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Lemma 3.35. (*) Let \( G \) be a group that acts on a set \( X \) . We denote by \( p : X \rightarrow X/G \) the projection. Let \( A \) and \( B \) be subsets of \( X \) . Then\n\n\[ p\left( A\right) \cap p\left( B\right) = \varnothing \Leftrightarrow \text{ for every }g \in G\text{ we have }{gA} \cap B = \varnothing ... | Proof of Lemma \( {3.35}\left( *\right) \) . We prove the second statement. The first statement is easily seen to be equivalent to the second statement. We have\n\n\[ p\left( A\right) \cap p\left( B\right) \neq \varnothing \Leftrightarrow \text{ there exist }a \in A\text{ and }b \in B\text{ with }p\left( a\right) = p\l... | Yes |
Theorem 3.38. (Rouché’s Theorem) Let \( U \subset \mathbb{C} \) be an open set, let \( f, g : U \rightarrow \mathbb{C} \) be two holomorphic functions and let \( z \in U \) and \( r \in {\mathbb{R}}^{ + } \) be such that \( {\bar{B}}_{r}\left( z\right) = \{ w \in \mathbb{C}\left| \right| w - z \mid \leq r\} \) is conta... | Proof. This theorem is proved in every self-respecting book on complex analysis, see e.g. Lan99, Theorem VI.1.6]. | No |
(1) The map\n\n\[ d : {\mathbb{C}}^{n}/{S}_{n} \times {\mathbb{C}}^{n}/{S}_{n} \rightarrow {\mathbb{R}}_{ \geq 0} \]\n\n\[ \left( {w, z}\right) \mapsto d\left( {w, z}\right) \]\n\nis a metric on \( {\mathbb{C}}^{n}/{S}_{n} \) . | Proof. We hand over the task providing the proof of the lemma to the surely very meticulous reader. | No |
Proposition 3.40. The real projective and the complex projective spaces are compact and Hausdorff. | In the proof of Proposition 3.40 it is convenient to have the following description of complex projective spaces.\n\n\( {}^{66} \) For example an inverse is given by the map \( \left( {{\mathbb{R}}^{n + 1}\smallsetminus \{ 0\} }\right) /\left( {\mathbb{R}\smallsetminus \{ 0\} }\right) \rightarrow {S}^{n}/\{ \pm 1\} \) ... | No |
Lemma 3.41. Let \( n \in {\mathbb{N}}_{0} \) . We make the identification\n\n\[ \n{S}^{{2n} + 1} = \left\{ {\left( {{z}_{1},\ldots ,{z}_{n + 1}}\right) \in {\mathbb{C}}^{n + 1}\left| \right| {z}_{1}\left| {{}^{2} + \cdots + }\right| {z}_{n + 1}{\left. \right| }^{2} = 1}\right\} .\n\]\n\nWe consider the action\n\n\[ \n{... | LEMMA 3.41. It follows easily from the definition of the quotient topology on the complex projective space \( {\mathbb{{CP}}}^{n} = \left( {{\mathbb{C}}^{n + 1}\smallsetminus \{ 0\} }\right) /\left( {\mathbb{C}\smallsetminus \{ 0\} }\right) \) and from Lemma 3.21 (3) together with Lemma 3.22 that both maps are continuo... | Yes |
(1) Let \( n \in \mathbb{N} \) . If we denote by \( \sim \) the equivalence relation on \( {\bar{B}}^{n} \) that is generated by \( P \sim - P \) for \( P \in {S}^{n - 1} = \partial {\bar{B}}^{n} \), then the map\n\n\[ \n{\mathbb{{RP}}}^{n} = {S}^{n}/\{ \pm 1\} \;\overset{ \cong }{ \rightarrow }\;{\bar{B}}^{n}/ \sim \n... | Proof. First note that one can easily verify that both maps are actually well-defined and that they are bijections. Next we explain why the two maps are continuous:\n\n(1) By Lemma 3.22 it suffices to show that the corresponding map \( {S}^{n} \rightarrow {\bar{B}}^{n}/ \sim \) is continuous. One can easily verify that... | Yes |
Lemma 3.45. Let \( X \) be a topological space and let \( A \) and \( B \) be two subsets of \( X \) with \( X = A \cup B \) . We denote by \( \varphi : A \cap B \rightarrow A \cap B \) the identity map. We suppose that one of the following two conditions hold:\n\n(1) Both subsets \( A \) and \( B \) are closed.\n\n(2)... | Proof. Let \( X \) be a topological space and let \( A \) and \( B \) be two subsets of \( X \) with \( X = A \cup B \) . We denote by \( \varphi : A \cap B \rightarrow A \cap B \) the identity map. In the following we consider the case that \( A \) and \( B \) are both closed subsets of \( X \) . The case that both \(... | Yes |
Lemma 3.47. (*) Let \( {X}_{1},{Y}_{1},{X}_{2},{Y}_{2} \) be four topological spaces and let \( f : {X}_{1} \rightarrow {X}_{2} \) and \( g : {Y}_{1} \rightarrow {Y}_{2} \) be maps. The corresponding map\n\n\[ f * g : {X}_{1} * {Y}_{1} \rightarrow {X}_{2} * {Y}_{2} \]\n\n\[ \left\lbrack \left( {x, t, y}\right) \right\r... | Proof (*). It is straightforward to verify that the map \( f * g \) is well-defined. It follows easily from Lemma 3.22, Lemma 3.21 (3) and Lemma 3.8 (3c) that the map \( f * g \) is continuous. | No |
Lemma 3.48. If \( X \) and \( Y \) are compact topological spaces, then the following two statements hold:\n\n(1) The join \( X * Y \) is compact. | Proof.\n\n(1) It follows immediately from our hypothesis and Proposition 3.12 together with Lemma 3.21\n\n(4) that the join \( X * Y = \left( {X \times \left\lbrack {0,1}\right\rbrack \times Y}\right) / \sim \) is compact. | Yes |
Lemma 3.49. Given \( P, Q \in {\mathbb{R}}^{n} \) we denote by \( \overline{PQ} = \{ x \cdot \left( {1 - t}\right) + y \cdot t \mid t \in \left\lbrack {0,1}\right\rbrack \} \) the Euclidean segment determined by \( P \) and \( Q \) . Let \( X \) and \( Y \) be two compact subsets of \( {\mathbb{R}}^{n} \) which have th... | Proof. It follows almost immediately from Lemma 3.48 and our hypothesis that the map \( \Theta \) is a homeomorphism. | No |
Given any \( m, n \in {\mathbb{N}}_{0} \) the map\n\n\[ \n{S}^{m} * {S}^{n} \rightarrow {S}^{m + n + 1} \]\n\n\[ \n\left\lbrack \left( {x, t, y}\right) \right\rbrack \mapsto \underset{ \in {\mathbb{R}}^{m + 1} \times {\mathbb{R}}^{n + 1} = {\mathbb{R}}^{m + n + 2}}{\underbrace{\left( x \cdot \cos \left( \frac{\pi t}{2}... | Proof. Using Lemma 3.48 (2) one can fairly easily verify that the given map is a homeomorphism. | No |
For any \( n \in \mathbb{N} \) the matrix groups \( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \) and \( \mathrm{{GL}}\left( {n,\mathbb{C}}\right) \) are topological groups. | Proof. As we pointed out on page 163, matrix multiplication is continuous. Furthermore, basic linear algebra, see e.g. HJ13, Chapter 0.8.2], says that taking the inverse is given by the map\n\n\( \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \rightarrow \mathrm{{GL}}\left( {n,\mathbb{R}}\right) \)\n\nBut this map is clearl... | Yes |
Lemma 3.55. Let \( G \) be a topological group.\n\n(1) Given any \( g \in G \) the two maps\n\n\[ \begin{aligned} {l}_{g} : G & \rightarrow G \\ h & \mapsto g \cdot h \end{aligned}\;\text{ and }\;\begin{aligned} {r}_{g} : G & \rightarrow G \\ h & \mapsto h \cdot g \end{aligned} \]\n\nare continuous.\n\n(2) If \( G \) i... | Proof (*).\n\n(1) The map \( {l}_{g} \) is the composition of the two maps\n\n\[ \begin{matrix} G & \rightarrow & G \times G \\ h & \mapsto & \left( {g, h}\right) \end{matrix}\;\text{ and }\;\begin{matrix} G \times G & \rightarrow & G \\ \left( {x, y}\right) & \mapsto & x \cdot y. \end{matrix} \]\n\nThe first map is co... | Yes |
