suppose that there exists a homeomorphism \( f : {\mathbb{R}}^{2} \rightarrow {\mathbb{R}}^{3} \) . Let \( P \in {\mathbb{R}}^{2} \) be a point. We put \( Q \mathrel{\text{:=}} f\left( P\right) \) . Then \( f \) restricts to a homeomorphism from \( {\mathbb{R}}^{2} \smallsetminus \{ P\} \) to \( {\mathbb{... | Yes |
Proposition 18.20. Let \( X \) and \( Y \) be topological spaces. Furthermore let \( { \sim }_{X} \) be an equivalence relation on \( X \) and let \( { \sim }_{Y} \) be an equivalence relation on \( Y \) . If \( F : X \times \left\lbrack {0,1}\right\rbrack \rightarrow Y \) is a homotopy such that \( F\left( {x, t}\righ... | Proof. We denote by \( p : X \rightarrow X/{ \sim }_{X} \) and \( q : Y \rightarrow Y/{ \sim }_{Y} \) the obvious projection maps. We have the following commutative diagram\n\n\n\nBy Lemma 5.15 (1) the map \( p \) is... | Yes |
Lemma 18.21. Let \( X \) be a topological space. Furthermore let \( A \subset B \) be two subsets of \( X \) . If \( F : B \times \left\lbrack {0,1}\right\rbrack \rightarrow X \) is a homotopy with \( F\left( {A \times \left\lbrack {0,1}\right\rbrack }\right) \subset A \), then the induced map\n\n\[ \varphi : \left( {B... | Proof. This statement follows immediately from Proposition 18.20 and the fact that quotients of topological spaces are defined via equivalence relations. | No |
Let \( X \) be a topological space and let \( A \) be a subset of \( X \) . If \( A \) is a deformation retract of \( X \), then the point \( A/A \) is a deformation retract of \( X/A \) . In particular \( X/A \) is contractible. | Proof. Let \( r : X \times \left\lbrack {0,1}\right\rbrack \rightarrow X \) be a deformation retraction from \( X \) to \( A \) . Since \( r \) is deformation retraction we obtain from Lemma 18.21 that the map\n\n\[ \left( {X/A}\right) \times \left\lbrack {0,1}\right\rbrack \rightarrow X/A \]\n\n\[ \left( {\left\lbrack... | Yes |
Lemma 18.23. Let \( f : X \rightarrow Y \) and \( g : X \rightarrow Z \) be maps between topological space. As on page 197 we define the pushout\n\n\[ Y{ \cup }_{X}Z \mathrel{\text{:=}} \left( {Y \sqcup Z}\right) / \sim \;\text{ where }f\left( x\right) \sim g\left( x\right) \text{ for all }x \in X. \] | Furthermore let \( W \) be another topological space. Suppose that we are given two homotopies \( G : Y \times \left\lbrack {0,1}\right\rbrack \rightarrow W \) and \( H : Z \times \left\lbrack {0,1}\right\rbrack \rightarrow W \) such that for any \( x \in X \) and any \( t \in \left\lbrack {0,1}\right\rbrack \) we have... | Yes |
Lemma 18.24. Let \( {\left\{ {X}_{i}\right\} }_{i \in I} \) be a family of smooth manifolds. We suppose that for each \( i \in I \) we are given points \( {x}_{i},{y}_{i} \in {X}_{i} \smallsetminus \partial {X}_{i} \) . If each \( {X}_{i} \) is connected, then there exists a homeomorphism\n\n\[ \mathop{\bigvee }\limits... | Proof. The lemma is an almost immediate consequence of Proposition 8.29. We leave it to the reader to fill in the details. | No |
Lemma 18.25. (*) Let \( {\left\{ {X}_{j}\right\} }_{j \in J} \) be a family of topological spaces and suppose that for each \( j \in J \) we are given a point \( {x}_{j} \in {X}_{j} \). Let \( k \in J \). (1) The natural inclusion \( {i}_{k} \) is an embedding. | Proof (*). (1) It follows easily from Lemma 3.3 and Lemma 3.21 (3) that the natural inclusion \( {i}_{k} \) is continuous. It is clear that the natural inclusion \( {i}_{k} \) is an injection. Evidently we have \( {p}_{k} \circ {i}_{k} = {\operatorname{id}}_{{X}_{k}} \). Since the natural projection \( {p}_{k} \) is co... | Yes |
Lemma 18.26. Let \( {\left\{ \left( {X}_{i},{x}_{i}\right) \right\} }_{i \in I} \) be a family of pointed topological spaces.\n\n(1) If \( I \) is finite and if each \( {X}_{i} \) is compact, then \( \mathop{\bigvee }\limits_{{i \in I}}{X}_{i} \) is compact. | (1) It follows immediately from Lemma 18.25 (3) and Lemma 3.3 (2) that the map \( \mathop{\bigsqcup }\limits_{{i \in I}}{p}_{i} : \mathop{\bigsqcup }\limits_{{i \in I}}{X}_{i} \rightarrow \mathop{\bigvee }\limits_{{i \in I}}{X}_{i} \) is continuous. Thus the desired statement follows from Lemma 3.3\n\n(4) together with... | No |
Lemma 18.27. (*) Let \( {\left\{ \left( {X}_{i},{x}_{i}\right) \right\} }_{i \in I} \) be a family of pointed topological spaces and let \( \left( {Z,{z}_{0}}\right) \) be a pointed topological space. Let \( {\left\{ {f}_{i} : \left( {X}_{i},{x}_{i}\right) \rightarrow \left( Z,{z}_{0}\right) \right\} }_{i \in I} \) and... | Proof. The \ | No |
Lemma 18.28. (*) Let \( \left( {A,{a}_{0}}\right) \) and \( \left( {B,{b}_{0}}\right) \) be two pointed topological spaces. We consider the wedge \( A \vee B = \left( {A,{a}_{0}}\right) \vee \left( {B,{b}_{0}}\right) \). If \( \left\{ {a}_{0}\right\} \) is a deformation retract of \( A \), then \( B = \left\{ {a}_{0}\r... | Proof (*). Let \( F : A \times \left\lbrack {0,1}\right\rbrack \rightarrow A \) be a deformation retraction from \( A \) to \( \left\{ {a}_{0}\right\} \). Let \( G \) be the trivial homotopy on \( B \). It follows again from Lemma 18.23 that we can combine the two homotopies to get the desired deformation retraction. | No |
Lemma 18.30. Let \( G = \left( {V, E, i, t}\right) \) be an abstract graph and let \( {G}^{\prime } = \left( {{V}^{\prime },{E}^{\prime },{i}^{\prime },{t}^{\prime }}\right) \) be a subgraph.\n\n(1) The inclusions \( {V}^{\prime } \rightarrow V \) and \( {E}^{\prime } \rightarrow E \) define a map \( \iota : {G}^{\prim... | Proof.\n\n(1) This statement follows immediately from the definitions.\n\n(2) By the discussion on page [481] we know that the map \( \left| \iota \right| \) is continuous. By Lemma 2.42 it remains to show that \( \left| \iota \right| \) is a closed map. We leave it to the reader to show, using Lemma 3.1 and Lemma 3.21... | No |
