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Lemma 2.26 (Extension Lemma for Smooth Functions). Suppose \( M \) is a smooth manifold with or without boundary, \( A \subseteq M \) is a closed subset, and \( f : A \rightarrow {\mathbb{R}}^{k} \) is a smooth function. For any open subset \( U \) containing \( A \), there exists a smooth function \( \widetilde{f} : M...
Proof. For each \( p \in A \), choose a neighborhood \( {W}_{p} \) of \( p \) and a smooth function \( {\widetilde{f}}_{p} : {W}_{p} \rightarrow {\mathbb{R}}^{k} \) that agrees with \( f \) on \( {W}_{p} \cap A \) . Replacing \( {W}_{p} \) by \( {W}_{p} \cap U \), we may assume that \( {W}_{p} \subseteq U \) . The fami...
Yes
Proposition 2.28 (Existence of Smooth Exhaustion Functions). Every smooth manifold with or without boundary admits a smooth positive exhaustion function.
Proof. Let \( M \) be a smooth manifold with or without boundary, let \( {\left\{ {V}_{j}\right\} }_{j = 1}^{\infty } \) be any countable open cover of \( M \) by precompact open subsets, and let \( \left\{ {\psi }_{j}\right\} \) be a smooth partition of unity subordinate to this cover. Define \( f \in {C}^{\infty }\le...
Yes
Lemma 3.1 (Properties of Derivations). Suppose \( a \in {\mathbb{R}}^{n}, w \in {T}_{a}{\mathbb{R}}^{n} \), and \( f, g \in \) \( {C}^{\infty }\left( {\mathbb{R}}^{n}\right) \) . (a) If \( f \) is a constant function, then \( {wf} = 0 \) . (b) If \( f\left( a\right) = g\left( a\right) = 0 \), then \( w\left( {fg}\right...
Proof. It suffices to prove (a) for the constant function \( {f}_{1}\left( x\right) \equiv 1 \), for then \( f\left( x\right) \equiv c \) implies \( {wf} = w\left( {c{f}_{1}}\right) = {cw}{f}_{1} = 0 \) by linearity. For \( {f}_{1} \), the product rule gives \[ w{f}_{1} = w\left( {{f}_{1}{f}_{1}}\right) = {f}_{1}\left(...
Yes
Proposition 3.2. Let \( a \in {\mathbb{R}}^{n} \). (a) For each geometric tangent vector \( {v}_{a} \in {\mathbb{R}}_{a}^{n} \), the map \( {\left. {D}_{v}\right| }_{a} : {C}^{\infty }\left( {\mathbb{R}}^{n}\right) \rightarrow \mathbb{R} \) defined by (3.1) is a derivation at a. (b) The map \( {v}_{a} \mapsto {\left. {...
Proof. The fact that \( {\left. {D}_{v}\right| }_{a} \) is a derivation at \( a \) is an immediate consequence of the product rule (3.2).\n\nTo prove that the map \( {\left. {v}_{a} \mapsto {D}_{v}\right| }_{a} \) is an isomorphism, we note first that it is linear, as is easily checked. To see that it is injective, sup...
Yes
For any \( a \in {\mathbb{R}}^{n} \), the \( n \) derivations \[ {\left. \frac{\partial }{\partial {x}^{1}}\right| }_{a},\ldots ,{\left. \frac{\partial }{\partial {x}^{n}}\right| }_{a}\text{ defined by }{\left. \frac{\partial }{\partial {x}^{i}}\right| }_{a}f = \frac{\partial f}{\partial {x}^{i}}\left( a\right) \] form...
Proof. Apply the previous proposition and note that \( {\left. \partial /\partial {x}^{i}\right| }_{a} = {\left. {D}_{{e}_{i}}\right| }_{a} \).
No
Lemma 3.4 (Properties of Tangent Vectors on Manifolds). Suppose \( M \) is a smooth manifold with or without boundary, \( p \in M, v \in {T}_{p}M \), and \( f, g \in {C}^{\infty }\left( M\right) \) .\n\n(a) If \( f \) is a constant function, then \( {vf} = 0 \) .\n\n(b) If \( f\left( p\right) = g\left( p\right) = 0 \),...
Exercise 3.5. Prove Lemma 3.4.
No
Proposition 3.6 (Properties of Differentials). Let \( M, N \), and \( P \) be smooth manifolds with or without boundary, let \( F : M \rightarrow N \) and \( G : N \rightarrow P \) be smooth maps, and let \( p \in M \). (a) \( d{F}_{p} : {T}_{p}M \rightarrow {T}_{F\left( p\right) }N \) is linear. (b) \( d{\left( \dot{G...
## - Exercise 3.7. Prove Proposition 3.6.
No
Proposition 3.8. Let \( M \) be a smooth manifold with or without boundary, \( p \in M \) , and \( v \in {T}_{p}M \) . If \( f, g \in {C}^{\infty }\left( M\right) \) agree on some neighborhood of \( p \), then \( {vf} = {vg} \) .
Proof. Let \( h = f - g \), so that \( h \) is a smooth function that vanishes in a neighborhood of \( p \) . Let \( \psi \in {C}^{\infty }\left( M\right) \) be a smooth bump function that is identically equal to 1 on the support of \( h \) and is supported in \( M \smallsetminus \{ p\} \) . Because \( \psi \equiv 1 \)...
Yes
Proposition 3.9 (The Tangent Space to an Open Submanifold). Let \( M \) be a smooth manifold with or without boundary, let \( U \subseteq M \) be an open subset, and let \( \iota : U \hookrightarrow M \) be the inclusion map. For every \( p \in U \), the differential \( d{\iota }_{p} : {T}_{p}U \rightarrow \) \( {T}_{p...
Proof. To prove injectivity, suppose \( v \in {T}_{p}U \) and \( d{\iota }_{p}\left( v\right) = 0 \in {T}_{p}M \) . Let \( B \) be a neighborhood of \( p \) such that \( \bar{B} \subseteq U \) . If \( f \in {C}^{\infty }\left( U\right) \) is arbitrary, the extension lemma for smooth functions guarantees that there exis...
Yes
Proposition 3.10 (Dimension of the Tangent Space). If \( M \) is an \( n \) -dimensional smooth manifold, then for each \( p \in M \), the tangent space \( {T}_{p}M \) is an \( n \) -dimensional vector space.
Proof. Given \( p \in M \), let \( \left( {U,\varphi }\right) \) be a smooth coordinate chart containing \( p \) . Because \( \varphi \) is a diffeomorphism from \( U \) onto an open subset \( \widehat{U} \subseteq {\mathbb{R}}^{n} \), it follows from Proposition 3.6(d) that \( d{\varphi }_{p} \) is an isomorphism from...
Yes
Lemma 3.11. Let \( \iota : {\mathbb{H}}^{n} \hookrightarrow {\mathbb{R}}^{n} \) denote the inclusion map. For any \( a \in \partial {\mathbb{H}}^{n} \), the differential \( d{\iota }_{a} : {T}_{a}{\mathbb{H}}^{n} \rightarrow {T}_{a}{\mathbb{R}}^{n} \) is an isomorphism.
Proof. Suppose \( a \in \partial {\mathbb{H}}^{n} \) . To show that \( d{\iota }_{a} \) is injective, assume \( d{\iota }_{a}\left( v\right) = 0 \) . Suppose \( f : {\mathbb{H}}^{n} \rightarrow \mathbb{R} \) is smooth, and let \( \widetilde{f} \) be any extension of \( f \) to a smooth function defined on all of \( {\m...
Yes
Proposition 3.12 (Dimension of Tangent Spaces on a Manifold with Boundary). Suppose \( M \) is an \( n \) -dimensional smooth manifold with boundary. For each \( p \in M \) , \( {T}_{p}M \) is an \( n \) -dimensional vector space.