Lemma 4.1. Let \( G = \left( {V, E, i, t}\right) \) be an abstract graph.\n\n(1) The obvious map \( V \rightarrow \left| G\right| \) is an inclusion and the image is a discrete subset of \( \left| G\right| \) . | (1) This statement follows fairly easily from the definitions of the quotient topology and the disjoint union topology. We leave it to the reader to fill in the details. | No |
Lemma 4.2. Let \( G = \left( {V, E, i, t}\right) \) be an abstract graph and let \( n \in {\mathbb{N}}_{0} \) . Suppose we are given the following data:\n\n(1) An injective map \( \alpha : V \rightarrow {\mathbb{R}}^{n} \) .\n\n(2) For each \( e \in E \) we are given a map \( {\beta }_{e} : \left\lbrack {0,1}\right\rbr... | Proof. It follows easily from our hypotheses that the map is well-defined and that it is an injection. Furthermore we obtain from Lemma 3.22 that the map is continuous. If \( G \) is finite, then it follows from Lemma \( \left| \overline{4.1}\right| \) together with Proposition \( \left| \overline{2.43}\right| \left( 3... | Yes |
Lemma 4.4. Let \( G = \left( {V, E, i, t}\right) \) be an abstract graph.\n\n(1) We define \( \varphi : E \rightarrow \mathcal{P}\left( V\right) \) by \( \varphi \left( e\right) = \{ i\left( e\right), t\left( e\right) \} \) . The triple \( \widetilde{G} = \left( {V, E,\varphi }\right) \) is an undirected abstract graph... | Proof.\n\n(1) This statement follows immediately from the definitions.\n\n(2) It follows fairly easily from the definitions that the map is well-defined and a bijection. Furthermore one can deduce easily from Lemma 3.28 that the map is continuous. Working directly with the definitions one sees that the map is open. In ... | Yes |
Lemma 4.5. Let \( G = \left( {V, E, i, t}\right) \) be an abstract graph or let \( G = \left( {V, E,\varphi }\right) \) be an undirected abstract graph. Let \( f : V \rightarrow X \) be a map to some set \( X \) such that for any \( e \in E \) the values of \( f \) on the endpoints of e agree. If \( G \) is connected, ... | Proof. This lemma follows almost immediately from the definitions. | No |
Lemma 4.6. Let \( G = \\left( {V, E, i, t}\\right) \) be an abstract graph or let \( G = \\left( {V, E,\\varphi }\\right) \) be an undirected abstract graph. If \( G \) is a tree and if \( G \) has at least one edge, then \( G \) admits at least two vertices of valence one. | Proof (*). We prove Lemma 4.6 for abstract graphs, the proof for undirected abstract graphs is verbatim the same. Thus let \( G = \\left( {V, E, i, t}\\right) \) be a tree that has at least one edge. Since \( G \) has at least one edge and since it is by definition connected, we see that the tree \( G \) does not have ... | Yes |
Corollary 4.7. Let \( G = \left( {V, E, i, t}\right) \) be an abstract graph and let \( \widetilde{v} \in V \) be a vertex. If \( G \) is a tree, then there exist distinct edges \( {e}_{1},\ldots ,{e}_{k} \) and distinct vertices \( {v}_{1},\ldots ,{v}_{k} \) with the following properties:\n\n(1) We have \( V = \left\{... | Proof. For abstract graphs we will provide the proof of the corollary in Exercise 4.5. For undirected abstract graphs the proof is verbatim the same. | No |
Proposition 4.8. Let \( G = \left( {V, E, i, t}\right) \) be a finite connected non-empty abstract graph.\n\n(1) The graph \( G \) admits a spanning tree.\n\n(2) Every spanning tree of \( G \) contains all vertices of \( G \) . | Proof. In the following we provide the proof for abstract graphs. The proof for undirected abstract graphs is verbatim the same. Thus let \( G = \left( {V, E, i, t}\right) \) be a finite connected abstract graph.\n\n(1) We pick a vertex \( v \in V \) . We define \( {T}_{0} = \{ v\} \) and view \( {T}_{0} \) as a subtre... | No |
Lemma 5.1. Let \( m, n \in \mathbb{N} \) . We have the obvious inclusion\n\n\[ \mathrm{M}\left( {m \times n,\mathbb{R}}\right) \rightarrow C\left( {{\mathbb{R}}^{n},{\mathbb{R}}^{m}}\right) = {\left( {\mathbb{R}}^{m}\right) }^{{\mathbb{R}}^{n}} \]\n\n\[ A \mapsto \left( \begin{array}{rrr} {\mathbb{R}}^{n} & \rightarrow... | Sketch of PROOF. This statement can be proved, with some effort, using Lemma 2.23. | No |
Proposition 5.4. Let \( X \) and \( Y \) be topological spaces. If \( Y \) is regionally compact, then the following two statements hold:\n\n(1) The evaluation map \( {}^{80} \)\n\n\[ e : {X}^{Y} \times Y \rightarrow X \]\n\n\[ \left( {\left( {f : Y \rightarrow X}\right), y}\right) \mapsto f\left( y\right) \]\n\nis con... | Proof \( \left( *\right) \) .\n\n(1) Let \( U \) be an open subset of \( X \) . We need to show that \( {e}^{-1}\left( U\right) \) is open. By Lemma 2.5 it suffices to show that given \( \left( {f, y}\right) \in {e}^{-1}\left( U\right) \) there exists a neighborhood \( W \) of \( \left( {f, y}\right) \) with \( W \subs... | Yes |
Lemma 5.5. (*) Let \( X \) and \( Y \) be topological spaces.\n\n(1) If \( \varphi : X \rightarrow \widetilde{X} \) is a continuous map, then the induced map\n\n\[ \n{\varphi }^{ * } : {X}^{Y} \rightarrow {\widetilde{X}}^{Y} \n\]\n\n\[ \n\left( {f : Y \rightarrow X}\right) \mapsto \left( {\varphi \circ f : Y \rightarro... | Proof \( \left( *\right) \) . To clarify the statements we replace the notation \( M\left( {K, U}\right) \) by the notation \( {M}_{{X}^{Y}}\left( {K, U}\right) \), with the obvious interpretation thereof.\n\n(1) We check continuity using the continuity criterion of Proposition 2.37. Thus let \( K \subset Y \) be a com... | No |
Proposition 5.6. Let \( Y \) be a topological space that is regionally compact. Furthermore let \( X \) and \( T \) be two topological spaces. Let \( H : Y \times T \rightarrow X \) be a map. Given \( t \in T \) we denote by \( {H}_{t} : Y \rightarrow X \) the map defined by \( {H}_{t}\left( y\right) \mathrel{\text{:=}... | Proof \( \left( *\right) \) . \( {}^{84} \) Note that the \ | No |
Lemma 5.8. Let \( X \) and \( Y \) be topological spaces. Furthermore let \( \mathcal{V} \) be a subbasis for \( X \) . We consider\n\n\[ \mathcal{C} = \left\{ {M\left( {K, V}\right) \in \mathcal{P}\left( {X}^{Y}\right) \mid K\text{ a compact subset of }Y\text{ and }V \in \mathcal{V}}\right\} .\n\]\nIf \( Y \) is Hausd... | Proof. Let \( K \) be a compact subset of \( Y \), let \( U \) be an open subset of \( X \) and let \( f \) be a point in \( M\left( {K, U}\right) \), i.e. \( f : Y \rightarrow X \) is a map with \( K \subset {f}^{-1}\left( U\right) \) . By Lemma 2.27 (2) we need\nto show that there exist finitely many \( {C}_{1},\ldot... | Yes |
Proposition 5.9. Let \( X, Y \) and \( Z \) be topological spaces. If \( Y \) is regionally compact, then the map \[ \Phi : {X}^{Y} \times {Y}^{Z} \rightarrow {X}^{Z} \] \[ \left( {f, g}\right) \mapsto f \circ g \] is continuous. | The proof of Proposition 5.9 rests on the following little lemma. Lemma 5.10. (*) | No |