Lemma 18.32. Let \( G = \left( {V, E, i, t}\right) \) be an abstract graph and let \( H = \left( {W, F}\right) \) be a nonempty subgraph. We continue with the above notation.\n\n(1) We have \( \chi \left( {G/H}\right) = \chi \left( G\right) - \chi \left( H\right) + 1 \) . | Proof.\n\n(1) We calculate that\n\n\[ \chi \left( {G/H}\right) = \# \left( {V/ \sim }\right) - \# \left( {E \smallsetminus F}\right) = \left( {\# V - \# W + 1}\right) - \left( {\# E - \# F}\right) = \chi \left( G\right) - \chi \left( H\right) + 1. \] | Yes |
there exists a path \( \gamma \) from \( {x}_{0} \) to \( {x}_{1} \) such that \( \left\lbrack {\gamma * {f}_{1} * \bar{\gamma }}\right\rbrack = \left\lbrack {f}_{0}\right\rbrack \in {\pi }_{1}\left( {X,{x}_{0}}\right) \) . | Proof. Note that the second statement is an immediate consequence of the first statement and the observation that if \( {x}_{0} = {x}_{1} \) any path from \( {x}_{0} \) to \( {x}_{1} = {x}_{0} \) is in fact a loop and thus defines an element in \( {\pi }_{1}\left( {X,{x}_{0}}\right) \) . Therefore it suffices to prove ... | Yes |
(1) Let \( S \) be a set and let \( \phi : S \rightarrow G \) be a map to an abelian group \( G \) . Then there exists a unique homomorphism \( \psi : {\mathbb{Z}}^{\left( S\right) } \rightarrow G \) that makes the following diagram commute | (1) We consider the map 315\n\n\[ \psi : {\mathbb{Z}}^{\left( S\right) } \rightarrow G \]\n\n\[ f = \mathop{\sum }\limits_{{i = 1}}^{n}{n}_{i}{s}_{i} \mapsto \psi \left( f\right) \mathrel{\text{:=}} \mathop{\sum }\limits_{{i = 1}}^{n}{n}_{i} \cdot \phi \left( {s}_{i}\right) . \]\n\nIt is straightforward to verify that ... | Yes |
Lemma 19.9. The abelian group \( \left( {\mathbb{Q}, + }\right) \) is torsion-free but not free abelian. | Proof. Clearly \( \left( {\mathbb{Q}, + }\right) \) is torsion-free. If it was free abelian, then there would exist an epimorphism to \( \mathbb{Z} \) . In Exercise 19.1 we will show that such an epimorphism cannot exist. | No |
Proposition 19.10. Let \( A \) and \( B \) be any two abelian groups. If \( C \) is a finitely generated abelian group, then\n\n\[ A \oplus C \cong B \oplus C\; \Rightarrow \;A \cong B. \] | Proof. We will consider a special case in Exercise 19.3 (b). The general case is proved in Cohn56, Walka56 | No |
The free product of two groups \( G \) and \( H \) is again a group. The neutral element of \( G * H \) is hereby given by the empty sequence () and the inverse of an element \( \left( {{x}_{1},\ldots ,{x}_{m}}\right) \in G * H \) is given by \[ {\left( {x}_{1},\ldots ,{x}_{m}\right) }^{-1} = \left( {{x}_{m}^{-1},\ldot... | Proof. It is clear that the empty sequence is a neutral element. It follows immediately from the definition of the product that the inverse of an element \( \left( {{x}_{1},\ldots ,{x}_{m}}\right) \in G * H \) is given by \( \left( {{x}_{m}^{-1},\ldots ,{x}_{1}^{-1}}\right) \) . It remains to show that the multiplicati... | No |
Lemma 19.12. Let \( G \) and \( H \) be two groups. The following two statements hold:\n\n(1) The maps\n\n\[ \n\begin{matrix} g & \mapsto & \left\{ \begin{array}{ll} \left( g\right) , & \text{ if }g \neq e, \\ \left( \right) , & \text{ if }g = e \end{array}\right. & \text{ and } & & & h & \mapsto & \left\{ \begin{array... | Proof. It is clear that the maps \( i \) and \( j \) are injective, furthermore it follows immediately from the definition of the group structure on \( G * H \) that the maps are group homomorphisms.\n\nNow let \( \alpha : G \rightarrow A \) and \( \beta : H \rightarrow A \) be two group homomorphisms. For an element \... | Yes |
Lemma 19.14. Let \( S \) be a set, let \( G \) be an arbitrary group and let \( g : S \rightarrow G \) be a map. Then there exists a unique homomorphism \( \varphi : \langle S\rangle \rightarrow G \) with \( \varphi \left( s\right) = g\left( s\right) \) for every \( s \in S \) . | For \( s \in S \) the map\n\n\[ \langle s\rangle \rightarrow G \]\n\n\[ {s}^{n} \mapsto g{\left( s\right) }^{n} \]\n\nis evidently the unique homomorphism \( {\varphi }_{s} : \langle s\rangle \rightarrow G \) with \( {\varphi }_{s}\left( s\right) = g\left( s\right) \) . It now follows from the obvious generalization of... | Yes |
Lemma 19.16. Let \( S \) and \( T \) be two sets.\n\n(1) If there exists an epimorphism \( \langle S\rangle \rightarrow \langle T\rangle \), then \( \# S \geq \# T \) . | Proof.\n\n(1) We first recall that we had just seen in the previous example that for any set \( U \) there exists a unique epimorphism \( {\psi }_{U} : \langle U\rangle \rightarrow {\mathbb{Z}}^{\left( U\right) } \) that sends \( u \in U \) to \( u \) viewed as an element in \( {\mathbb{Z}}^{\left( U\right) } \) .\n\nN... | Yes |
Lemma 19.18. Let \( \Phi : G \rightarrow H \) and \( \Psi : H \rightarrow G \) be two group homomorphisms such that \( \Phi \circ \Psi = {\operatorname{id}}_{H} \) and such that \( \Psi \) is an epimorphism. Then \( \Phi \) is an isomorphism. | Proof. It follows from \( \Phi \circ \Psi = {\operatorname{id}}_{H} \) that \( \Phi \) is an epimorphism. It remains to show that \( \Phi \) is a monomorphism. So let \( g \in \ker \left( \Phi \right) \) . Since \( \Psi \) is an epimorphism there exists an \( h \in H \) with \( \Psi \left( h\right) = g \) . Since \( \P... | Yes |