Proof. Let \( p \in M \) be arbitrary. If \( p \) is an interior point, then because Int \( M \) is an open submanifold of \( M \), Proposition 3.9 implies that \( {T}_{p}\left( {\operatorname{Int}M}\right) \cong {T}_{p}M \) . Since Int \( M \) is a smooth \( n \) -manifold without boundary, its tangent spaces all have...
Yes
Proposition 3.13 (The Tangent Space to a Vector Space). Suppose \( V \) is a finite-dimensional vector space with its standard smooth manifold structure. For each point \( a \in V \), the map \( v \mapsto {\left. {D}_{v}\right| }_{a} \) defined by (3.5) is a canonical isomorphism from \( V \) to \( {T}_{a}V \), such th...
Proof. Once we choose a basis for \( V \), we can use the same argument as in the proof of Proposition 3.2 to show that \( {\left. {D}_{v}\right| }_{a} \) is indeed a derivation at \( a \), and that the map \( v \mapsto {\left. {D}_{v}\right| }_{a} \) is an isomorphism.\n\nNow suppose \( L : V \rightarrow W \) is a lin...
Yes
Proposition 3.14 (The Tangent Space to a Product Manifold). Let \( {M}_{1},\ldots ,{M}_{k} \) be smooth manifolds, and for each \( j \), let \( {\pi }_{j} : {M}_{1} \times \cdots \times {M}_{k} \rightarrow {M}_{j} \) be the projection onto the \( {M}_{j} \) factor. For any point \( p = \left( {{p}_{1},\ldots ,{p}_{k}}\...
Proof. See Problem 3-2.
No
Proposition 3.15. Let \( M \) be a smooth \( n \) -manifold with or without boundary, and let \( p \in M \) . Then \( {T}_{p}M \) is an \( n \) -dimensional vector space, and for any smooth chart \( \left( {U,\left( {x}^{i}\right) }\right) \) containing \( p \), the coordinate vectors \( \partial /{\left. \partial {x}^...
Thus, a tangent vector \( v \in {T}_{p}M \) can be written uniquely as a linear combination\n\n\[ v = {\left. {v}^{i}\frac{\partial }{\partial {x}^{i}}\right| }_{p} \]\n\nwhere we use the summation convention as usual, with an upper index in the denominator being considered as a lower index, as explained on p. 52. The ...
Yes
The transition map between polar coordinates and standard coordinates in suitable open subsets of the plane is given by \( \left( {x, y}\right) = \left( {r\cos \theta, r\sin \theta }\right) \). Let \( p \) be the point in \( {\mathbb{R}}^{2} \) whose polar coordinate representation is \( \left( {r,\theta }\right) = \le...
Applying (3.11) to the coordinate vectors, we find\n\n\[ {\left. \frac{\partial }{\partial r}\right| }_{p} = {\left. \cos \left( \frac{\pi }{2}\right) \frac{\partial }{\partial x}\right| }_{p} + {\left. \sin \left( \frac{\pi }{2}\right) \frac{\partial }{\partial y}\right| }_{p} = {\left. \frac{\partial }{\partial y}\ri...
Yes
Proposition 3.20. If \( M \) is a smooth \( n \) -manifold with or without boundary, and \( M \) can be covered by a single smooth chart, then \( {TM} \) is diffeomorphic to \( M \times {\mathbb{R}}^{n} \) .
Proof. If \( \left( {U,\varphi }\right) \) is a global smooth chart for \( M \), then \( \varphi \) is, in particular, a diffeomorphism from \( U = M \) to an open subset \( \widehat{U} \subseteq {\mathbb{R}}^{n} \) or \( {\mathbb{H}}^{n} \) . The proof of the previous proposition showed that the natural coordinate cha...
Yes
Proposition 3.21. If \( F : M \rightarrow N \) is a smooth map, then its global differential \( {dF} : {TM} \rightarrow {TN} \) is a smooth map.
Proof. From the local expression (3.9) for \( d{F}_{p} \) in coordinates, it follows that \( {dF} \) has the following coordinate representation in terms of natural coordinates for \( {TM} \) and \( {TN} \) :\n\n\[ \n{dF}\left( {{x}^{1},\ldots ,{x}^{n},{v}^{1},\ldots ,{v}^{n}}\right) = \left( {{F}^{1}\left( x\right) ,\...
Yes
Proposition 3.23. Suppose \( M \) is a smooth manifold with or without boundary and \( p \in M \) . Every \( v \in {T}_{p}M \) is the velocity of some smooth curve in \( M \) .
Proof. First suppose that \( p \in \operatorname{Int}M \) (which includes the case \( \partial M = \varnothing \) ). Let \( \left( {U,\varphi }\right) \) be a smooth coordinate chart centered at \( p \), and write \( v = {\left. {v}^{i}\partial /\partial {x}^{i}\right| }_{p} \) in terms of the coordinate basis. For suf...
Yes
Proposition 3.24 (The Velocity of a Composite Curve). Let \( F : M \rightarrow N \) be a smooth map, and let \( \gamma : J \rightarrow M \) be a smooth curve. For any \( {t}_{0} \in J \), the velocity at \( t = {t}_{0} \) of the composite curve \( F \circ \gamma : J \rightarrow N \) is given by\n\n\[ \n{\left( F \circ ...
Proof. Just go back to the definition of the velocity of a curve:\n\n\[ \n{\left( F \circ \gamma \right) }^{\prime }\left( {t}_{0}\right) = d\left( {F \circ \gamma }\right) \left( {\left. \frac{d}{dt}\right| }_{{t}_{0}}\right) = {dF} \circ {d\gamma }\left( {\left. \frac{d}{dt}\right| }_{{t}_{0}}\right) = {dF}\left( {{\...
Yes
Proposition 4.1. Suppose \( F : M \rightarrow N \) is a smooth map and \( p \in M \) . If \( d{F}_{p} \) is surjective, then \( p \) has a neighborhood \( U \) such that \( {\left. F\right| }_{U} \) is a submersion. If \( d{F}_{p} \) is injective, then \( p \) has a neighborhood \( U \) such that \( {\left. F\right| }_...
Proof. If we choose any smooth coordinates for \( M \) near \( p \) and for \( N \) near \( F\left( p\right) \) , either hypothesis means that Jacobian matrix of \( F \) in coordinates has full rank at \( p \) . Example 1.28 shows that the set of \( m \times n \) matrices of full rank is an open subset of \( \mathrm{M}...
Yes
Theorem 4.5 (Inverse Function Theorem for Manifolds). Suppose \( M \) and \( N \) are smooth manifolds, and \( F : M \rightarrow N \) is a smooth map. If \( p \in M \) is a point such that \( d{F}_{p} \) is invertible, then there are connected neighborhoods \( {U}_{0} \) of \( p \) and \( {V}_{0} \) of \( F\left( p\rig...
Proof. The fact that \( d{F}_{p} \) is bijective implies that \( M \) and \( N \) have the same dimension, say \( n \) . Choose smooth charts \( \left( {U,\varphi }\right) \) centered at \( p \) and \( \left( {V,\psi }\right) \) centered at \( F\left( p\right) \) , with \( F\left( U\right) \subseteq V \) . Then \( \wid...
Yes
Proposition 4.8. Suppose \( M \) and \( N \) are smooth manifolds (without boundary), and \( F : M \rightarrow N \) is a map.\n\n(a) \( F \) is a local diffeomorphism if and only if it is both a smooth immersion and a smooth submersion.\n\n(b) If \( \dim M = \dim N \) and \( F \) is either a smooth immersion or a smoot...