Lemma 5.10. (*) Let \( Y \) be a topological space that is regionally compact. If \( K \) is a compact subset that is contained in an open subset \( U \) of \( Y \), then there exists a compact subset \( L \) with \( K \subset L \subset L \subset U \) . | Proof (*). Let \( x \in K \) . Since \( Y \) is regionally compact there exists a compact neighborhood \( {L}_{x} \subset U \) of \( x \) . Since \( K \) is compact there exist \( {x}_{1},\ldots ,{x}_{m} \in K \) with \( K \subset {L}_{{x}_{1}} \cup \cdots \cup {L}_{{x}_{m}} \) . We set \( L \mathrel{\text{:=}} {L}_{{x... | No |
Corollary 5.11. Let \( X \) be a topological space. If \( X \) is compact and Hausdorff, then the group\n\n\[ \n\operatorname{Homeo}\left( X\right) \mathrel{\text{:=}} \text{the set of all self-homeomorphisms of}X\text{,}\n\]\n\nequipped with the subspace topology coming from \( {X}^{X} \), is a topological group. | Proof (*). Let \( X \) be a topological space that is compact and Hausdorff. First note that by Lemma 2.73 we know that \( X \) is regionally compact. Thus it follows from Proposition 5.9 that the multiplication map \( \operatorname{Homeo}\left( X\right) \times \operatorname{Homeo}\left( X\right) \rightarrow \operatorn... | Yes |
Let \( X \) be a topological space.\n\n(1) Let \( \widetilde{X} \) and \( Y \) be topological spaces. If \( Y \) is regionally compact, then the map\n\n\[ \Psi : {X}^{Y} \times {\widetilde{X}}^{Y} \rightarrow {\left( X \times \widetilde{X}\right) }^{Y} \]\n\n\[ \left( {f,\widetilde{f}}\right) \mapsto \left( \begin{alig... | Proof.\n\n(1) By Proposition 2.37 it suffices to show that for any compact subset \( K \subset Y \) and any open subset \( W \subset \widetilde{X} \times \widetilde{X} \) the preimage \( {\Psi }^{-1}\left( {{M}_{{\left( X\underline{ \times }\widetilde{X}\right) }^{Y}}\left( {K, W}\right) }\right) \) is an open subset o... | Yes |
Lemma 5.13. Let \( X \) and \( Y \) be two topological spaces. Suppose we are given an equivalence relation \( \sim \) on \( X \) . We denote by \( { \sim }_{ \times } \) the equivalence relation on \( X \times Y \) that is generated by \( \left( {x, y}\right) { \sim }_{ \times }\left( {{x}^{\prime }, y}\right) \) when... | Proof (*). It follows basically immediately from the definition of \( { \sim }_{ \times } \) that the map \( \varphi \) is a bijection. We will walk cautiously through the continuity argument.\n\nWe denote by \( p : X \times Y \rightarrow \left( {X \times Y}\right) /{ \sim }_{ \times } \) and \( q : X \rightarrow X/ \s... | Yes |
(1) Let \( X \) be a topological space and let \( \sim \) be an equivalence relation on \( X \) . The projection map \( X \rightarrow X/ \sim \) is a quotient map. | (1) This statement follows immediately from the definition of the quotient topology on the quotient \( X/ \sim \) . | No |
Lemma 5.17. Suppose we are given a commutative diagram of maps between topological spaces:\n\n\n\nThe following two statements hold:\n\n(1) If \( p \) and \( f \) are quotient maps, then \( \varphi \) is also a quoti... | Proof.\n\n(1) Suppose that \( p \) and \( f \) are quotient maps. Let \( V \subset Z \) be an open subset. We have the following implications:\n\n\( V \) is open \( \Leftrightarrow {f}^{-1}\left( V\right) \) is open \( \Leftrightarrow {p}^{-1}\left( {{\varphi }^{-1}\left( V\right) }\right) \) is open \( \Leftrightarrow... | Yes |
Lemma 5.18. Let \( X \) and \( Y \) be two topological spaces. Suppose we are given an equivalence relation \( \sim \) on \( X \) . We denote by \( { \sim }_{ \times } \) the equivalence relation on \( X \times Y \) that is generated by \( \left( {x, y}\right) { \sim }_{ \times }\left( {{x}^{\prime }, y}\right) \) when... | Proof (*). As we pointed out in the proof of Lemma 5.13, it is basically clear that \( \varphi \) is a bijection. By Lemma 5.15 it remains to show that \( \varphi \) is a quotient map. We denote by \( p : X \times Y \rightarrow \left( {X \times Y}\right) /\overline{{ \sim }_{ \times }\text{ and }}q : X \rightarrow X/ \... | Yes |
Lemma 5.19. Let \( X \) and \( Y \) be two topological spaces, let \( A \subset X \) be a subset and let \( f : A \rightarrow Y \) be a map. Using the notation from page 200, we see that the map\n\n\[ \left( {X \times \left\lbrack {0,1}\right\rbrack }\right) { \cup }_{f \times {\mathrm{{id}}}_{\left\lbrack 0,1\right\rb... | Proof. Once one unravels the definitions it is clear that this statement follows immediately  | No |
Proposition 3.52. Let \( X, Y \) and \( Z \) be non-empty topological spaces that are compact and Hausdorff. Then there exists a homeomorphism\n\n\[ \left( {X * Y}\right) * Z\overset{ \cong }{ \rightarrow }X * \left( {Y * Z}\right) \] | Proof. We define the join \( X * Y * Z \) as follows: we consider\n\n\[ \{ \left( {r, x, s, y, t, z}\right) \in \left\lbrack {0,1}\right\rbrack \times X \times \left\lbrack {0,1}\right\rbrack \times Y \times \left\lbrack {0,1}\right\rbrack \times Z \mid r + s + t = 1\} / \sim \]\n\nwhere \ | No |
Proposition 5.20. (*) Let \( X \) be a topological space, let \( A \) be a subset of \( X \) and let \( Y \) be a topological space that is regionally compact. Then the map\n\n\[ \varphi : \left( {X/A}\right) \times Y \rightarrow \left( {X \times Y}\right) /\left( {A \times Y}\right) \]\n\n\[ \left( {\left\lbrack x\rig... | Proof (*). We denote by \( p : X \times Y \rightarrow \left( {X \times Y}\right) /\left( {A \times Y}\right) \) and \( q : X \rightarrow X/A \) the two obvious projection maps. We have the following commutative diagram\n\n be a map between two sets and let \( Y \) be another set. Let \( \widetilde{W} \) be a subset of \( X \times Y \) that is saturated with respect to the map \( f \times {\operatorname{id}}_{Y} : X \times Y \rightarrow Z \times Y \) . Let \( K \subset Y \) . We consider the set\n... | Proof. We write \( h = f \times {\operatorname{id}}_{Y} \) . By the above we need to show that \( {f}^{-1}\left( {f\left( U\right) }\right) \subset U \) . Thus let \( x \in {f}^{-1}\left( {f\left( U\right) }\right) \subset X \) . We have\n\n\[ \nh\left( {\{ x\} \times K}\right) = \left( {f \times {\operatorname{id}}_{Y... | Yes |
Lemma 5.22. (*) Let \( X \) and \( Y \) be topological spaces, let \( x \in X \) and let \( K \subset Y \) be a compact subset. Let \( \widetilde{W} \subset X \times Y \) be an open subset that contains \( \{ x\} \times K \) . There exist open neighborhoods \( U \subset X \) of \( x \) and \( V \subset Y \) of \( K \) ... | Proof of Lemma \( §{5.22}\left( *\right) \) . Since \( \widetilde{W} \subset X \times Y \) is open and since \( \{ x\} \times K \subset \widetilde{W} \) we can find for any \( k \in K \) open neighborhoods \( {A}_{k} \) of \( x \in X \) and \( {B}_{k} \) of \( k \in Y \) such that \( {A}_{k} \times {B}_{k} \subset \wid... | Yes |