Theorem 19.19. (Grushko-Neumann Theorem) Given any two finitely generated groups \( A \) and \( B \) we have\n\n\[ d\left( {A * B}\right) = d\left( A\right) + d\left( B\right) \] | Proof. The proof is evidently given in the above two references, but it can also be found in any self-respecting book on combinatorial group theory, see e.g. [LS77, Corollay IV.1.9] and [MKS76, p. 192]. | No |
Theorem 19.21. Let \( G \) be a finitely generated group.\n\n(1) The group \( G \) is isomorphic to the free product \( G = {A}_{1} * \cdots * {A}_{k} \) of non-trivial indecomposable groups \( {}^{341} \)\n\n(2) The isomorphism types of \( {A}_{1},\ldots ,{A}_{k} \) are unique up to permutation. More precisely, if \( ... | Proof.\n\n(1) This statement follows easily from induction on \( d\left( G\right) \) and the Grushko-Neumann Theorem 19.19. Indeed, if \( d\left( G\right) = 0 \) then there is nothing to prove. Next suppose that the statement is known for all groups \( G \) with \( d\left( G\right) < k \) . Let \( G \) be a group with ... | Yes |
Lemma 20.1. Let \( X \) be a topological space and let \( U, V \subset X \) be two open subsets with \( X = U \cup V \) and such that \( U \cap V \neq \varnothing \) . We choose a base point \( {x}_{0} \in U \cap V \) . If \( U \cap V \) is path-connected, then the homomorphism \[ {\pi }_{1}\left( {U,{x}_{0}}\right) * ... | Proof of Lemma 20.1 Let \( X \) be a topological space and let \( U, V \subset X \) be two open subsets with \( X = U \cup V \) and such that \( U \cap V \neq \varnothing \) . We choose a base point \( {x}_{0} \in U \cap V \) . It suffices to prove the following claim.\n\nClaim. Every loop \( s : \left\lbrack {0,1}\rig... | No |
Lemma 20.4. Let \( k \in {\mathbb{N}}_{0} \) . The map\n\n\[ \left\langle {{x}_{1},\ldots ,{x}_{k}}\right\rangle \rightarrow {\pi }_{1}\left( {\mathop{\bigvee }\limits_{{i = 1}}^{k}{S}^{1}, * }\right) \]\n\nthat is given by\n\n\[ {x}_{i} \mapsto \left\lbrack \begin{array}{rrr} \left\lbrack {0,1}\right\rbrack & \rightar... | Proof. Similar to the above example the lemma follows from applying Proposition 20.3 altogether \( k - 1 \) times. | No |
(1) Let \( G = \left( {V, E, i, t}\right) \) be a finite connected non-empty abstract graph and let \( T \) be a maximal tree of \( G \) . Let \( {e}_{1},\ldots ,{e}_{n} \) be the edges not contained in \( T \) . We pick some \( v \in V \) . For \( j = 1,\ldots, n \) we pick a path \( {\alpha }_{j} : \left\lbrack {0,1}... | (1) We consider the maps isomorphism induced by the homeomorphism on page 560 \( \begin{array}{l} \left\langle {{x}_{1},\ldots ,{x}_{n}}\right\rangle \overset{{\Theta }_{G}}{ \rightarrow }{\pi }_{1}\left( {\left| G\right|, v}\right) \rightarrow {\pi }_{1}\left( {\left| G\right| /\left| T\right|, v}\right) \overset{ \do... | Yes |
Lemma 20.6. Let \( X \) be a path-connected topological space and let \( {x}_{0} \in X \) . Furthermore let \( {X}_{k}, k \in \mathbb{N} \) be a sequence of subsets of \( X \) such that the following hold:\n\n(1) each \( {X}_{k} \) is open,\n\n(2) each \( {X}_{k} \) is simply connected,\n\n(3) the sequence \( {X}_{k} \... | The proof of the lemma, which will make essential use of Lemma 2.41, is the content of Exercise 20.2. | No |
Lemma 20.9. Let \( g \in {\mathbb{N}}_{0} \) . We denote by \( {\sum }_{g} \) the surface of genus \( g \) .\n\n(1) For any point \( P \) the complement \( {\sum }_{g} \smallsetminus \{ P\} \) is homotopy equivalent to the wedge of \( {2g} \) circles. In particular\n\n\[ \n{\pi }_{1}\left( {{\sum }_{g}\smallsetminus \{... | Proof of Lemma 20.9. We prove the lemma for the surface \( \sum \) of genus 2 . The proof for \( g \geq 3 \) is almost identical. So let \( \sum = {E}_{8}/ \sim \) as sketched in Figure 390,\n\n(1) We denote by \( O \) the origin in \( {E}_{8} \subset \mathbb{C} \) . We first show that \( \sum \smallsetminus \{ O\} \) ... | Yes |
Theorem 20.10. (Seifert-van Kampen Theorem for topological manifolds) Let \( M \) be an \( m \) -dimensional topological manifold and let \( R, S \subset M \) be two \( m \) -dimensional submanifolds such that the following hold:\n\n(1) \( M = R \cup S \) ,\n\n(2) \( R \cap S \) is a component of \( \partial R \) and i... | We will of course prove Theorem 20.10 by reducing it to Theorem 20.2 As we will see, the proof is very similar to the proof of Proposition 20.3 where we determined the fundamental group of a wedge of two spaces.\n\nThe proof of Theorem 20.10 relies on the following lemma which we will use on several occasions.\n\nLemma... | No |
Lemma 20.11. (*) Let \( X \) be a topological space. Furthermore let \( A \) and \( B \) be two subsets with \( X = A \cup B \) . If \( A \cap B \) is a deformation retract of \( B \) and if \( A \) and \( B \) are closed subsets of \( X \), then \( A \) is a deformation retract of \( X \) . | Proof (*). Lemma 20.11 can also be viewed as a consequence of Lemma 3.45 and the slightly scary Lemma 18.23. Since our situation is fairly straightforward we also give a direct, fairly elementary proof.\n\nWe pick a deformation retraction \( F : B \times \left\lbrack {0,1}\right\rbrack \rightarrow B \) from \( B \) to ... | Yes |
Proposition 20.12. Let \( n \geq 3 \) and let \( M \) and \( {M}^{\prime } \) be two oriented connected non-empty \( n \) -dimensional smooth manifolds. Then \( {}^{362} \n\n\[ \n{\pi }_{1}\left( {M\# {M}^{\prime }}\right) \cong {\pi }_{1}\left( M\right) * {\pi }_{1}\left( {M}^{\prime }\right) .\n\] | Proof of Proposition 20.12. Let \( M \) and \( {M}^{\prime } \) be two oriented connected non-empty \( n \) -dimensional smooth manifolds and let \( \varphi : {\bar{B}}^{n} \rightarrow M \) and \( {\varphi }^{\prime } : {\bar{B}}^{n} \rightarrow {M}^{\prime } \) be two smooth embeddings where \( \varphi \) is orientati... | Yes |