Proof. Suppose first that \( F \) is a local diffeomorphism. Given \( p \in M \), there is a neighborhood \( U \) of \( p \) such that \( F \) is a diffeomorphism from \( U \) to \( F\left( U\right) \) . It then follows from Proposition 3.6(d) that \( d{F}_{p} : {T}_{p}M \rightarrow {T}_{F\left( p\right) }N \) is an is...
Yes
Theorem 4.12 (Rank Theorem). Suppose \( M \) and \( N \) are smooth manifolds of dimensions \( m \) and \( n \), respectively, and \( F : M \rightarrow N \) is a smooth map with constant rank \( r \) . For each \( p \in M \) there exist smooth charts \( \left( {U,\varphi }\right) \) for \( M \) centered at \( p \) and ...
Proof. Because the theorem is local, after choosing smooth coordinates we can replace \( M \) and \( N \) by open subsets \( U \subseteq {\mathbb{R}}^{m} \) and \( V \subseteq {\mathbb{R}}^{n} \) . The fact that \( {DF}\left( p\right) \) has rank \( r \) implies that its matrix has some \( r \times r \) submatrix with ...
Yes
Corollary 4.13. Let \( M \) and \( N \) be smooth manifolds, let \( F : M \rightarrow N \) be a smooth map, and suppose \( M \) is connected. Then the following are equivalent:\n\n(a) For each \( p \in M \) there exist smooth charts containing \( p \) and \( F\left( p\right) \) in which the coordinate representation of...
Proof. First suppose \( F \) has a linear coordinate representation in a neighborhood of each point. Since every linear map has constant rank, it follows that the rank of \( F \) is constant in a neighborhood of each point, and thus by connectedness it is constant on all of \( M \) . Conversely, if \( F \) has constant...
Yes
Theorem 4.14 (Global Rank Theorem). Let \( M \) and \( N \) be smooth manifolds, and suppose \( F : M \rightarrow N \) is a smooth map of constant rank.\n\n(a) If \( F \) is surjective, then it is a smooth submersion.\n\n(b) If \( F \) is injective, then it is a smooth immersion.\n\n(c) If \( F \) is bijective, then it...
Proof. Let \( m = \dim M, n = \dim N \), and suppose \( F \) has constant rank \( r \) . To prove (a), assume that \( F \) is not a smooth submersion, which means that \( r < n \) . By the rank theorem, for each \( p \in M \) there are smooth charts \( \left( {U,\varphi }\right) \) for \( M \) centered at \( p \) and \...
Yes
Theorem 4.15 (Local Immersion Theorem for Manifolds with Boundary). Suppose \( M \) is a smooth \( m \) -manifold with boundary, \( N \) is a smooth \( n \) -manifold, and \( F : M \rightarrow N \) is a smooth immersion. For any \( p \in \partial M \), there exist a smooth boundary chart \( \left( {U,\varphi }\right) \...
Proof. By choosing initial smooth charts for \( M \) and \( N \), we may assume that \( M \) and \( N \) are open subsets of \( {\mathbb{H}}^{m} \) and \( {\mathbb{R}}^{n} \), respectively, and also that \( p = 0 \in {\mathbb{H}}^{m} \) , and \( F\left( p\right) = 0 \in {\mathbb{R}}^{n} \) . By definition of smoothness...
Yes
Consider the curve \( \beta : \left( {-\pi ,\pi }\right) \rightarrow {\mathbb{R}}^{2} \) defined by\n\n\[ \beta \left( t\right) = \left( {\sin {2t},\sin t}\right) \text{.} \]\n\nIts image is a set that looks like a figure-eight in the plane (Fig. 4.3), sometimes called a lemniscate. (It is the locus of points \( \left(...
It is easy to see that \( \beta \) is an injective smooth immersion because \( {\beta }^{\prime }\left( t\right) \) never vanishes; but it is not a topological embedding, because its image is compact in the subspace topology, while its domain is not.
Yes
Example 4.20 (A Dense Curve on the Torus). Let \( {\mathbb{T}}^{2} = {\mathbb{S}}^{1} \times {\mathbb{S}}^{1} \subseteq {\mathbb{C}}^{2} \) denote the torus, and let \( \alpha \) be any irrational number. The map \( \gamma : \mathbb{R} \rightarrow {\mathbb{T}}^{2} \) given by\n\n\[ \gamma \left( t\right) = \left( {{e}^...
Consider the set \( \gamma \left( \mathbb{Z}\right) = \{ \gamma \left( n\right) : n \in \mathbb{Z}\} \) . It follows from Dirichlet’s approximation theorem (see below) that for every \( \varepsilon > 0 \), there are integers \( n, m \) such that \( \left| {{\alpha n} - m}\right| < \varepsilon \) . Using the fact that \...
Yes
Lemma 4.21 (Dirichlet’s Approximation Theorem). Given \( \alpha \in \mathbb{R} \) and any positive integer \( N \), there exist integers \( n, m \) with \( 1 \leq n \leq N \) such that \( \left| {{n\alpha } - m}\right| < 1/N \) .
Proof. For any real number \( x \), let \( f\left( x\right) = x - \lfloor x\rfloor \), where \( \lfloor x\rfloor \) is the greatest integer less than or equal to \( x \) . Since the \( N + 1 \) numbers \( \{ f\left( {i\alpha }\right) : i = 0,\ldots, N\} \) all lie in the interval \( \lbrack 0,1) \), by the pigeonhole p...
Yes
Proposition 4.22. Suppose \( M \) and \( N \) are smooth manifolds with or without boundary, and \( F : M \rightarrow N \) is an injective smooth immersion. If any of the following holds, then \( F \) is a smooth embedding.
Proof. If \( F \) is open or closed, then it is a topological embedding by Theorem A.38, so it is a smooth embedding. Either (b) or (c) implies that \( F \) is closed: if \( F \) is proper, then it is closed by Theorem A.57, and if \( M \) is compact, then \( F \) is closed by the closed map lemma. Finally, assume \( M...
Yes
Theorem 4.25 (Local Embedding Theorem). Suppose \( M \) and \( N \) are smooth manifolds with or without boundary, and \( F : M \rightarrow N \) is a smooth map. Then \( F \) is a smooth immersion if and only if every point in \( M \) has a neighborhood \( U \subseteq M \) such that \( {\left. F\right| }_{U} : U \right...
Proof. One direction is immediate: if every point has a neighborhood on which \( F \) is a smooth embedding, then \( F \) has full rank everywhere, so it is a smooth immersion.\n\nConversely, suppose \( F \) is a smooth immersion, and let \( p \in M \) . We show first that \( p \) has a neighborhood on which \( F \) is...
Yes
Theorem 4.26 (Local Section Theorem). Suppose \( M \) and \( N \) are smooth manifolds and \( \pi : M \rightarrow N \) is a smooth map. Then \( \pi \) is a smooth submersion if and only if every point of \( M \) is in the image of a smooth local section of \( \pi \) .
Proof. First suppose that \( \pi \) is a smooth submersion. Given \( p \in M \), let \( q = \) \( \pi \left( p\right) \in N \) . By the rank theorem, we can choose smooth coordinates \( \left( {{x}^{1},\ldots ,{x}^{m}}\right) \) centered at \( p \) and \( \left( {{y}^{1},\ldots ,{y}^{n}}\right) \) centered at \( q \) i...
Yes
Proposition 4.28 (Properties of Smooth Submersions). Let \( M \) and \( N \) be smooth manifolds, and suppose \( \pi : M \rightarrow N \) is a smooth submersion. Then \( \pi \) is an open map, and if it is surjective it is a quotient map.
Proof. Suppose \( W \) is an open subset of \( M \) and \( q \) is a point of \( \pi \left( W\right) \) . For any \( p \in W \) such that \( \pi \left( p\right) = q \), there is a neighborhood \( U \) of \( q \) on which there exists a smooth local section \( \sigma : U \rightarrow M \) of \( \pi \) satisfying \( \sigm...