Lemma 5.24. (*) Let \( X \) and \( Y \) be topological spaces and let \( B \) be a compact subset of \( Y \) . Let \( x \in X \) . If \( W \) is an open subset of \( X \times Y \) that contains \( \{ x\} \times B \), then there exists an open subset \( U \) of \( X \) and an open subset \( \widetilde{V} \subset Y \) wi... | Proof (*). Let \( W \) be an open subset of \( X \times Y \) that contains \( \{ x\} \times B \) . By definition of the product topology we can find for each \( b \in B \) an open neighborhood \( {U}_{b} \) of \( x \) in \( X \) and an open neighborhood \( {V}_{b} \) of \( b \) in \( Y \) such that \( {U}_{b} \times {V... | Yes |
Lemma 6.2. (*) Let \( X \) be a topological space and let \( {\left\{ {U}_{i}\right\} }_{i \in I} \) be an open cover of \( X \) . If \( X \) is second-countable, then there exists a countable subset \( J \subset I \) with \( X = \mathop{\bigcup }\limits_{{j \in J}}{U}_{j} \) . | Proof (*). Let \( \mathcal{B} \) be a countable basis for the topology of \( X \) . We write\n\n\[ \n{\mathcal{B}}^{\prime } \mathrel{\text{:=}} \left\{ {V \in \mathcal{B} \mid V \subset {U}_{i}\text{ for some }i \in I}\right\} .\n\]\n\nGiven \( V \in {\mathcal{B}}^{\prime } \) we pick \( j \in I \) with \( V \subset {... | Yes |
Lemma 6.3. Let \( n \in \mathbb{N} \) . Every subset of \( {\mathbb{R}}^{n} \) is second-countable and Hausdorff. | Proof. Let \( A \) be a subset of \( {\mathbb{R}}^{n} \) . By Lemma 6.1 we know that \( A \) is second-countable. Furthermore, it follows from Proposition 2.11 and \( \overline{\text{ Lemma }} \) 2.12 that \( A \) is also Hausdorff. | No |
Lemma 6.4. (*) Let \( X \) be a topological space.\n\n(1) If \( X \) admits a finite atlas, then \( X \) is second-countable.\n\n(2) If \( X \) is compact and if it admits an atlas, then \( X \) is second-countable. | Proof (*). The first statement is an immediate consequence of Lemma 6.1 (4). The second statement in turn is an immediate consequence of the first statement since every compact topological space that admits an atlas also admits a finite atlas. | Yes |
Proposition 6.5. Let \( M \) be an \( m \) -dimensional topological manifold with possibly nonempty boundary and let \( N \) be an \( n \) -dimensional topological manifold with empty boundary. Then the following hold:\n\n(1) The product \( M \times N \) is an \( \left( {m + n}\right) \) -dimensional topological manifo... | Proof (*).\n\n(1) First note that it follows from Proposition 3.12 (1) that \( M \times N \) is Hausdorff and it follows from Lemma 6.1 that \( M \times N \) is second-countable. One can now easily use products of charts to show that \( M \times N \) is an \( \left( {m + n}\right) \) -dimensional topological manifold.\... | No |
Lemma 6.6. (*) Let \( M \) and \( N \) be two topological manifolds.\n\n(1) If \( f : M \rightarrow N \) is a local homeomorphism, then \( f\left( {\partial M}\right) = \left( {\partial N}\right) \cap f\left( M\right) \) . | Proof \( \left( *\right) \) . The lemma follows from the elementary observation that the restriction of a chart for a point on a topological manifold has the same type as the original chart. We leave it to the reader to fill in the minuscule details. | No |
Lemma 6.7. (*) Let \( M \) be a topological manifold.\n\n(1) Any union of components of \( \partial M \) is a closed subset of \( M \) . In particular \( \partial M \) itself is a closed subset of \( M \) . | Proof \( \\left( *\\right) \) . Let \( M \) be a topological manifold.\n\n(1) First note that, basically by definition, the set of points in \( M \) admitting a chart of type (i) is an open subset of \( M \) . It follows from the definition of \( \\partial M \) that \( M \\smallsetminus \\partial M \) is an open subset... | Yes |
(1) For every \( g \in {\mathbb{N}}_{0} \) the surface \( {\sum }_{g} \) of genus \( g \) is a closed 2-dimensional topological manifold. | Proof (*). To simplify the discussion we will only prove Statement (1) of the proposition. We leave it to the reader to modify the argument to deal with the non-orientable case.\n\nWe start out this proof with the discussion of the two cases \( g = 0 \) and \( g = 1 \) that we had basically dealt with already. First re... | No |
(1) Every n-dimensional topological manifold is locally homeomorphic to an open convex subset of \( {\mathbb{R}}^{n} \) or \( {H}_{n} \) . | Let \( M \) be an \( n \) -dimensional topological manifold and let \( P \in M \) . First assume that \( P \notin \partial M \) . This means that there exists a chart \( \Phi : U \rightarrow V \) to some open subset \( V \) of \( {\mathbb{R}}^{n} \) . We set \( Q = \Phi \left( P\right) \) . Since \( V \) is open there ... | Yes |
Lemma 6.10. Let \( M \) be a topological manifold and let \( \Phi : U \rightarrow V \) be a chart. Let \( A \subset V \) be a subset that is compact. Then the corresponding subset \( {\Phi }^{-1}\left( A\right) \) is a closed subset of \( M. \) | Proof. Let \( M \) be a topological manifold and let \( \Phi : U \rightarrow V \) be a chart. We denote by \( \Psi = {\Phi }^{-1} : V \rightarrow U \) be the inverse map. Let \( A \subset V \) be a subset that is compact. It follows from Lemma 2.40 that \( \Psi \left( A\right) \) is a compact subset of \( M \) . But \(... | Yes |
Lemma 6.11. Let \( M \) be a topological manifold. If \( N \subset M \) is a \( k \) -dimensional submanifold, then it is also a \( k \) -dimensional topological manifold. | Proof. It follows from Lemmas 2.12 and 6.1 that \( N \) is Hausdorff and second-countable. It is clear from the definitions that \( N \) admits a \( k \) -dimensional atlas. Thus \( N \) is indeed a \( k \) -dimensional topological manifold. | Yes |
Proposition 6.12. Let \( a < b \) be two real numbers and let \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) be a function. The following two statements are equivalent:\n\n(1) The function \( f : \left\lbrack {a, b}\right\rbrack \rightarrow \mathbb{R} \) is smooth in the above sense.\n\n(2) All deriv... | Proof. The \ | No |
(1) Suppose that we are given real numbers \( a \leq b < c \leq d \) and \( \epsilon > 0 \) . There exists a smooth function \( f : \left\lbrack {a, d}\right\rbrack \rightarrow \left\lbrack {0,1}\right\rbrack \) with the following properties:\n\n(a) \( {\left. f\right| }_{\left\lbrack a, b\right\rbrack } \equiv 0 \) an... | (1) This statement is proved in any self-respecting analysis course, we refer to [Kön04, Chapter 9] for details. | No |
Proposition 6.16. Let \( U \subset {\mathbb{R}}^{n} \) be an open convex subset that contains the origin. Furthermore let \( f : U \rightarrow {\mathbb{R}}^{m} \) be a smooth map with \( f\left( 0\right) = 0 \) . There exists a smooth map \( A : U \rightarrow \mathrm{M}\left( {m \times n,\mathbb{R}}\right) \) such that... | Proof. By considering the coordinate functions of a given map \( f = \left( {{f}_{1},\ldots ,{f}_{m}}\right) : U \rightarrow {\mathbb{R}}^{m} \) separately we see that it suffices to deal with the case \( m = 1 \) . Thus let \( f : U \rightarrow \mathbb{R} \) be a smooth map with \( f\left( 0\right) = 0 \) . Note that ... | Yes |