Lemma 21.1. Let \( G \) be a group and let \( H \) be a normal subgroup. Let \( \varphi : G \rightarrow \Gamma \) be a group homomorphism with the property that \( H \subset \ker \left( \varphi \right) \) . Then there exists a unique homomorphism \( \psi : G/H \rightarrow \Gamma \) such that the following diagram commu... | Proof. For \( {gH} \in G/H \) we put \( \psi \left( {gH}\right) \mathrel{\text{:=}} \varphi \left( g\right) \) . Since \( H \subset \ker \left( \varphi \right) \) this definition is independent of the choice of the representatives \( {g}_{1}^{364} \) It is straightforward to verify that \( \psi \) is a homomorphism and... | No |
Lemma 21.4. Let \( G \) be a group and let \( A \subset G \) be a subset. Let \( \alpha : G \rightarrow \Gamma \) be a group homomorphism such that \( \alpha \left( a\right) = e \in \Gamma \) for all \( a \in A \). Then there exists a unique group homomorphism \( \beta : G/\langle \langle A\rangle \rangle \rightarrow \... | Proof. Let \( G \) be a group and let \( A \subset G \) be a subset. Furthermore let \( \alpha : G \rightarrow \Gamma \) be a group homomorphism such that \( \alpha \left( a\right) = e \in \Gamma \) for all \( a \in A \). This means that \( A \subset \ker \left( \alpha \right) \). Since \( \ker \left( \alpha \right) \)... | Yes |
Lemma 21.5. Let \( \left\langle {{x}_{1},\ldots ,{x}_{k} \mid {r}_{1},\ldots ,{r}_{l}}\right\rangle \) be a presentation.\n\n(1) Given any \( a, b \in \left\langle {{x}_{1},\ldots ,{x}_{k}}\right\rangle \), given any \( i \in \{ 1,\ldots, l\} \) and given any \( \epsilon \in \{ - 1,1\} \) we have\n\n\[ a \cdot {r}_{i}^... | Proof. The first statement is just a special case of Remark (1) on page 626 and the second statement is just a special case of Remark (2) on page 626. | Yes |
(1) There exists a unique homomorphism\n\n\[ \n\beta : \pi = \langle x, y\rangle /\langle \langle \left\lbrack {x, y}\right\rbrack \rangle \rangle = \langle x, y \mid \left\lbrack {x, y}\right\rbrack \rangle \rightarrow {\mathbb{Z}}^{2} \n\]\n\nwith \( \beta \left( x\right) = \left( {1,0}\right) \) and \( \beta \left( ... | Proof. By Lemma 19.14 there exists a unique homomorphism \( \alpha : \langle x, y\rangle \rightarrow {\mathbb{Z}}^{2} \) which satisfies \( \alpha \left( x\right) = \left( {1,0}\right) \) and \( \alpha \left( y\right) = \left( {0,1}\right) \) . Note that\n\n\[ \n\alpha \left( \left\lbrack {x, y}\right\rbrack \right) = ... | Yes |
In Exercise 21.4 we will show that the equalities \( y{x}^{-1} = {x}^{-1}y \) and \( {x}^{-1}{y}^{-1} = {y}^{-1}{x}^{-1} \) hold in \( \pi = \langle x, y \mid \left\lbrack {x, y}\right\rbrack \rangle \) . This means that in \( \pi \) an arbitrary product of powers of \( x \) and \( y \) can be reordered to equal a prod... | For example in \( \pi \) we have the equality\n\n\[ \n{x}^{3}{y}^{2}{x}^{-1}y = {x}^{3}y \cdot y{x}^{-1} \cdot y\underset{ \uparrow }{ = }{x}^{3}y \cdot {x}^{-1}y \cdot y = {x}^{3} \cdot y{x}^{-1} \cdot {y}^{2}\underset{ \uparrow }{ = }{x}^{3} \cdot {x}^{-1}y \cdot {y}^{2} = {x}^{2}{y}^{3}.\n\]\n\n\[ \n\text{since}y{x}... | No |
Lemma 21.8. Let \( \theta : G \rightarrow H \) be a group homomorphism and let \( X \) be a generating set for \( G \) . The subgroup of \( G * H \) that is generated by \( \left\{ {g \cdot \theta \left( {g}^{-1}\right) \mid g \in G}\right\} \) equals that the subgroup of \( G * H \) that is normally generated by \( \l... | Proof of Lemma 21.8, We set \( N \mathrel{\text{:=}} \left\{ {x \cdot \theta \left( {x}^{-1}\right) \mid x \in X}\right\} \) . We only need to show that given any \( g \in G \) we have \( g \cdot \theta \left( {g}^{-1}\right) \in \langle \langle N\rangle \rangle \) . This statement follows easily by induction on the le... | Yes |
Lemma 21.9. Let \( G \) and \( H \) be two groups, let \( \theta : G \rightarrow H \) be an isomorphism and let \( \alpha : G \rightarrow \pi \) be an isomorphism. We consider the homomorphism \[ \Psi : G * H \rightarrow \pi \] defined by \( \alpha \) on \( G \) and by \( \alpha \circ {\theta }^{-1} \) on \( H \) . The... | Proof of Lemma 21.9. We set \( M \mathrel{\text{:=}} \left\{ {g \cdot \theta \left( {g}^{-1}\right) \mid g \in G}\right\} \) . We need to show that the kernel of the homomorphism \( \Psi \) is generated by \( M \) . Let \( {g}_{1} \cdot {h}_{1}\cdots \cdots {g}_{k} \cdot {h}_{k} \in \ker \left( \Psi \right) \subset G *... | Yes |
Lemma 21.10. Let\n\n\\[ \n\\phi : \\left\\langle {{x}_{1},\\ldots ,{x}_{k} \\mid {r}_{1}\\left( {{x}_{1},\\ldots ,{x}_{k}}\\right) ,\\ldots ,{r}_{l}\\left( {{x}_{1},\\ldots ,{x}_{k}}\\right) }\\right\\rangle \\rightarrow \\pi \n\\]\n\nbe a presentation. For pairwise different \\( {y}_{1},\\ldots ,{y}_{k} \\) the presen... | Proof of Lemma 21.10. We consider the Tietze transformations\n\n\\[ \n\\left\\langle {{x}_{1},\\ldots ,{x}_{k} \\mid {r}_{1}\\left( {{x}_{1},\\ldots ,{x}_{k}}\\right) ,\\ldots }\\right\\rangle \\overset{\\left( 2\\right) }{ \\leftrightarrow }\\left\\langle {{x}_{1},\\ldots ,{x}_{k},{y}_{1} \\mid {r}_{1}\\left( {{x}_{1}... | Yes |