Yes
Theorem 4.29 (Characteristic Property of Surjective Smooth Submersions). Suppose \( M \) and \( N \) are smooth manifolds, and \( \pi : M \rightarrow N \) is a surjective smooth submersion. For any smooth manifold \( P \) with or without boundary, a map \( F : N \rightarrow P \) is smooth if and only if \( F \circ \pi ...
Proof. If \( F \) is smooth, then \( F \circ \pi \) is smooth by composition. Conversely, suppose that \( F \circ \pi \) is smooth, and let \( q \in N \) be arbitrary. Since \( \pi \) is surjective, there is a point \( p \in {\pi }^{-1}\left( q\right) \), and then the local section theorem guarantees the existence of a...
Yes
Theorem 4.30 (Passing Smoothly to the Quotient). Suppose \( M \) and \( N \) are smooth manifolds and \( \pi : M \rightarrow N \) is a surjective smooth submersion. If \( P \) is a smooth manifold with or without boundary and \( F : M \rightarrow P \) is a smooth map that is constant on the fibers of \( \pi \), then th...
Proof. Because a surjective smooth submersion is a quotient map, Theorem A. 30 shows that there exists a unique continuous map \( \widetilde{F} : N \rightarrow P \) satisfying \( \widetilde{F} \circ \pi = F \). It is smooth by Theorem 4.29.
Yes
Proposition 4.36 (Local Section Theorem for Smooth Covering Maps). Suppose \( E \) and \( M \) are smooth manifolds with or without boundary, and \( \pi : E \rightarrow M \) is a smooth covering map. Given any evenly covered open subset \( U \subseteq M \), any \( q \in U \) , and any \( p \) in the fiber of \( \pi \) ...
Proof. Suppose \( U \subseteq M \) is evenly covered, \( q \in U \), and \( p \in {\pi }^{-1}\left( q\right) \) . Let \( {\widetilde{U}}_{0} \) be the component of \( {\pi }^{-1}\left( U\right) \) containing \( p \) . Since the restriction of \( \pi \) to \( {\widetilde{U}}_{0} \) is a diffeomorphism onto \( U \), the ...
Yes
Proposition 4.46. Suppose \( E \) and \( M \) are nonempty connected smooth manifolds with or without boundary. If \( \pi : E \rightarrow M \) is a proper local diffeomorphism, then \( \pi \) is a smooth covering map.
Proof. Because \( \pi \) is a local diffeomorphism, it is an open map, and because it is proper, it is a closed map (Theorem A.57). Thus \( \pi \left( E\right) \) is both open and closed in \( M \) . Since it is obviously nonempty, it is all of \( M \), so \( \pi \) is surjective.\n\n![eac462f6-ce8a-4286-98b4-9fe2c2939...
Yes
Proposition 5.1 (Open Submanifolds). Suppose \( M \) is a smooth manifold. The embedded submanifolds of codimension 0 in \( M \) are exactly the open submanifolds.
Proof. Suppose \( U \subseteq M \) is an open submanifold, and let \( \iota : U \hookrightarrow M \) be the inclusion map. Example 1.26 showed that \( U \) is a smooth manifold of the same dimension as \( M \), so it has codimension 0 . In terms of the smooth charts for \( U \) constructed in Example 1.26, \( \iota \) ...
Yes
Proposition 5.2 (Images of Embeddings as Submanifolds). Suppose \( M \) is a smooth manifold with or without boundary, \( N \) is a smooth manifold, and \( F : N \rightarrow M \) is a smooth embedding. Let \( S = F\left( N\right) \) . With the subspace topology, \( S \) is a topological manifold, and it has a unique sm...
Proof. If we give \( S \) the subspace topology that it inherits from \( M \), then the assumption that \( F \) is an embedding means that \( F \) can be considered as a homeomorphism from \( N \) onto \( S \), and thus \( S \) is a topological manifold. We give \( S \) a smooth structure by taking the smooth charts to...
Yes
Proposition 5.3 (Slices of Product Manifolds). Suppose \( M \) and \( N \) are smooth manifolds. For each \( p \in N \), the subset \( M \times \{ p\} \) (called a slice of the product manifold) is an embedded submanifold of \( M \times N \) diffeomorphic to \( M \) .
Proof. The set \( M \times \{ p\} \) is the image of the smooth embedding \( x \mapsto \left( {x, p}\right) \) .
No
Proposition 5.4 (Graphs as Submanifolds). Suppose \( M \) is a smooth m-manifold (without boundary), \( N \) is a smooth \( n \) -manifold with or without boundary, \( U \subseteq M \) is open, and \( f : U \rightarrow N \) is a smooth map. Let \( \Gamma \left( f\right) \subseteq M \times N \) denote the graph of \( f ...
Proof. Define a map \( {\gamma }_{f} : U \rightarrow M \times N \) by\n\n\[ {\gamma }_{f}\left( x\right) = \left( {x, f\left( x\right) }\right) . \]\n\n(5.2)\n\nIt is a smooth map whose image is \( \Gamma \left( f\right) \) . Because the projection \( {\pi }_{M} : M \times N \rightarrow M \) satisfies \( {\pi }_{M} \ci...
Yes
Proposition 5.5. Suppose \( M \) is a smooth manifold with or without boundary and \( S \subseteq M \) is an embedded submanifold. Then \( S \) is properly embedded if and only if it is a closed subset of \( M \) .
Proof. If \( S \) is properly embedded, then it is closed by Theorem A.57. Conversely, if \( S \) is closed in \( M \), then Proposition A.53(c) shows that the inclusion map \( S \hookrightarrow M \) is proper.
Yes
Corollary 5.6. Every compact embedded submanifold is properly embedded.
Proof. Compact subsets of Hausdorff spaces are closed.
No
Proposition 5.7 (Global Graphs Are Properly Embedded). Suppose \( M \) is a smooth manifold, \( N \) is a smooth manifold with or without boundary, and \( f : M \rightarrow N \) is a smooth map. With the smooth manifold structure of Proposition 5.4, \( \Gamma \left( f\right) \) is properly embedded in \( M \times N \) ...
Proof. In this case, the projection \( {\pi }_{M} : M \times N \rightarrow M \) is a smooth left inverse for the embedding \( {\gamma }_{f} : M \rightarrow M \times N \) defined by (5.2). Thus \( {\gamma }_{f} \) is proper by Proposition A.53.
Yes
Example 5.9 (Spheres as Submanifolds). For any \( n \geq 0,{\mathbb{S}}^{n} \) is an embedded sub-manifold of \( {\mathbb{R}}^{n + 1} \)
because it is locally the graph of a smooth function: as we showed in Example 1.4, the intersection of \( {\mathbb{S}}^{n} \) with the open subset \( \left\{ {x : {x}^{i} > 0}\right\} \) is the graph of the smooth function\n\n\[ \n{x}^{i} = f\left( {{x}^{1},\ldots ,{x}^{i - 1},{x}^{i + 1},\ldots ,{x}^{n + 1}}\right) , ...
Yes
Theorem 5.11. If \( M \) is a smooth \( n \) -manifold with boundary, then with the subspace topology, \( \partial M \) is a topological \( \left( {n - 1}\right) \) -dimensional manifold (without boundary), and has a smooth structure such that it is a properly embedded submanifold of \( M \) .
Proof. See Problem 5-2.
No
Theorem 5.12 (Constant-Rank Level Set Theorem). Let \( M \) and \( N \) be smooth manifolds, and let \( \Phi : M \rightarrow N \) be a smooth map with constant rank \( r \) . Each level set of \( \Phi \) is a properly embedded submanifold of codimension \( r \) in \( M \).