Proposition 6.17. Let \( U \subset {\mathbb{R}}^{m} \) and \( V \subset {\mathbb{R}}^{n} \) be open sets.\n\n(1) If \( f : U \rightarrow V \) is a diffeomorphism, then for any \( P \in U \) the differential \( \mathrm{D}{f}_{P} \) is an invertible matrix.\n\n(2) If \( U \) and \( V \) are diffeomorphic and non-empty, t... | Proof.\n\n(1) Let \( f : U \rightarrow V \) be a diffeomorphism between open sets \( U \subset {\mathbb{R}}^{m} \) and \( V \subset {\mathbb{R}}^{n} \) . We denote by \( g : V \rightarrow U \) the inverse of \( f \) . Let \( P \in U \) . It follows from the chain rule, i.e. from Proposition 6.15, that the matrices \( \... | Yes |
Theorem 6.18. (Inverse Function Theorem) Let \( f : U \rightarrow V \) be a smooth map between two open subsets of \( {\mathbb{R}}^{n} \) and let \( P \in U \) be a point. If \( \mathrm{D}{f}_{P} \) is invertible, then \( f \) is a local diffeomorphism at \( P \) . | Proof. This theorem is proved in most courses on multivariable real analysis. A proof is for example given in [Lee02, Corollary C.34]. | No |
Theorem 6.19. (Smooth Invariance of Domain Theorem) Let \( U \subset {\mathbb{R}}^{n} \) be an open set and let \( f : U \rightarrow {\mathbb{R}}^{n} \) be a smooth map such that the differential \( \mathrm{D}{f}_{P} \) is invertible at every point \( P \) in the open set \( U \) . Then the following two statements hol... | Proof. This proposition is a straightforward consequence of the Inverse Function Theorem 6.18. We leave it to the reader to fill in the few remaining details or alternatively we refer to [Lee02, Corollary C.36] for a proof. | No |
Proposition 6.21. The following topological spaces admit a smooth atlas:\n\n(1) open subsets of \( {\mathbb{R}}^{n} \),\n\n(2) all closed balls \( {\bar{B}}^{n} \),\n\n(3) all spheres \( {S}^{n} \subset {\mathbb{R}}^{n + 1} \),\n\n(4) the n-dimensional torus \( {\left( {S}^{1}\right) }^{n} \),\n\n(5) the surfaces of ge... | Proof (*). It is straightforward to verify that the atlases for examples (1) to (4) that we provided in Section 6.2 are smooth. It remains to consider the (non-orientable) surfaces of genus \( g \geq 2 \) . The atlas that we provided in the proof of Proposition 6.8 looks more dubious. In fact all transition maps are co... | Yes |
Lemma 6.23. Let \( f : M \rightarrow N \) be a map between two smooth manifolds.\n\n(1) If there exists an open cover \( {\left\{ {U}_{i}\right\} }_{i \in I} \) of \( M \) such that each restriction \( {\left. f\right| }_{{U}_{i}} : {U}_{i} \rightarrow N \) is smooth, then \( f \) itself is smooth.\n\n(2) If \( f : M \... | Proof.\n\n(1) This statement follows from the observation that smoothness is an open condition.\n\n(2) This statement is a straightforward consequence of Lemma 2.35 and the well-known fact that given an open set \( U \subset {\mathbb{R}}^{n} \), any smooth map \( U \rightarrow \overline{{\mathbb{R}}^{m}} \) is also con... | Yes |
Proposition 6.26. Let \( M \) be a non-empty \( m \) -dimensional smooth manifold and let \( N \) be an \( n \) -dimensional smooth manifold. If \( M \) and \( N \) are diffeomorphic, then \( m = n \) . | Proof. The proposition follows easily from Proposition 6.17 by considering charts. | No |
Proposition 6.27. Let \( X \) be an \( n \) -dimensional smooth manifold, i.e. \( X \) is an \( n \) -dimensional topological manifold equipped with a smooth atlas. The following statements hold:\n\n(1) Every point on \( X \) admits either a chart of type (i) from the given smooth atlas or it admits a chart of type (ii... | Proof (*). Let \( X \) be an \( n \) -dimensional topological manifold that is equipped with a smooth atlas \( {\left\{ {\Phi }_{i} : {U}_{i} \rightarrow {V}_{i}\right\} }_{i \in I} \) .\n\n(1) With a little bit of effort this statement can be deduced from Theorem 6.19. We refer to Lee02, Theorem 1.46] for the full det... | No |
Lemma 6.28. Let \( M \) be an \( n \) -dimensional smooth manifold and let \( X \) be an \( n \) -dimensional submanifold. If \( X \) is a closed subset of \( M \) (e.g. if \( X \) is a compact subset), then the following statements hold:\n\n(1) \( {\partial }_{0}X \) and \( {\partial }_{1}X \) are unions of components... | Proof (*). Let \( X \) be an \( n \) -dimensional submanifold of \( M \) that is a closed subset of \( M \) .\n\n(1) It follows easily from Proposition 6.27 (1) that \( {\partial }_{1}X = \partial X \cap \partial M \) is an open subset of \( \partial X \) . Since \( X \) is a closed subset of \( M \) it also follows fr... | Yes |
Proposition 6.30. (*) Let \( M \) be an n-dimensional smooth manifold and let \( W \) be an \( n \) -dimensional submanifold of \( M \) with \( W \cap \partial M = \varnothing \) . If \( W \) is compact, then the following two statements hold:\n\n(1) The smooth manifold \( M \) is decomposed into the submanifolds \( M ... | Proof \( \\left( *\\right) \) .\n\n(1) First note that it is a consequence of Lemma 2.17 and the fact that smooth manifolds are Hausdorff that \( W \) is a closed subset of \( M \) . The statement is now a fairly immediate consequence of Lemma 6.28. We leave it to the reader to fill in the details.\n\n(2) This statemen... | No |
Proposition 6.31. (*) Given any n-dimensional topological manifold \( M \) there exists a countable basis \( {\left\{ {U}_{i}\right\} }_{i \in \mathbb{N}} \) of the topology with the following properties:\n\n(1) Each \( {U}_{i} \) is open and each \( {U}_{i} \) is precompact.\n\n(2) Given any \( i \in \mathbb{N} \) the... | SKETCH OF PROOF (*).\n\n(1) If \( M \) is an open subset of \( {\mathbb{R}}^{n} \), then the statement follows immediately from Lemma 2.7, If \( M \) is an open subset of the upper half-space \( {H}_{n} \), then the statement follows from a modest variation on the argument of Lemma 2.7.\n\n(2) By Lemma 6.2 we know that... | No |
Lemma 6.33. (*) Let \( X \) be a topological space and let \( G \) be a group that acts freely, properly and continuously on \( X \) . If \( X \) is Hausdorff, then for every \( x \in X \) there exists an open neighborhood \( U \) such that \( {gU} \cap U = \varnothing \) for all \( g \neq e \) . | Proof of Lemma 6.33 (*) . Let \( X \) be a topological space that is Hausdorff. Furthermore let \( G \) be a group that acts freely, properly and continuously on \( X \) . Finally let \( x \in X \) . Since \( G \) acts freely we know that for every non-trivial \( g \in G \) we have \( {gx} \neq x \) . Since the action ... | No |
Lemma 6.34. (*) Let \( X \) be a topological space that is Hausdorff. Furthermore let \( G \) be a group that acts continuously on \( X \). We denote by \( p : X \rightarrow X/G \) the projection. If \( U \subset X \) is an open subset such that the map \( p : U \rightarrow X/G \) is injective, then the map \( p : U \r... | Proof of Lemma 6.34 (*) . By Lemma 3.30 the projection \( p : X \rightarrow X/G \) is continuous. Thus the restriction of \( p \) to \( U \) is also continuous. By our hypothesis the map \( p : U \rightarrow p\left( U\right) \) is injective and evidently it is surjective. Thus the map is a bijection. By Lemma 3.30 (2) ... | No |