Lemma 21.12. Let \( X = \left\{ {{x}_{1},\ldots ,{x}_{k}}\right\} \) be a finite set and furthermore let \( r = \left\{ {{r}_{1},\ldots ,{r}_{l}}\right\} \) be a finite subset of \( \langle X\rangle = \left\langle {{x}_{1},\ldots ,{x}_{k}}\right\rangle \) . Suppose that we can write each \( {r}_{i}\left( {{x}_{1},\ldot... | Proof. It follows from Lemma 21.13 that there exists in fact a unique homomorphism\n\n\[ \left\langle {{x}_{1},\ldots ,{x}_{k} \mid {r}_{1},\ldots ,{r}_{l}}\right\rangle \rightarrow \left\langle {{x}_{1},\ldots ,{x}_{k - 1}, t \mid {s}_{1},\ldots ,{s}_{l}}\right\rangle \]\n\nthat has the property that \( {x}_{i} \mapst... | Yes |
Lemma 21.13. Let \( \pi \) be a group with a finite presentation \( \pi = \left\langle {{x}_{1},\ldots ,{x}_{k} \mid {r}_{1},\ldots ,{r}_{l}}\right\rangle \) . For any group \( G \) the map\n\n\[ \operatorname{Hom}\left( {\pi, G}\right) \rightarrow \left\{ {\operatorname{all}\left( {{g}_{1},\ldots ,{g}_{k}}\right) \in ... | Proof. This lemma follows immediately from Lemma 21.4 and Lemma 19.14. | No |
Proposition 21.14. Every non-trivial normal subgroup of a free group is infinitely generated. | Proof. A purely group theoretic proof is given in [LS77, Proposition I.3.12]. Alternatively a proof using topological methods is provided in [dlH00, p. 45]. | No |
Lemma 21.16. (*) Every retract of a finitely presented group is also finitely presented. | Proof (*). Let \( H \) be a group, let \( G \) be a finitely presented group and let \( \varphi : H \rightarrow G \) and \( \rho : G \rightarrow H \) be group homomorphisms such that \( \rho \circ \varphi = {\operatorname{id}}_{H} \) . We need to show that \( H \) is also finitely presented.\n\nTo do so we pick a finit... | Yes |
Lemma 21.19. Let \( \pi \) be a group. Then the following hold:\n\n(1) The group \( \pi /\left\lbrack {\pi ,\pi }\right\rbrack \) is abelian. | (1) For any two elements \( x, y \in \pi \) we have\n\n\[ \n{xy}\left\lbrack {\pi ,\pi }\right\rbrack = {yx} \cdot {x}^{-1}{y}^{-1}{xy}\left\lbrack {\pi ,\pi }\right\rbrack = {yx} \cdot \underset{ \in \left\lbrack {\pi ,\pi }\right\rbrack }{\underbrace{\left\lbrack {x}^{-1},{y}^{-1}\right\rbrack }}\left\lbrack {\pi ,\p... | Yes |
Proposition 21.21. Let \( \alpha : G \rightarrow A \) and \( \beta : G \rightarrow B \) be two group homomorphisms. Then there exists a triple \[ \left( {A{ * }_{G}B,\varphi : A \rightarrow A{ * }_{G}B,\psi : B \rightarrow A{ * }_{G}B}\right) \] where \( A{ * }_{G}B \) is a group and \( \varphi \) and \( \psi \) are ho... | Proof. Let \( \alpha : G \rightarrow A \) and \( \beta : G \rightarrow B \) be two group homomorphisms. As usual we view \( A \) and \( B \) as subgroups of \( A * B \) . We put \[ A{ * }_{G}B \mathrel{\text{:=}} A * B/\left\langle \left\langle {\{ \alpha \left( g\right) \beta {\left( g\right) }^{-1}{\} }_{g \in G}}\ri... | Yes |
Lemma 21.23. Let\n\n\[ A = \left\langle {{x}_{1},\ldots ,{x}_{k} \mid {r}_{1},\ldots ,{r}_{p}}\right\rangle \]\n\nand\n\n\[ B = \left\langle {{y}_{1},\ldots ,{y}_{l} \mid {s}_{1},\ldots ,{s}_{q}}\right\rangle \]\n\nbe two finitely presented groups. Furthermore let \( \alpha : G \rightarrow A \) and \( \beta : G \righta... | Proof of Lemma 21.23. We need to show that\n\n\[ \underset{\text{definition of }{A}^{ * }{}_{G}B}{\underbrace{\frac{A * B}{\left\langle {\left\{ \alpha \left( g\right) \beta {\left( g\right) }^{-1}\right\} }_{g \in G}\right\rangle }}} = \underset{\text{right-hand side of the proposition }}{\underbrace{\frac{A * B}{\lef... | Yes |
Theorem 22.2. (Seifert-van Kampen Theorem for Topological Manifolds) Let \( M \) be an \( m \) -dimensional topological manifold and let \( R, S \subset M \) be two \( m \) -dimensional submanifolds such that the following hold:\n\n(1) \( M = R \cup S \) ,\n\n(2) \( R \cap S \) is a component of \( \partial R \) and it... | Example. Given \( n \in {\mathbb{N}}_{0} \) we denote as usual by \( {S}_{ > 0}^{n} \) the upper hemisphere of \( {S}^{n} \) and we denote by \( {S}_{ \leq 0}^{n} \) the lower hemisphere of \( {S}^{n} \) . We denote by \( {S}_{ = 0}^{n} = {S}_{ \geq 0}^{n} \cap {S}_{ \leq 0}^{n} \) the \ | No |
(1) For any \( g \in {\mathbb{N}}_{0} \) we have\n\n\[{\pi }_{1}\left( {\text{surface}\;{\sum }_{g}\;\text{of}\;\text{genus}\;g}\right) \; \cong \;\langle {x}_{1},{y}_{1},\ldots ,{x}_{g},{y}_{g} \mid \left\lbrack {{x}_{1},{y}_{1}}\right\rbrack \cdots \cdots \left\lbrack {{x}_{g},{y}_{g}}\right\rbrack \rangle\]\n\nand w... | (1) Let \( g \in {\mathbb{N}}_{0} \) . For \( g = 0 \) we obtain the fundamental group from Proposition 14.14 and for \( g = 1 \) we obtain the presentation from the discussion on page 656 . For \( \overline{g \geq 2} \) the presentation follows from a straightforward generalization of the calculation preceding the pro... | No |
(1) For any \( g \in \mathbb{N} \) we have\n\n\( {\pi }_{1} \) (non-orientable surface \( {N}_{g} \) of genus \( g \) ) \( \cong \left\langle {{x}_{1},\ldots ,{x}_{g} \mid {x}_{1}^{2}\cdots \cdots {x}_{g}^{2}}\right\rangle \)\n\nand we have\n\n\( {\pi }_{1}{\left( \text{ non-orientable surface }{N}_{g}\text{ of genus }... | (1) Let \( g \in \mathbb{N} \) . Recall that the non-orientable surface of genus one is by definition the real projective plane. In Corollary 16.18 we saw that \( {\pi }_{1}\left( {\mathbb{{RP}}}^{2}\right) \cong {\mathbb{Z}}_{2} \cong \left\langle {x \mid {x}^{2}}\right\rangle \) . The case \( g \geq 2 \) follows from... | Yes |
Lemma 22.8. We have\n\n\[ \n{\pi }_{1}\left( {{\mathbb{{RP}}}^{2}\# {\mathbb{{RP}}}^{2}}\right) \cong \left\langle {x, y \mid {x}^{2} = {y}^{2}}\right\rangle .\n\] | Proof. In order to distinguish the two copies of \( {\mathbb{{RP}}}^{2} = {\bar{B}}^{2}/ \sim \) we write \( P = {\mathbb{{RP}}}^{2} \) and \( Q = {\mathbb{{RP}}}^{2} \) . The proof is of course a variation on the proof of Proposition 20.12. We have\n\n\[ \n{\pi }_{1}\left( {{\mathbb{{RP}}}^{2}\# {\mathbb{{RP}}}^{2}}\r... | Yes |