Proof. Write \( m = \dim M, n = \dim N \), and \( k = m - r \) . Let \( c \in N \) be arbitrary, and let \( S \) denote the level set \( {\Phi }^{-1}\left( c\right) \subseteq M \) . From the rank theorem, for each \( p \in S \) there are smooth charts \( \left( {U,\varphi }\right) \) centered at \( p \) and \( \left( {...
Yes
Corollary 5.13 (Submersion Level Set Theorem). If \( M \) and \( N \) are smooth manifolds and \( \Phi : M \rightarrow N \) is a smooth submersion, then each level set of \( \Phi \) is a properly embedded submanifold whose codimension is equal to the dimension of \( N \) .
Proof. Every smooth submersion has constant rank equal to the dimension of its codomain.
No
Corollary 5.14 (Regular Level Set Theorem). Every regular level set of a smooth map between smooth manifolds is a properly embedded submanifold whose codimension is equal to the dimension of the codomain.
Proof. Let \( \Phi : M \rightarrow N \) be a smooth map and let \( c \in N \) be a regular value. The set \( U \) of points \( p \in M \) where \( \operatorname{rank}d{\Phi }_{p} = \dim N \) is open in \( M \) by Proposition 4.1, and contains \( {\Phi }^{-1}\left( c\right) \) because of the assumption that \( c \) is a...
Yes
Now we can give a much easier proof that \( {\mathbb{S}}^{n} \) is an embedded submanifold of \( {\mathbb{R}}^{n + 1} \) . The sphere is a regular level set of the smooth function \( f : {\mathbb{R}}^{n + 1} \rightarrow \mathbb{R} \) given by \( f\left( x\right) = {\left| x\right| }^{2} \), since \( d{f}_{x}\left( v\ri...
The sphere is a regular level set of the smooth function \( f : {\mathbb{R}}^{n + 1} \rightarrow \mathbb{R} \) given by \( f\left( x\right) = {\left| x\right| }^{2} \), since \( d{f}_{x}\left( v\right) = 2\mathop{\sum }\limits_{i}{x}^{i}{v}^{i} \), which is surjective except at the origin.
No
Proposition 5.16. Let \( S \) be a subset of a smooth \( m \) -manifold \( M \). Then \( S \) is an embedded \( k \) -submanifold of \( M \) if and only if every point of \( S \) has a neighborhood \( U \) in \( M \) such that \( U \cap S \) is a level set of a smooth submersion \( \Phi : U \rightarrow {\mathbb{R}}^{m ...
Proof. First suppose \( S \) is an embedded \( k \) -submanifold. If \( \left( {{x}^{1},\ldots ,{x}^{m}}\right) \) are slice coordinates for \( S \) on an open subset \( U \subseteq M \), the map \( \Phi : U \rightarrow {\mathbb{R}}^{m - k} \) given in coordinates by \( \Phi \left( x\right) = \left( {{x}^{\widehat{k} +...
Yes
Let \( H \) be the half-plane \( \{ \left( {r, z}\right) : r > 0\} \) , and suppose \( C \subseteq H \) is an embedded 1-dimensional submanifold. The surface of revolution determined by \( C \) is the subset \( {S}_{C} \subseteq {\mathbb{R}}^{3} \) given by\n\n\[ \n{S}_{C} = \left\{ {\left( {x, y, z}\right) : \left( {\...
A computation shows that the Jacobian matrix of \( \Phi \) is\n\n\[ \n{D\Phi }\left( {x, y, z}\right) = \left( {\frac{x}{r}\frac{\partial \varphi }{\partial r}\left( {r, z}\right) \frac{y}{r}\frac{\partial \varphi }{\partial r}\left( {r, z}\right) \frac{\partial \varphi }{\partial z}\left( {r, z}\right) }\right) ,\n\]\...
Yes
Proposition 5.18 (Images of Immersions as Submanifolds). Suppose \( M \) is a smooth manifold with or without boundary, \( N \) is a smooth manifold, and \( F : N \rightarrow \) \( M \) is an injective smooth immersion. Let \( S = F\left( N\right) \) . Then \( S \) has a unique topology and smooth structure such that i...
Proof. The proof is very similar to that of Proposition 5.2, except that now we also have to define the topology on \( S \) . We give \( S \) a topology by declaring a set \( U \subseteq S \) to be open if and only if \( {F}^{-1}\left( U\right) \subseteq N \) is open, and then give it a smooth structure by taking the s...
Yes
Proposition 5.21. Suppose \( M \) is a smooth manifold with or without boundary, and \( S \subseteq M \) is an immersed submanifold. If any of the following holds, then \( S \) is embedded.\n\n(a) \( S \) has codimension 0 in \( M \) .\n\n(b) The inclusion map \( S \subseteq M \) is proper.\n\n(c) \( S \) is compact.
Proof. Problem 5-3.
No
Proposition 5.22 (Immersed Submanifolds Are Locally Embedded). If \( M \) is a smooth manifold with or without boundary, and \( S \subseteq M \) is an immersed submanifold, then for each \( p \in S \) there exists a neighborhood \( U \) of \( p \) in \( S \) that is an embedded submanifold of \( M \) .
Proof. Theorem 4.25 shows that each \( p \in S \) has a neighborhood \( U \) in \( S \) such that the inclusion \( {\left. \iota \right| }_{U} : U \hookrightarrow M \) is an embedding.
Yes
Example 5.25 (Graph Parametrizations). Suppose \( U \subseteq {\mathbb{R}}^{n} \) is an open subset and \( f : U \rightarrow {\mathbb{R}}^{k} \) is a smooth function. The map \( {\gamma }_{f} : U \rightarrow {\mathbb{R}}^{n} \times {\mathbb{R}}^{k} \) given by \( {\gamma }_{f}\left( u\right) = \left( {u, f\left( u\righ...
For example, the map \( F : {\mathbb{B}}^{2} \rightarrow {\mathbb{R}}^{3} \) given by\n\n\[ F\left( {u, v}\right) = \left( {u, v,\sqrt{1 - {u}^{2} - {v}^{2}}}\right) \]\n\nis a smooth local parametrization of \( {\mathbb{S}}^{2} \) whose image is the open upper hemisphere, and whose inverse is one of the graph coordina...
No
Theorem 5.27 (Restricting the Domain of a Smooth Map). If \( M \) and \( N \) are smooth manifolds with or without boundary, \( F : M \rightarrow N \) is a smooth map, and \( S \subseteq M \) is an immersed or embedded submanifold (Fig. 5.9), then \( {\left. F\right| }_{S} : S \rightarrow N \) is smooth.
Proof. The inclusion map \( \iota : S \hookrightarrow M \) is smooth by definition of an immersed sub-manifold. Since \( {\left. F\right| }_{S} = F \circ \iota \), the result follows.
Yes
Theorem 5.29 (Restricting the Codomain of a Smooth Map). Suppose \( M \) is a smooth manifold (without boundary), \( S \subseteq M \) is an immersed submanifold, and \( F : N \rightarrow M \) is a smooth map whose image is contained in \( S \) (Fig. 5.10). If \( F \) is continuous as a map from \( N \) to \( S \), then...
Proof. Let \( p \in N \) be arbitrary and let \( q = F\left( p\right) \in S \) . Proposition 5.22 guarantees that there is a neighborhood \( V \) of \( q \) in \( S \) such that \( {\left. \iota \right| }_{V} : V \hookrightarrow M \) is a smooth embedding. Thus there exists a smooth chart \( \left( {W,\psi }\right) \) ...
Yes
Corollary 5.30 (Embedded Case). Let \( M \) be a smooth manifold and \( S \subseteq M \) be an embedded submanifold. Then every smooth map \( F : N \rightarrow M \) whose image is contained in \( S \) is also smooth as a map from \( N \) to \( S \) .