Lemma 6.35. Let \( n \in \mathbb{N} \) . If \( A \in \mathrm{{GL}}\left( {n,\mathbb{Z}}\right) \) is a matrix, then the map\n\n\[ f\left( A\right) : {\mathbb{R}}^{n}/{\mathbb{Z}}^{n} \rightarrow {\mathbb{R}}^{n}/{\mathbb{Z}}^{n} \]\n\n\[ v \mapsto {Av} \]\n\n is a diffeomorphism. It is orientation-preserving if and onl... | Proof. We will provide the proof in Exercise 6.18. | No |
Proposition 6.36. Let \( M \) be an open subset of \( {\mathbb{R}}^{n} \) or of \( {H}_{n} \). For any \( P \in M \) the map\n\n\[ \n{\mathbb{R}}^{n} \rightarrow {\mathrm{T}}_{P}M \n\]\n\n\[ \nv \mapsto \left( \begin{aligned} {{C}^{\infty }\left( {M,\mathbb{R}}\right) } & \rightarrow \mathbb{R} \\ f & \mapsto {\left. \... | Proof. The proof of this proposition is an amusing exercise in real analysis. A detailed proof for the case that \( M \) is an open subset of \( {\mathbb{R}}^{n} \) is given in [Lee02, Proposition 3.2]. The case that \( M \) is an open subset of \( {H}_{n} \) follows fairly easily from the definition of a smooth map on... | No |
Proposition 6.37. Let \( M \) and \( N \) be two smooth manifolds.\n\n(1) If \( f : M \rightarrow N \) is a smooth map, then for any \( P \in M \) the map\n\n\[ \mathrm{D}{f}_{P} = {f}_{ * } : {\mathrm{T}}_{P}M \rightarrow {\mathrm{T}}_{f\left( P\right) }N \]\n\n\[ v \mapsto \left( \begin{aligned} {{C}^{\infty }\left( ... | Sketch of Proof. The first three statements follow fairly easily from the definitions. Using charts one can reduce the proof of Statement (4) to the case that \( M \) is an open subset of \( {\mathbb{R}}^{n} \) or \( {H}_{n} \) . But we had dealt with this case in Proposition 6.36. We refer to [Lee02, Proposition 3.10]... | No |
Lemma 6.38. (*) If \( W \subset {\mathbb{R}}^{n} \) is a \( k \) -dimensional vector space, then \( W \) is a \( k \) -dimensional submanifold of \( {\mathbb{R}}^{n} \) and for any \( P \in W \) we have \( {\mathrm{V}}_{P}W = W \) . | Proof (*). It follows from elementary linear algebra that \( W \) is a \( k \) -dimensional submani-fold of \( {\mathbb{R}}^{n} \) . We leave it as an elementary exercise to the reader to verify that for any \( P \in W \) we have \( {\mathrm{V}}_{P}W = W \) . | No |
Let \( M \) be an \( n \) -dimensional submanifold of \( {\mathbb{R}}^{k} \) and let \( P \in M \) . The map\n\n\[ \n{\Phi }_{M} : {\mathrm{V}}_{P}M \rightarrow {\mathrm{T}}_{P}M \n\]\n\n\[ \n{\gamma }^{\prime }\left( 0\right) \mapsto \left( \begin{aligned} {{C}^{\infty }\left( {M,\mathbb{R}}\right) \rightarrow \mathbb... | (2) First note that it follows almost immediately from the definitions that the given diagram does indeed commute. It remains to show that the horizontal maps are isomorphisms. Using charts it suffices to show this for open subsets of \( {\mathbb{R}}^{n} \) and of \( {H}_{n} \) . But in this special case that is an imm... | No |
Theorem 6.40. (Inverse Mapping Theorem) (*) Let \( f : M \rightarrow N \) be a smooth map between two smooth manifolds.\n\n(1) Let \( P \in M \) such that \( \mathrm{D}{f}_{P} \) is invertible.\n\n(a) If \( P \notin \partial M \), then \( f \) is a local diffeomorphism at \( P \) .\n\n(b) If \( P \in \partial M \) and ... | (1) (a) This can be reduced fairly easily to the Inverse Function Theorem 6.18 that deals with maps between open subsets of some \( {\mathbb{R}}^{n} \) . We refer to [Lee02, Theorem 4.5] for more details.\n\n(b) This statement can also be deduced from the Inverse Function Theorem 6.18 and the definition of a smooth map... | No |
Lemma 6.41. Let \( {V}_{0},\ldots ,{V}_{k} \) be oriented vector spaces and let \( {f}_{i} : {V}_{i} \rightarrow {V}_{i + 1}, i = 0,\ldots, k - 1 \) be isomorphisms. Then the map \( {f}_{k - 1} \circ \cdots \circ {f}_{0} : {V}_{0} \rightarrow {V}_{k} \) \( \Leftrightarrow \;\begin{matrix} \text{ the number of }{f}_{i}\... | Proof. The lemma is an elementary exercise in linear algebra that is left to the reader. | No |
Lemma 6.42. Let \( U, V \) and \( W \) be a finite-dimensional real vector spaces and suppose we are given maps \[ U \subset {}^{\varphi } \rightarrow V\xrightarrow[]{\psi }W \] such that \( \varphi \) is injective, \( \psi \) is surjective and \( \operatorname{im}\left( \varphi \right) = {\left. \ker \left( \psi \righ... | Sketch of PROOF. It is clear that the given vectors define a basis for \( V \) . Furthermore note that any other choice of a positive basis for \( U \), choice of positive basis for \( W \) and right-inverse \( W \rightarrow V \) leads to a base change matrix of the form \[ \left( \begin{matrix} A & X \\ 0 & B \end{mat... | No |
Lemma 6.43. Let \( M \) be a \( k \) -dimensional smooth manifold with \( k \geq 1 \) . Suppose that for each \( P \in M \) we are given an orientation of \( {\mathrm{T}}_{P}M \) . The following statements are equivalent:\n\n(1) The orientations of the tangent spaces defined an orientation \( M \) .\n\n(2) Given any \(... | Proof (*). The implication \( \left( 1\right) \Rightarrow \left( 2\right) \) is a tautology. The reverse implication \( \left( 2\right) \Rightarrow \left( 1\right) \) follows from the fact that a map that is locally constant, is actually constant. We leave it to the reader to fill in the details. | No |
Lemma 6.44. Let \( f : X \rightarrow Y \) be a diffeomorphism between two open subsets of \( {\mathbb{R}}^{n} \). Given \( P \in X \) we have \( {}^{116} \)\n\n\[ \n{f}_{ * } = \mathrm{D}{f}_{P} : {\mathrm{T}}_{P}X \rightarrow {\mathrm{T}}_{f\left( P\right) }Y\text{ is orientation-preserving }\; \Leftrightarrow \;\det ... | Proof. By Proposition 6.39 we have the following commutative diagram \n\nwhere the horizontal maps are isomorphisms and where the horizontal maps are orientation-preserving by definition of the orientations of the or... | Yes |
Lemma 6.45. (*) Let \( \\left( {M,\\mathcal{A}}\\right) \) be a smooth manifold. Suppose that all transition maps between charts in \( \\mathcal{A} \) are orientation-preserving. Given \( P \\in M \) we define\n\norientation of \( {\\mathrm{T}}_{P}M \\mathrel{\\text{:=}} {\\left( \\mathrm{D}{\\Phi }_{P}\\right) }^{-1} ... | Sketch of PROOF (*) . The first statement follows easily from the hypothesis that the transition maps are orientation-preserving. The second statement is a consequence of the observation in Footnote 114. | No |
Lemma 6.46. Let \( n \in \mathbb{N} \) and let \( M \) be an \( n \)-dimensional smooth manifold.\n\n(1) If two orientations on \( M \) agree at a point and if \( M \) is path-connected, then the two orientations agree everywhere. | (1) Suppose we are given two orientations on \( M \) . We denote by \( U \) the set of all points where the orientations agree and we denote by \( V \) the set of all points where the orientations disagree. It follows easily from the definitions that \( U \) and \( V \) are open. If the orientations agree at a point we... | No |