Is the fundamental group of the real projective space \( {\mathbb{{RP}}}^{2} \) torsion-free? | We recall that the fundamental group of the real projective space \( {\mathbb{{RP}}}^{2} \) is isomorphic to \( {\mathbb{Z}}_{2} \), therefore it is torsion and in particular it is not torsion-free. | Yes |
Lemma 22.11. Let \( W \) be a connected \( n \) -dimensional smooth manifold. Suppose we are given \( k \in {\mathbb{N}}_{0} \) and a submanifold of the form \( {\bar{B}}^{n - k} \times K \) where \( K \) is diffeomorphic to \( {S}^{k} \). (1) If \( n \geq k + 3 \), then for every base point \( {w}_{0} \in W \smallsetm... | Proof of Lemma 22.11. We write \( X = W \smallsetminus \left( {{B}^{n - k} \times K}\right) \) . Note that it follows from Proposition 15.11 that it suffices to prove the statement for some base point \( {w}_{0} \in {S}^{n - k - 1} \times \) \( K \) . First we consider the case that \( n \geq k + 3 \) . Next note that ... | Yes |
Lemma 22.12. Let \( W \) be an n-dimensional smooth manifold and suppose we are given a boundary component of the form \( {S}^{n - 2} \times C \) where \( C \) admits a diffeomorphism \( \varphi : {S}^{1} \rightarrow C \). We set \[ X\; \mathrel{\text{:=}} \;\left( {W \sqcup \left( {{S}^{n - 2} \times {\overset{―}{B}}^... | Proof. Note that it follows from Proposition 15.11 that it suffices to prove the statement for some base point \( {w}_{0} \in {S}^{n - 2} \times C \). Next note that we have the following two isomorphisms: \[ {\pi }_{1}\left( {W,{w}_{0}}\right) /\langle \langle C\rangle \rangle \underset{ \uparrow }{\overset{ \cong }{ ... | Yes |
Lemma 23.1. Let \( \sum \) be a closed connected non-empty 2-dimensional smooth manifold and let \( n \in {\mathbb{N}}_{0} \). (1) The topological space \( \sum \) minus \( n \) open disks is a compact connected topological sub-manifold of \( \sum \) with \( n \) boundary components, which are given by the boundaries o... | SKETCH OF PROOF. (1),(2) Let \( {\varphi }_{1},\ldots ,{\varphi }_{n} : {\bar{B}}^{2} \rightarrow \sum \) be smooth embeddings with disjoint images. Note that we obtain from applying Proposition 8.2 (2) iteratively altogether \( n \) times that \( \sum \smallsetminus \left( {{\varphi }_{1}\left( {B}^{2}\right) \cup \cd... | No |
Lemma 23.3. Let \( F \) be a closed connected non-empty 2-dimensional smooth manifold and let \( n \in {\mathbb{N}}_{0} \) . If we set \( M \) to be \( F \) minus \( n \) open disks, then \( \widehat{M} \) is homeomorphic to \( F \) . | Proof (*). Let \( {\varphi }_{1},\ldots ,{\varphi }_{n} : {\bar{B}}^{2} \rightarrow F \) be smooth embeddings with disjoint images. We need to show that for \( M \mathrel{\text{:=}} F \smallsetminus \left( {{\varphi }_{1}\left( {B}^{2}\right) \cup \cdots \cup {\varphi }_{n}\left( {B}^{2}\right) }\right) \) there exists... | Yes |
Proposition 23.7. For \( g \in {\mathbb{N}}_{0} \) and \( k \in \mathbb{N} \) we have the following table of invariants: | Proof. The proposition is an immediate consequence of Propositions 22.3 and 22.7, together with Lemmas 23.1, 23.2 and 23.3. | No |
Proposition 23.8. Let \( M \) and \( N \) be two compact connected non-empty 2-dimensional topological (respectively smooth) manifolds. The following are equivalent:\n\n(1) \( M \) and \( N \) are homeomorphic (respectively diffeomorphic),\n\n(2) (a) \( {\pi }_{1}{\left( M\right) }_{\mathrm{{ab}}} \cong {\pi }_{1}{\lef... | Proof. In the topological setting this proposition follows immediately from Proposition 23.7 and the Surface Classification Theorem 23.4. In the smooth setting the proposition follows from the above, together with the fact that on page 300 we showed that the \( {\sum }_{g, m} \) are orientable and that in Lemma 6.48 we... | Yes |
Lemma 23.12. Let \( g \in {\mathbb{N}}_{0} \) and let \( n \in \mathbb{N} \). (1) We denote the boundary components of \( {\sum }_{g, n} \) by \( {C}_{1},\ldots ,{C}_{n} \). Given any \( i \in \{ 1,\ldots, n\} \) there exists a topological graph \( G \subset {\sum }_{g, n} \) which contains all \( {C}_{j} \) with \( i ... | SKETCH OF PROOF. (1) In Figure 427 we show that, starting from any boundary component \( {C}_{i} \) of \( {\sum }_{g, n} \), we can push \ | No |
The unique boundary component of \( {\sum }_{g,1} \) is not a retract of \( {\sum }_{g,1} \) . | Proof. Let \( g \in {\mathbb{N}}_{0} \) . (1) The case \( g = 0 \) was dealt with in Lemma 15.7. Now we assume that \( g \geq 1 \) . We denote by \( C \) the unique boundary component of \( \sum \mathrel{\text{:=}} {\sum }_{g,1} \) . We had just seen in Lemma 23.11 that \( {\pi }_{1}\left( C\right) \rightarrow {\pi }_{... | Yes |
Lemma 23.14. Let \( \sum \) be a compact connected 2-dimensional smooth manifold. We denote the boundary components of \( \sum \) by \( {C}_{1},\ldots ,{C}_{n} \) . Given any permutation \( \sigma \in {S}_{n} \) there exists a diffeomorphism \( f \) of \( \sum \) with \( f\left( {C}_{i}\right) = {C}_{\sigma \left( i\ri... | Proof. In the proof we deal with the setting that \( \sum \) is orientable. The case that \( \sum \) is nonorientable is dealt with in almost the same way. We pick an orientation for \( \sum \) and we equip the boundary components \( {C}_{1},\ldots ,{C}_{n} \) with the corresponding orientations. Furthermore we pick or... | Yes |