Proof. Since \( S \subseteq M \) has the subspace topology, a continuous map \( F : N \rightarrow M \) whose image is contained in \( S \) is automatically continuous into \( S \), by the characteristic property of the subspace topology (Proposition A.17(a)).
No
Theorem 5.31. Suppose \( M \) is a smooth manifold and \( S \subseteq M \) is an embedded submanifold. The subspace topology on \( S \) and the smooth structure described in Theorem 5.8 are the only topology and smooth structure with respect to which \( S \) is an embedded or immersed submanifold.
Proof. Suppose \( S \subseteq M \) is an embedded \( k \) -dimensional submanifold. Theorem 5.8 shows that it satisfies the local \( k \) -slice condition, so it is an embedded submani-fold with the subspace topology and the smooth structure of Theorem 5.8. Suppose there were some other topology and smooth structure on...
Yes
Theorem 5.32. Suppose \( M \) is a smooth manifold and \( S \subseteq M \) is an immersed sub-manifold. For the given topology on \( S \), there is only one smooth structure making \( S \) into an immersed submanifold.
Proof. See Problem 5-14.
No
Theorem 5.33. If \( M \) is a smooth manifold and \( S \subseteq M \) is a weakly embedded sub-manifold, then \( S \) has only one topology and smooth structure with respect to which it is an immersed submanifold.
Proof. See Problem 5-16.
No
Lemma 5.34 (Extension Lemma for Functions on Submanifolds). Suppose \( M \) is a smooth manifold, \( S \subseteq M \) is a smooth submanifold, and \( f \in {C}^{\infty }\left( S\right) \) . (a) If \( S \) is embedded, then there exist a neighborhood \( U \) of \( S \) in \( M \) and a smooth function \( \widetilde{f} \...
Proof. Problem 5-17.
No
Proposition 5.37. Suppose \( M \) is a smooth manifold, \( S \subseteq M \) is an embedded sub-manifold, and \( p \in S \). As a subspace of \( {T}_{p}M \), the tangent space \( {T}_{p}S \) is characterized by\n\n\[ \n{T}_{p}S = \left\{ {v \in {T}_{p}M : {vf} = 0\text{ whenever }f \in {C}^{\infty }\left( M\right) \text...
Proof. First suppose \( v \in {T}_{p}S \subseteq {T}_{p}M \). This means, more precisely, that \( v = d{\iota }_{p}\left( w\right) \) for some \( w \in {T}_{p}S \), where \( \iota : S \rightarrow M \) is inclusion. If \( f \) is any smooth real-valued function on \( M \) that vanishes on \( S \), then \( f \circ \iota ...
Yes
Proposition 5.41. Suppose \( M \) is a smooth \( n \) -dimensional manifold with boundary, \( p \in \partial M \), and \( \left( {x}^{i}\right) \) are any smooth boundary coordinates defined on a neighborhood of \( p \) . The inward-pointing vectors in \( {T}_{p}M \) are precisely those with positive \( {x}^{n} \) -com...
## - Exercise 5.42. Prove Proposition 5.41.
No
Proposition 5.43. Every smooth manifold with boundary admits a boundary defining function.
Proof. Let \( \left\{ \left( {{U}_{\alpha },{\varphi }_{\alpha }}\right) \right\} \) be a collection of smooth charts whose domains cover \( M \) . For each \( \alpha \), define a smooth function \( {f}_{\alpha } : {U}_{\alpha } \rightarrow \lbrack 0,\infty ) \) as follows: if \( {U}_{\alpha } \) is an interior chart, ...
Yes
Consider the subset \( S = \{ \left( {x, y}\right) : y = \left| x\right| \} \subseteq {\mathbb{R}}^{2} \). It is easy to check that \( S \smallsetminus \{ \left( {0,0}\right) \} \) is an embedded 1-dimensional submanifold of \( {\mathbb{R}}^{2} \), so if \( S \) itself is a smooth submanifold at all, it must be 1-dimen...
Writing \( \gamma \left( t\right) = \left( {x\left( t\right), y\left( t\right) }\right) \), we see that \( y\left( t\right) \) takes a global minimum at \( t = 0 \), so \( {y}^{\prime }\left( 0\right) = 0 \). On the other hand, because every point \( \left( {x, y}\right) \in S \) satisfies \( {x}^{2} = {y}^{2} \), we h...
Yes
Proposition 5.46. Suppose \( M \) is a smooth manifold without boundary and \( D \subseteq M \) is a regular domain. The topological interior and boundary of \( D \) are equal to its manifold interior and boundary, respectively.
Proof. Suppose \( p \in D \) is arbitrary. If \( p \) is in the manifold boundary of \( D \), Theorem 4.15 shows that there exist a smooth boundary chart \( \left( {U,\varphi }\right) \) for \( D \) centered at \( p \) and a smooth chart \( \left( {V,\psi }\right) \) for \( M \) centered at \( p \) in which \( F \) has...
Yes
Proposition 5.47. Suppose \( M \) is a smooth manifold and \( f \in {C}^{\infty }\left( M\right) \). (a) For each regular value \( b \) of \( f \), the sublevel set \( {f}^{-1}(\left( {-\infty, b\rbrack }\right) \) is a regular domain in \( M \). (b) If \( a \) and \( b \) are two regular values of \( f \) with \( a < ...
Proof. Problem 5-21.
No
Theorem 5.48. If \( M \) is a smooth manifold and \( D \subseteq M \) is a regular domain, then there exists a defining function for \( D \) . If \( D \) is compact, then \( f \) can be taken to be a smooth exhaustion function for \( M \) .
Proof. Problem 5-22.
No
Lemma 6.2. Suppose \( A \subseteq {\mathbb{R}}^{n} \) is a compact subset whose intersection with \( \{ c\} \times {\mathbb{R}}^{n - 1} \) has \( \left( {n - 1}\right) \) -dimensional measure zero for every \( c \in \mathbb{R} \) . Then \( A \) has \( n \) -dimensional measure zero.
Proof. Choose an interval \( \left\lbrack {a, b}\right\rbrack \subseteq \mathbb{R} \) such that \( A \subseteq \left\lbrack {a, b}\right\rbrack \times {\mathbb{R}}^{n - 1} \) . For each \( c \in \) \( \left\lbrack {a, b}\right\rbrack \), let \( {A}_{c} \subseteq {\mathbb{R}}^{n - 1} \) denote the compact subset \( \lef...
Yes
Proposition 6.3. Suppose \( A \) is an open or closed subset of \( {\mathbb{R}}^{n - 1} \) or \( {\mathbb{H}}^{n - 1} \), and \( f : A \rightarrow \mathbb{R} \) is a continuous function. Then the graph of \( f \) has measure zero in \( {\mathbb{R}}^{n} \) .
Proof. First assume \( A \) is compact. We prove the theorem in this case by induction on \( n \) . When \( n = 1 \), it is trivial because the graph of \( f \) is at most a single point. To prove the inductive step, we appeal to Lemma 6.2. For each \( c \in \mathbb{R} \), the intersection of the graph of \( f \) with ...
Yes
Corollary 6.4. Every proper affine subspace of \( {\mathbb{R}}^{n} \) has measure zero in \( {\mathbb{R}}^{n} \) .
Proof. Let \( S \subseteq {\mathbb{R}}^{n} \) be a proper affine subspace. Suppose first that \( \dim S = n - 1 \) . Then there is at least one coordinate axis, say the \( {x}^{i} \) -axis, that is not parallel to \( S \), and in that case \( S \) is the graph of an affine function of the form \( {x}^{i} = F\left( {{x}...