Proposition 6.47. Let \( G \) be a group that acts freely, properly and smoothly on an \( n \) -dimensional smooth manifold \( M \ ).\n\n(1) If \( M \) is oriented and if for each \( g \in G \) the map\n\n\[ M \rightarrow M \]\n\n\[ P \mapsto g \cdot P \]\n\n is orientation-preserving, then \( M/G \) admits an orientat... | Proof. We leave it to the reader to write down the fairly straightforward proof of the proposition. | No |
Lemma 6.48. The following smooth manifolds are non-orientable:\n\n(1) the Möbius band,\n\n(2) the projective plane \( {\mathbb{{RP}}}^{2} \) ,\n\n(3) the Klein bottle, and\n\n(4) any non-orientable surface of genus \( \geq 1 \) . | Proof. Let \( X = \mathbb{R} \times \left( {-1,1}\right) \) and \( G = \mathbb{Z} \) . Very similar to the definition on page 188 we consider the action\n\n\[ \mathbb{Z} \times \left( {\mathbb{R} \times \left( {-1,1}\right) }\right) \rightarrow \mathbb{R} \times \left( {-1,1}\right) \]\n\n\[ \left( {n,\left( {x, y}\rig... | Yes |
Lemma 6.50. Let \( n \geq 2 \) and let \( M \) be an oriented \( n \) -dimensional smooth manifold. Let \( P \in \partial M \) . We denote by \( w \in {\mathrm{T}}_{P}M \) a tangent vector which points outward in the sense of the definition on page 291. We say that a basis \( {v}_{1},\ldots ,{v}_{n - 1} \) for \( {\mat... | Proof. We leave the fairly elementary verification of the lemma to the reader. Alternatively full details are given in Lee02, Proposition 15.24]. | No |
Proposition 6.51. (*) Let \( M \) be a smooth manifold with possibly non-empty boundary and let \( N \) be a smooth manifold with empty boundary. Given \( Q \in N \) and \( P \in M \) we consider the inclusion maps\n\n\[ \n\\begin{aligned} {i}_{Q} : M & \\rightarrow M \\times N \\\\\nx & \\mapsto \\left( {x, Q}\\right)... | (1),(2) By Proposition 6.5 we already know that \( M \\times N \) is a topological manifold. Furthermore it is very easy to see that the products of charts from the smooth atlases for \( M \) and for \( N \) define a smooth atlas for \( M \\times N \) which has the properties specified in (2).\n\n(3) The elementary pro... | No |
Lemma 6.52. Let \( M \) be an \( m \) -dimensional smooth manifold, let \( X \subset M \) a proper sub-manifold of codimension \( k \) and let \( Y \) be a proper submanifold of \( M \) of codimension \( l \) .\n\n(1) If \( P \in M \) is a transverse intersection point of \( X \) and \( Y \), then there exists a chart ... | Proof.\n\n(1) The first statement is proved in [Kos93, Theorem IV.1.6] or implicitly also in [Lee02], Theorem 6.30 (b)]. | No |
Theorem 6.53. (Regular Value Theorem) Let \( M \) be an \( m \) -dimensional smooth manifold, let \( N \) be an \( n \) -dimensional smooth manifold without boundary, let \( f : M \rightarrow N \) be a smooth map and let \( s \in N \) be a regular value of \( f \). (1) The preimage \( X \mathrel{\text{:=}} {f}^{-1}\lef... | Sketch of PROOF. The first statement is proved in basically every book on smooth manifolds, see e.g. [Miln65a, p. 11] or [Lee02, Corollary 5.14]. The second statement is unfortunately not written down explicitly in the literature, but it can be proved without too much effort by modifying the proof of (1). The third sta... | No |
Lemma 6.54. Let \( f : M \rightarrow N \) be a smooth map between two smooth manifolds of the same dimension.\n\n(1) For any regular point \( x \in M \smallsetminus \partial M \) the map \( f \) is a local diffeomorphism around \( x \) . | (1) Let \( x \in M \smallsetminus \partial M \) be a regular value of \( f \) . This means that \( \mathrm{D}{f}_{x} : {\mathrm{T}}_{f\left( x\right) }M \rightarrow {\mathrm{T}}_{y}N \) is an epimorphism. Since \( M \) and \( N \) have the same dimension this implies that \( \mathrm{D}{f}_{x} \) is an isomorphism. It f... | Yes |
Lemma 6.55. Let \( n \in \mathbb{N} \) . The set\n\n\[ \mathrm{O}\left( n\right) = \left\{ {A \in \mathrm{M}\left( {n \times n,\mathbb{R}}\right) \mid {A}^{T}A = \mathrm{{id}}}\right\} \]\n\nof orthogonal matrices (and thus also its component \( \mathrm{{SO}}\left( n\right) \) ) is a \( \frac{1}{2}n\left( {n - 1}\right... | Proof. We start with a preamble: Throughout the proof we use Proposition 6.39 to identify the abstract tangent spaces \( {\mathrm{T}}_{P}M \) with the visual tangent spaces \( {\mathrm{V}}_{P}M \) . Now we proceed to the actual proof of the lemma. First we prove the desired statements for \( \mathrm{O}\left( n\right) \... | Yes |
Theorem 6.56. (Submersion Theorem) Let \( M \) be a compact \( m \) -dimensional smooth manifold, let \( C \) be an \( n \) -dimensional submanifold of \( {\mathbb{R}}^{n} \) and let \( f : M \rightarrow C \) be a surjective map. If \( C \) is a convex subset of \( {\mathbb{R}}^{n} \) and if \( f \) is a submersion, th... | Proof. This theorem follows from the proof of [Dun18, Theorem 8.5.10] or alternatively the proof of Ebe07, Theorem 4.1]. | No |
Lemma 6.61. Let \( M \) be a topological manifold. There exists a family of compact subsets \( {\left\{ {C}_{n}\right\} }_{n \in \mathbb{N}} \) of \( M \) and a family of open subsets \( {\left\{ {D}_{n + \frac{1}{2}}\right\} }_{n \in \mathbb{N}} \) of \( M \) such that the following two conditions are satisfied:\n\n(1... | Lemma 6.2 that there exists a family \( {\left\{ {A}_{n}\right\} }_{n \in \mathbb{N}} \) of compact subsets such that \( \bar{M} = \mathop{\bigcup }\limits_{{n \in \mathbb{N}}}{A}_{n} \) . It suffices to prove the following claim.\n\nClaim. There exists a sequence of compact subsets \( {C}_{1},{C}_{2},\ldots \) of \( M... | Yes |
Proposition 6.62. Let \( M \) be an \( m \) -dimensional smooth manifold.\n\n(1) (a) The union of countably many subsets of measure zero is again a subset of measure zero.\n\n(b) The intersection of countably many subsets of full measure is again a subset of full measure.\n\n(2) (a) Given any subset \( X \) of measure ... | Sketch of Proof.\n\n(1) Let \( {\left\{ {X}_{i}\right\} }_{i \in I} \) be a countable family of subsets of measure zero of \( {\mathbb{R}}^{n} \) . Using the fact that \( \mathop{\sum }\limits_{{i \in \mathbb{N}}}\frac{1}{{2}^{i}} = 1 \) it is straightforward to show that \( \mathop{\bigcup }\limits_{{i \in I}}{X}_{i} ... | No |
Theorem 6.63. (Sard’s Theorem) If \( f : M \rightarrow N \) is a smooth map between smooth manifolds (possibly with boundary) \( {}^{127} \), then the set of critical values of \( f \) is of measure zero in \( N \) . | Proof. The theorem is a straightforward consequence of [Lee02, Theorem 6.10]. | No |