Let \( \sum \) be a compact oriented connected 2-dimensional smooth manifold.\n\n(1) Let \( C \) and \( D \) be two distinct boundary components of \( \sum \) . Gluing \( C \) to \( D \) via an orientation-reversing diffeomorphism increases the genus by one and it decreases the number of boundary components by two.\n\n... | SKETCH OF PROOF.\n\n(1) By the Surface Classification Theorem 23.4 we can assume that \( \sum = {\sum }_{g, n} \) for some \( g \in {\mathbb{N}}_{0} \) and some \( n \in {\mathbb{N}}_{ \geq 2} \) . We denote by \( M \) the result of gluing the boundary components \( C \) and \( D \) via an orientation-reversing diffeom... | No |
Proposition 23.16. Let \( \sum \) be a compact oriented connected 2-dimensional smooth manifold and let \( C \) and \( D \) be two closed oriented curves in \( \sum \smallsetminus \partial \sum \) .\n\n(1) If \( C \) and \( D \) are non-separating, then \( C \) and \( D \) are equivalent.\n\n(2) Suppose \( C \) and \( ... | Sketch of PROOF. Let \( \sum \) be a compact oriented connected 2-dimensional smooth manifold.\n\n(1) Let \( C \) and \( D \) be two closed oriented non-separating curves on \( \sum \) . We denote by \( M \) respectively \( N \) the smooth manifold obtained from \( \sum \) by cutting along \( C \) respectively \( D \) ... | No |
Lemma 24.1. Let \( X \) be a topological space. The tip of the cone \( \operatorname{Cone}\left( X\right) \) is a deformation retract of \( \operatorname{Cone}\left( X\right) \) . | Proof. If \( X = \varnothing \), then it follows from the definition of \( \operatorname{Cone}\left( \varnothing \right) = \{ * \} \) that the cone equals the tip. Thus it remains to consider the case of a non-empty topological space. We consider the \ | No |
(1) If \( X \) is compact, then \( \operatorname{Cone}\left( X\right) \) and the suspension \( \sum \left( X\right) \) are also compact. | (1) This statement follows from Lemma 3.21 (4) and Proposition 3.12. | No |
Lemma 24.3. If \( X \) is a path-connected non-empty topological space, then its suspension \( \sum \left( X\right) \) is simply connected. | Proof (*). Let \( X \) be a path-connected non-empty topological space. In the following we denote by \( p : X \times \left\lbrack {-1,1}\right\rbrack \rightarrow \sum \left( X\right) \) the obvious projection map. We consider the subsets \( U = p\left( {X \times \left\lbrack {-1,\frac{1}{2}}\right) }\right) \) and \( ... | No |
(1) If \( f : X \rightarrow Y \) is a map between topological spaces, then \( \sum \left( f\right) : \sum \left( X\right) \rightarrow \sum \left( Y\right) \) is also continuous. | (1) We consider the following commutative diagram\n\n\n\nBy Lemma 3.8 (2b) we know that the top horizontal map is continuous. By Lemma 3.21 (3) the left and right vertical maps are continuous. It follows that the dia... | Yes |
Lemma 24.5. We consider the topological space \( X = \left\lbrack {-1,1}\right\rbrack \), the subset \( A = \left( {-1,1}\right) \) and the inclusion map \( i : A \rightarrow X \) . The induced map \( \sum \left( i\right) : \sum \left( A\right) \rightarrow \sum \left( X\right) \) is not an embedding. | Proof. We will provide the proof in Exercise 24.10. | No |
Lemma 24.6. (*) Let \( f : X \rightarrow Y \) be a map between topological spaces.\n\n(3) The natural inclusions \( X \rightarrow \operatorname{Cyl}\left( f\right) \) and \( Y \rightarrow \operatorname{Cyl}\left( f\right) \) are embeddings, in particular they are continuous. | Proof. We leave the proofs of the first two and of the last statement as voluntary exercises to the reader. Finally we will provide the proof of the third statement in Exercise 24.6. | No |
Lemma 24.7. (*) Let \( X \) be a topological space and let \( A \subset X \) be a subset. We denote by \( i : A \rightarrow X \) the inclusion map and we consider the map\n\nequipped with the subspace topology\n\ncoming from the product \( X \times \left\lbrack {0,1}\right\rbrack \)\n\n\( \Theta : \operatorname{Cyl}\le... | Proof.\n\n(1) It follows easily from Lemma 3.22 that the given map \( \Theta \) is continuous. Furthermore it is basically clear that \( \Theta \) is a bijection.\n\n(2) Suppose that \( A \) is a closed subset of \( X \) . Note that \( A \times \{ 1\} \) is a closed subset of \( A \times \left\lbrack {0,1}\right\rbrack... | Yes |
Lemma 24.8. Let \( f : X \rightarrow Y \) be a map between topological spaces.\n\n(1) The subspace \( Y \) is a deformation retract of \( \operatorname{Cyl}\left( {f : X \rightarrow Y}\right) \), in fact an explicit deformation retraction is given by the following map\n\n\[ H : \operatorname{Cyl}\left( {f : X \rightarr... | Proof.\n\n(1) We need to show that the map \( H \) is continuous. First note that the two maps\n\n\[ \begin{aligned} \left( {X \times \left\lbrack {0,1}\right\rbrack }\right) \times \left\lbrack {0,1}\right\rbrack & \rightarrow \operatorname{Cyl}\left( f\right) \\ \left( {Q, t}\right) & \mapsto \left\lbrack \left( {Q, ... | Yes |
Lemma 24.9. (*) Let \( f : A \rightarrow X \) be a map between topological spaces.\n\n(1) We denote by \( i : A \rightarrow \operatorname{Cone}\left( A\right) \) the map that is given by \( a \mapsto \left\lbrack \left( {a,1}\right) \right\rbrack \) . The mapping cone \( \operatorname{Cone}\left( {f : A \rightarrow X}\... | SKETCH OF PROOF.\n\n(1) This statement follows basically immediately from the definitions.\n\n(2) This statement follows immediately from (1) together with Lemma 3.43\n\n(3) This statement also follows immediately from (1) together with Lemma 3.43.\n\n(4) It is clear that the given map is a bijection. One can easily ve... | No |