Yes
Proposition 6.5. Suppose \( A \subseteq {\mathbb{R}}^{n} \) has measure zero and \( F : A \rightarrow {\mathbb{R}}^{n} \) is a smooth map. Then \( F\left( A\right) \) has measure zero.
Proof. By definition, for each \( p \in A, F \) has an extension to a smooth map, which we still denote by \( F \), on a neighborhood of \( p \) in \( {\mathbb{R}}^{n} \) . Shrinking this neighborhood if necessary, we may assume that there is an open ball \( U \) containing \( p \) such that \( F \) extends smoothly to...
Yes
Lemma 6.6. Let \( M \) be a smooth \( n \) -manifold with or without boundary and \( A \subseteq M \) . Suppose that for some collection \( \left\{ \left( {{U}_{\alpha },{\varphi }_{\alpha }}\right) \right\} \) of smooth charts whose domains cover \( A,{\varphi }_{\alpha }\left( {A \cap {U}_{\alpha }}\right) \) has mea...
Proof. Let \( \left( {V,\psi }\right) \) be an arbitrary smooth chart. We need to show that \( \psi \left( {A \cap V}\right) \) has measure zero. Some countable collection of the \( {U}_{\alpha } \) ’s covers \( A \cap V \) . For each such \( {U}_{\alpha } \), we have\n\n\[ \psi \left( {A \cap V \cap {U}_{\alpha }}\rig...
Yes
Proposition 6.8. Suppose \( M \) is a smooth manifold with or without boundary and \( A \subseteq M \) has measure zero in \( M \) . Then \( M \smallsetminus A \) is dense in \( M \) .
Proof. If \( M \smallsetminus A \) is not dense, then \( A \) contains a nonempty open subset of \( M \), which implies that there is a smooth chart \( \left( {V,\psi }\right) \) such that \( \psi \left( {A \cap V}\right) \) contains a nonempty open subset of \( {\mathbb{R}}^{n} \) (where \( n = \dim M \) ). Because \(...
Yes
Theorem 6.9. Suppose \( M \) and \( N \) are smooth \( n \) -manifolds with or without boundary, \( F : M \rightarrow N \) is a smooth map, and \( A \subseteq M \) is a subset of measure zero. Then \( F\left( A\right) \) has measure zero in \( N \) .
Proof. Let \( \left\{ \left( {{U}_{i},{\varphi }_{i}}\right) \right\} \) be a countable cover of \( M \) by smooth charts. We need to show that for each smooth chart \( \left( {V,\psi }\right) \) for \( N \), the set \( \psi \left( {F\left( A\right) \cap V}\right) \) has measure zero in \( {\mathbb{R}}^{n} \) . Note th...
Yes
Corollary 6.11. Suppose \( M \) and \( N \) are smooth manifolds with or without boundary, and \( F : M \rightarrow N \) is a smooth map. If \( \dim M < \dim N \), then \( F\left( M\right) \) has measure zero in \( N \) .
Proof. In this case, each point of \( M \) is a critical point for \( F \) .
No
Corollary 6.12. Suppose \( M \) is a smooth manifold with or without boundary, and \( S \subseteq M \) is an immersed submanifold with or without boundary. If \( \dim S < \dim M \) , then \( S \) has measure zero in \( M \) .
Proof. Apply Corollary 6.11 to the inclusion map \( S \hookrightarrow M \) .
No
Lemma 6.13. Suppose \( M \subseteq {\mathbb{R}}^{N} \) is a smooth \( n \) -dimensional submanifold with or without boundary. For any \( v \in {\mathbb{R}}^{N} \smallsetminus {\mathbb{R}}^{N - 1} \), let \( {\pi }_{v} : {\mathbb{R}}^{N} \rightarrow {\mathbb{R}}^{N - 1} \) be the projection with kernel \( \mathbb{R}v \)...
Proof. In order for \( {\left. {\pi }_{v}\right| }_{M} \) to be injective, it is necessary and sufficient that \( p - q \) never be parallel to \( v \) when \( p \) and \( q \) are distinct points in \( M \) . Similarly, in order for \( {\left. {\pi }_{v}\right| }_{M} \) to be a smooth immersion, it is necessary and su...
Yes
Corollary 6.17. Suppose \( M \) is a compact smooth \( n \) -manifold with or without boundary. If \( N \geq {2n} + 1 \), then every smooth map from \( M \) to \( {\mathbb{R}}^{N} \) can be uniformly approximated by embeddings.
Proof. Assume \( N \geq {2n} + 1 \), and let \( f : M \rightarrow {\mathbb{R}}^{N} \) be a smooth map. By the Whitney embedding theorem, there is a smooth embedding \( F : M \rightarrow {\mathbb{R}}^{{2n} + 1} \) . The map \( G = f \times F : M \rightarrow {\mathbb{R}}^{N} \times {\mathbb{R}}^{{2n} + 1} \) is also a sm...
Yes
Theorem 6.18 (Whitney Immersion Theorem). Every smooth n-manifold with or without boundary admits a smooth immersion into \( {\mathbb{R}}^{2n} \) .
Proof. See Problem 6-2 for the case \( \partial M = \varnothing \), and Problem 9-14 for the general case.
No
Corollary 6.22. If \( M \) is a smooth manifold with or without boundary and \( \delta : M \rightarrow \) \( \mathbb{R} \) is a positive continuous function, there is a smooth function \( e : M \rightarrow \mathbb{R} \) such that \( 0 < e\left( x\right) < \delta \left( x\right) \) for all \( x \in M \) .
Proof. Use the Whitney approximation theorem to construct a smooth function \( e : M \rightarrow \mathbb{R} \) that satisfies \( \left| {e\left( x\right) - \frac{1}{2}\delta \left( x\right) }\right| < \frac{1}{2}\delta \left( x\right) \) for all \( x \in M \) .
Yes
Theorem 6.23. If \( M \subseteq {\mathbb{R}}^{n} \) is an embedded \( m \) -dimensional submanifold, then \( {NM} \) is an embedded \( n \) -dimensional submanifold of \( T{\mathbb{R}}^{n} \approx {\mathbb{R}}^{n} \times {\mathbb{R}}^{n} \) .
Proof. Let \( {x}_{0} \) be any point of \( M \), and let \( \left( {U,\varphi }\right) \) be a slice chart for \( M \) in \( {\mathbb{R}}^{n} \) centered at \( {x}_{0} \) . Write \( \widehat{U} = \varphi \left( U\right) \subseteq {\mathbb{R}}^{n} \), and write the coordinate functions of \( \varphi \) as \( \left( {{u...
Yes
Proposition 6.25. Let \( M \subseteq {\mathbb{R}}^{n} \) be an embedded submanifold. If \( U \) is any tubular neighborhood of \( M \), there exists a smooth map \( r : U \rightarrow M \) that is both a retraction and a smooth submersion.
Proof. Let \( {NM} \subseteq T{\mathbb{R}}^{n} \) be the normal bundle of \( M \), and let \( {M}_{0} \subseteq {NM} \) be the set \( {M}_{0} = \{ \left( {x,0}\right) : x \in M\} \) . By definition of a tubular neighborhood, there is an open subset \( V \subseteq {NM} \) containing \( {M}_{0} \) such that \( E : V \rig...
Yes
Theorem 6.26 (Whitney Approximation Theorem). Suppose \( N \) is a smooth manifold with or without boundary, \( M \) is a smooth manifold (without boundary), and \( F : N \rightarrow M \) is a continuous map. Then \( F \) is homotopic to a smooth map. If \( F \) is already smooth on a closed subset \( A \subseteq N \),...
Proof. By the Whitney embedding theorem, we may as well assume that \( M \) is a properly embedded submanifold of \( {\mathbb{R}}^{n} \). Let \( U \) be a tubular neighborhood of \( M \) in \( {\mathbb{R}}^{n} \), and let \( r : U \rightarrow M \) be the smooth retraction given by Proposition 6.25. For any \( x \in M \...