Proposition 6.64. Let \( W \) be an \( n \) -dimensional smooth manifold with compact boundary.\n\nThere exists a sequence \( {X}_{1},{X}_{2},\ldots \) of n-dimensional smooth submanifolds of \( W \) with the following four properties:\n\n(1) The sequence is nested, i.e. for each \( i \in \mathbb{N} \) we have \( {X}_{... | Proof. Let \( W \) be an \( n \) -dimensional smooth manifold. If \( W \) is empty, then there is nothing to show, so we might as well assume that \( W \) is non-empty. We pick a point \( {w}_{0} \in W \) . By\n\n---\n\n\( {}^{127} \) We stress that \( M \) and \( N \) are allowed to have boundary, since (a) we need th... | No |
Lemma 7.3. Let \( M \) be a 1-dimensional topological manifold and let \( \Phi : U \rightarrow \left( {-2,2}\right) \) be a chart. Let \( \left\lbrack {a, b}\right\rbrack \subset \left( {-2, - 2}\right) \) be a compact interval. Then \( {\Phi }^{-1}\left( \left\lbrack {a, b}\right\rbrack \right) \) is a closed subset o... | Proof. Since \( \Phi \) is a homeomorphism the map \( {\Phi }^{-1} : \left( {-2,2}\right) \rightarrow U \) is also continuous. Since \( \left\lbrack {a, b}\right\rbrack \) is compact it follows from Lemma \( {2.40} \) that \( {\Phi }^{-1}\left( \left\lbrack {a, b}\right\rbrack \right) \) is also compact. But by | No |
Theorem 7.5. Let \( M \) be a non-empty connected 1-dimensional smooth manifold. Then the diffeomorphism type of \( M \) is given by the following table\n\n<table><thead><tr><th></th><th>without boundary</th><th>with non-empty boundary</th></tr></thead><tr><td>compact</td><td>the circle \( {S}^{1} \)</td><td>the closed... | Proof. It is not that difficult to modify the proof of Theorem 7.5 to prove the desired result. One just needs need to be a little more cautious with the various constructions of maps.\n\nFor compact smooth manifolds we will also provide a different proof, using \ | No |
Lemma 7.7. Let \( \left( {X, \leq }\right) \) be a totally ordered set. The topological space \( X \) that is given by the order topology is Hausdorff. | Proof. Let \( \left( {X, \leq }\right) \) be a totally ordered set. We start out with the following claim.\n\nClaim. Given any \( x \in X \) the sets\n\n\[ \left( {-\infty, x}\right) \mathrel{\text{:=}} \{ a \in X \mid a < x\} \;\text{ and }\;\left( {x,\infty }\right) \mathrel{\text{:=}} \{ b \in X \mid x < b\} \]\n\na... | Yes |
Lemma 7.9. The well-ordered set \( {\omega }_{1} \) does not have a maximal element. | Proof. Let \( x \in {\omega }_{1} \) . We need to show that \( x \) is not a maximal element. By one of the two defining properties of \( {\omega }_{1} \) we know that the section \( {S}_{x} = \{ y \in X \mid y < x\} \) is countable. It follows from Lemma 1.7 (4) that \( \{ y \in X \mid y \leq x\} = {S}_{x} \cup \{ x\}... | Yes |
Theorem 7.12. Let \( X \) be a topological space that is Hausdorff and path-connected.\n\n(1) If \( X \) is locally homeomorphic to an open subset of \( \mathbb{R} \), then \( X \) is homeomorphic to one of the following: \( {S}^{1},\mathbb{R},{\mathbb{L}}^{ + } \) and \( \mathbb{L} \) .\n\n(2) If \( X \) is locally ho... | Proof.\n\n(1) This statement is proved in proved in Knes58.\n\n(2) This statement can be deduced easily from (1) by first removing the \ | No |
Theorem 7.13. (Principle of Transfinite Induction) Let \( \\left( {X, \\leq }\\right) \) be a well-ordered set. If \( A \\subset X \) is an inductive subset of \( X \), then \( A = X \) . | Proof. Let \( A \\subset X \) be an inductive subset. We suppose that \( A \\neq X \) . Since \( X \) is well-ordered the set \( X \\smallsetminus A \) has a minimal element \( y \) . Since \( y \) is the minimal element of \( X \\smallsetminus A \) we see that for every \( x < y \) we have \( x \\in A \) . But this me... | Yes |
Lemma 7.14. Let \( \\left( {X, \\leq }\\right) \) be a well-ordered set and let \( x \\in X \) such that the corresponding section \( {S}_{x} \) is countable. Precisely one of the following three statements holds:\n\n(1) \( x \) is the minimal element of \( X \) .\n\n(2) There exists an immediate predecessor of \( x \)... | Proof. Let \( \\left( {X, \\leq }\\right) \) be a well-ordered set and let \( x \\in X \) such that the corresponding section \( {S}_{x} = \\{ w \\in X \\mid w < x\\} \) is countable. We only need to consider the case that \( x \\in X \) is not the minimal element. This means that the section \( {S}_{x} \) is non-empty... | Yes |
Let \( a < b < c \) be three elements of \( X \) . The interval \( \widehat{\lbrack a, c)} \) has the order type of \( \widehat{\lbrack 0,1)} \) if and only if each of the two intervals \( \lbrack a, b) \) and \( \lbrack b, c) \) has the order type of \( \lbrack 0,1) \) . | Let \( a < b < c \) be three elements of \( X \) . First suppose that there exist order-preserving bijections \( f : \lbrack a, b) \rightarrow \lbrack 0,1) \) and \( g : \lbrack b, c) \rightarrow \lbrack 0,1) \) . It is clear that the map\n\n\[ \lbrack a, c) \rightarrow \lbrack 0,1) \]\n\n\[ t \mapsto \left\{ \begin{ar... | Yes |
Lemma 7.16. As above we denote by \( * \) the minimal element of \( {\omega }_{1} \) and let \( x \in {\omega }_{1} \smallsetminus \{ * \} \) . (1) The half-open interval \( \lbrack \left( {*,0}\right) ,\left( {x,0}\right) ) \subset {\mathbb{L}}_{ \geq 0} \) has the order type of \( \lbrack 0,1) \) . (2) The open inter... | Proof. We consider the set \[ B \mathrel{\text{:=}} \left\{ {b \in {\omega }_{1}\smallsetminus \{ * \} \mid \text{ the set }\lbrack \left( {*,0}\right) ,\left( {b,0}\right) )\text{ has order type of }\lbrack 0,1)}\right\} . \] We want to show that \( B = {\omega }_{1} \smallsetminus \{ * \} \) . Note that it follows fr... | Yes |
Proposition 8.2. Let \( M \) and \( N \) be smooth manifolds of the same dimension and furthermore let \( \varphi : N \rightarrow M \smallsetminus \partial M \) be a map.\n\n(1) If \( \varphi : N \rightarrow M \smallsetminus \partial M \) is an immersion, then \( \varphi \left( {N \smallsetminus \partial N}\right) \) i... | (1) Let \( \varphi : N \rightarrow M \smallsetminus \partial M \) be an immersion. It is not difficult to deduce from Theorem 6.19 that \( \varphi \left( {N \smallsetminus \partial N}\right) \) is an open subset of \( M \) . We leave it to the reader to fill in the details. | No |
Lemma 8.3. Given \( \lambda \in \mathbb{R} \) we consider the following map\n\n\[ \n{\varphi }_{\lambda } : \mathbb{R} \rightarrow {\mathbb{R}}^{2}/{\mathbb{Z}}^{2} \n\]\n\n\[ \nt \mapsto \left\lbrack \left( {t,\lambda \cdot t}\right) \right\rbrack .\n\]\n\nThe following statements hold:\n\n(1) The map \( {\varphi }_{\... | Proof. Statements (1), (2) and (4) are amusing exercises. Statement (3) requires the simple number theoretic fact that given any \( \alpha \in \mathbb{R} \) and any \( N \in \mathbb{N} \) there exist \( n, m \in \mathbb{Z} \) with \( 1 \leq n \leq N \) and with \( \left| {{n\alpha } - m}\right| < \frac{1}{N}1 \) . We l... | No |
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