(1) The maps\n\n\[ \n\\left( {f : X \\rightarrow Y}\\right) \\mapsto \\operatorname{Cyl}\\left( {f : X \\rightarrow Y}\\right) \n\]\n\n\[ \n\\left( {\\varphi : X \\rightarrow \\widetilde{X},\\psi : Y \\rightarrow \\widetilde{Y}}\\right) \\mapsto \\left( \\begin{aligned} \\mathrm{{Cyl}}\\left( {f : X \\rightarrow Y}\\ri... | Proof \\( \\left( *\\right) \\) . One needs to convince oneself that the map between the mapping cylinders and the mapping cones is in fact continuous. For the mapping cylinders that is an immediate consequence of Lemma 3.44 (5). For the mapping cones first note that the same argument as in the proof of Lemma 24.4 (1) ... | No |
Lemma 24.12. (*) Let \( {f}_{0},{f}_{1} : A \rightarrow X \) be maps between topological spaces. Furthermore let \( H : A \times \left\lbrack {0,1}\right\rbrack \rightarrow X \) be a homotopy between \( {f}_{0} \) and \( {f}_{1} \) . Let \( j \in \{ 0,1\} \) . We denote by \( {i}_{j} : A \rightarrow A \times \{ j\} \) ... | Proof of Lemma 24.12 (*). Let \( {f}_{0},{f}_{1} : A \rightarrow X \) be maps between topological spaces. Furthermore let \( H : \overline{A \times \left\lbrack {0,1}\right\rbrack } \rightarrow X \) be a homotopy between \( {f}_{0} \) and \( {f}_{1} \) . For notational convenience we only deal with the case \( j = 0 \)... | Yes |
Corollary 24.13. (*) Let \( f : A \rightarrow X \) be a map between topological spaces and let \( g : X \rightarrow Y \) be a homotopy equivalence. The map \[ \operatorname{Cyl}\left( {f : A \rightarrow X}\right) \rightarrow \operatorname{Cyl}\left( {g \circ f : A \rightarrow Y}\right) \] \[ \left\lbrack P\right\rbrack... | Proof. The corollary is a straightforward consequence of Proposition 24.11. We leave it to the reader to fill in the details. | No |
Lemma 24.14. (*) Let \( f : A \rightarrow X \) be a map between topological spaces and let \( {a}_{0} \in A \) . We write \( {x}_{0} \mathrel{\text{:=}} f\left( {a}_{0}\right) \) . If \( A \) is path-connected, then the inclusion induces an isomorphism\n\n\[ \n{\pi }_{1}\left( {X,{x}_{0}}\right) /{f}_{ * }\left( {{\pi ... | Proof. We will provide the proof in Exercise 24.12. | No |
Proposition 24.15. (*) Let \( X \) be a topological space and let \( A \) be a subset of \( X \). We denote by \( i : A \rightarrow X \) the inclusion map. We use Lemma 24.9 (2) to view \( X \) as a subset of Cone \( \left( {i : A \rightarrow X}\right) \). (1) If \( A \) is contractible, then given any \( {a}_{0} \in A... | Proof. We first prove Statement (1). Thus let \( {a}_{0} \in A \). Since \( A \) is contractible it follows from Lemma 18.15 (2) that there exists a map \( r : A \times \left\lbrack {0,1}\right\rbrack \rightarrow A \) such that \( {r}_{0} = \mathrm{{id}} \) and such that \( {r}_{1}\left( a\right) = {a}_{0} \) for all \... | Yes |
Lemma 24.17. Let \( X \) be a topological space. If \( Y \) is a regionally compact topological space (e.g. if \( Y = \left\lbrack {0,1}\right\rbrack \) ), then the map\n\n\[ \varphi : \sum \left( X\right) \times Y \rightarrow \sum \left( {X \times Y}\right) \]\n\n\[ \left( {\left\lbrack \left( {x, t}\right) \right\rbr... | Proof. We denote by \( p : \left( {X \times Y}\right) \times \left\lbrack {-1,1}\right\rbrack \rightarrow \sum \left( {X \times Y}\right) \) and \( q : X \times \left\lbrack {-1,1}\right\rbrack \rightarrow \sum \left( X\right) \) the two obvious projection maps. We have the following commutative diagram\n\n![448f61af-e... | Yes |
Lemma 24.18. Let \( X \) be a topological space and let \( f : X \rightarrow X \) be a map. The natural maps\n\n\[ \begin{aligned} \alpha : X \times \left\lbrack {0,1}\right\rbrack & \rightarrow \operatorname{Tor}\left( {X, f}\right) \\ \left( {x, t}\right) & \mapsto \left\lbrack \left( {x, t}\right) \right\rbrack \end... | Proof. First note that it follows from Lemma 3.43 (3) that \( \Psi \) is continuous. Next we consider the natural map\n\n\[ \Phi : \left( {X \times \left\lbrack {0,1}\right\rbrack }\right) / \sim \rightarrow \left( {X \times \left\lbrack {0,1}\right\rbrack }\right) { \cup }_{X\times \{ 0,1\} }X \]\n\n\[ \left\lbrack \l... | Yes |
Lemma 24.19. (*) Let \( X \) be a topological space and let \( f : X \rightarrow X \) be a map.\n\n(1) The map\n\n\[ \Theta : \left( {X \times \mathbb{R}}\right) /\left( {x, r}\right) \overset{ \downarrow }{ \sim }\left( {f\left( x\right), r + 1}\right) \rightarrow \operatorname{Tor}\left( {X, f}\right) = \left( {X \ti... | Proof. First note that it is straightforward to write down an inverse to \( \Theta \) . The second statement follows from elementary arguments. We leave it to the reader to fill in the details. | No |
Lemma 24.20. Let \( X \) be a topological space and let \( f : X \rightarrow X \) be a homeomorphism.\n\n(1) If \( X \) is Hausdorff, then the mapping torus \( \operatorname{Tor}\left( {X, f}\right) \) is Hausdorff. | Proof.\n\n(1) This statement will be proved in Exercise 24.13. | No |
Lemma 24.22. Let \( M \) be an \( n \) -dimensional smooth manifold and let \( f : M \rightarrow M \) be a diffeomorphism. The following statements hold:\n\n(1) The mapping torus \( \operatorname{Tor}\left( {M, f}\right) \) is naturally an \( \left( {n + 1}\right) \) -dimensional smooth manifold such that the obvious i... | Proof (*). First consider the case that \( M \) has no boundary. By Proposition 6.51 (4) we know that \( M \times \left\lbrack {0,1}\right\rbrack \) is an \( \left( {n + 1}\right) \) -dimensional smooth manifold with boundary given by \( M \times \{ 0,1\} \) . Note that we obtain the mapping torus \( \operatorname{Tor}... | No |
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