Yes
Corollary 6.27 (Extension Lemma for Smooth Maps). Suppose \( N \) is a smooth manifold with or without boundary, \( M \) is a smooth manifold, \( A \subseteq N \) is a closed subset, and \( f : A \rightarrow M \) is a smooth map. Then \( f \) has a smooth extension to \( N \) if and only if it has a continuous extensio...
Proof. If \( F : N \rightarrow M \) is a continuous extension of \( f \) to all of \( N \), the Whitney approximation theorem guarantees the existence of a smooth map \( \widetilde{F} \) (homotopic to \( F \), in fact, though we do not need that here) that agrees with \( f \) on \( A \) ; in other words, \( \widetilde{...
No
Lemma 6.28. If \( N \) and \( M \) are smooth manifolds with or without boundary, smooth homotopy is an equivalence relation on the set of all smooth maps from \( N \) to \( M \) .
Proof. Reflexivity and symmetry are proved just as for ordinary homotopy. To prove transitivity, suppose \( F, G, K : N \rightarrow M \) are smooth maps, and \( {H}_{1},{H}_{2} : N \times I \rightarrow M \) are smooth homotopies from \( F \) to \( G \) and \( G \) to \( K \), respectively. Let \( \varphi : \left\lbrack...
Yes
Theorem 6.29. Suppose \( N \) is a smooth manifold with or without boundary, \( M \) is a smooth manifold, and \( F, G : N \rightarrow M \) are smooth maps. If \( F \) and \( G \) are homotopic, then they are smoothly homotopic. If \( F \) and \( G \) are homotopic relative to some closed subset \( A \subseteq N \), th...
Proof. Suppose \( F, G : N \rightarrow M \) are smooth, and let \( H : N \times I \rightarrow M \) be a homotopy from \( F \) to \( G \) (relative to \( A \), which may be empty). We wish to show that \( H \) can be replaced by a smooth homotopy.\n\nDefine \( \bar{H} : N \times \mathbb{R} \rightarrow M \) by\n\n\[ \bar...
Yes
Theorem 6.30. Suppose \( N \) and \( M \) are smooth manifolds and \( S \subseteq M \) is an embedded submanifold.\n\n(a) If \( F : N \rightarrow M \) is a smooth map that is transverse to \( S \), then \( {F}^{-1}\left( S\right) \) is an embedded submanifold of \( N \) whose codimension is equal to the codimension of ...
Proof. The second statement follows easily from the first, simply by taking \( F \) to be the inclusion map \( {S}^{\prime } \hookrightarrow M \), and noting that a composition of smooth embeddings \( S \cap {S}^{\prime } \hookrightarrow S \hookrightarrow M \) is again a smooth embedding.\n\nTo prove (a), let \( m \) d...
Yes
Theorem 6.32 (Global Characterization of Graphs). Suppose \( M \) and \( N \) are smooth manifolds and \( S \subseteq M \times N \) is an immersed submanifold. Let \( {\pi }_{M} \) and \( {\pi }_{N} \) denote the projections from \( M \times N \) onto \( M \) and \( N \), respectively. The following are equivalent.\n\n...
Proof. Problem 6-15.
No
Corollary 6.33 (Local Characterization of Graphs). Suppose \( M \) and \( N \) are smooth manifolds, \( S \subseteq M \times N \) is an immersed submanifold, and \( \left( {p, q}\right) \in S \) . If \( S \) intersects the submanifold \( \{ p\} \times N \) transversely at \( \left( {p, q}\right) \), then there exist a ...
Proof. The hypothesis guarantees that \( d{\left( {\pi }_{M}\right) }_{\left( p, q\right) } : {T}_{\left( p, q\right) }S \rightarrow {T}_{p}M \) is an isomorphism, so \( {\left. {\pi }_{M}\right| }_{S} \) restricts to a diffeomorphism from a neighborhood \( V \) of \( \left( {p, q}\right) \) in \( S \) to a neighborhoo...
Yes
Proposition 6.34. If \( \left\{ {{F}_{s} : s \in S}\right\} \) is a smooth family of maps from \( N \) to \( M \) and \( S \) is connected, then for any \( {s}_{1},{s}_{2} \in S \), the maps \( {F}_{{s}_{1}},{F}_{{s}_{2}} : N \rightarrow M \) are homotopic.
Proof. Because \( S \) is connected, it is path-connected. If \( \gamma : \left\lbrack {0,1}\right\rbrack \rightarrow S \) is any path from \( {s}_{1} \) to \( {s}_{2} \), then \( H\left( {x, s}\right) = F\left( {x,\gamma \left( s\right) }\right) \) is a homotopy from \( {F}_{{s}_{1}} \) to \( {F}_{{s}_{2}} \) .
Yes
Theorem 6.36 (Transversality Homotopy Theorem). Suppose \( M \) and \( N \) are smooth manifolds and \( X \subseteq M \) is an embedded submanifold. Every smooth map \( f : N \rightarrow M \) is homotopic to a smooth map \( g : N \rightarrow M \) that is transverse to \( X \) .
Proof. The crux of the proof is constructing a smooth map \( F : N \times S \rightarrow M \) that is transverse to \( X \), where \( S = {\mathbb{B}}^{k} \) for some \( k \) and \( {F}_{0} = f \) . It then follows from the parametric transversality theorem that there is some \( s \in S \) such that \( {F}_{s} : N \righ...
Yes
Theorem 7.5. Every Lie group homomorphism has constant rank.
Proof. Let \( F : G \rightarrow H \) be a Lie group homomorphism, and let \( e \) and \( \widetilde{e} \) denote the identity elements of \( G \) and \( H \), respectively. Suppose \( {g}_{0} \) is an arbitrary element of \( G \) . We will show that \( d{F}_{{g}_{0}} \) has the same rank as \( d{F}_{e} \) . The fact th...
Yes
Corollary 7.6. A Lie group homomorphism is a Lie group isomorphism if and only if it is bijective.
Proof. The global rank theorem shows that a bijective Lie group homomorphism is a diffeomorphism.
Yes
Theorem 7.9 (Uniqueness of the Universal Covering Group). For any connected Lie group \( G \), the universal covering group is unique in the following sense: if \( \widetilde{G} \) and \( {\widetilde{G}}^{\prime } \) are simply connected Lie groups that admit smooth covering maps \( \pi : \widetilde{G} \rightarrow G \)...
Proof. See Problem 7-5.
No
For each \( n \), the map \( {\varepsilon }^{n} : {\mathbb{R}}^{n} \rightarrow {\mathbb{T}}^{n} \) given by\n\n\[ \n{\varepsilon }^{n}\left( {{x}^{1},\ldots ,{x}^{n}}\right) = \left( {{e}^{{2\pi i}{x}^{1}},\ldots ,{e}^{{2\pi i}{x}^{n}}}\right) \n\]\n\nis a Lie group homomorphism and a smooth covering map (see Example 7...
Since \( {\mathbb{R}}^{n} \) is simply connected, this shows that the universal covering group of \( {\mathbb{T}}^{n} \) is the additive Lie group \( {\mathbb{R}}^{n} \).
No
Proposition 7.11. Let \( G \) be a Lie group, and suppose \( H \subseteq G \) is a subgroup that is also an embedded submanifold. Then \( H \) is a Lie subgroup.
Proof. We need only check that multiplication \( H \times H \rightarrow H \) and inversion \( H \rightarrow H \) are smooth maps. Because multiplication is a smooth map from \( G \times G \) into \( G \), its restriction is clearly smooth from \( H \times H \) into \( G \) (this is true even if \( H \) is merely immers...
Yes