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I'm trying to prove the following claim from Lee's Intro to Smooth Manifolds.

$\textbf{Claim:}$ For smooth manifolds M and N, show that $F:M \rightarrow N$ is smooth if and only if $F^*(C^{\infty}(N)) \subseteq C^{\infty}(M).$

The textbook defines a map $F:M \rightarrow N$ between manifolds to be smooth if for every $p \in M,$ there exist smooth charts $(U, \phi)$ containing p and $(V, \psi)$ containing $F(p)$ such that $F(U) \subseteq V$ and the composite map $\psi \circ F \circ \phi^{-1}$ is smooth in the sense of ordinary calculus (i.e. $\psi \circ F \circ \phi^{-1} \in C^{\infty}(\mathbb{R}^n)).$ The problem defines $C(M)$ as the algebra of continuous functions $f:M \rightarrow \mathbb{R}$, and for any continuous map $F:M \rightarrow N$, it defines $F^*:C(N) \rightarrow C(M)$ by $F^*(f)=f \circ F$.

One direction is easy: Suppose $F:M \rightarrow N$ is smooth, and take any $f \in C^{\infty}(N)$. Then $F^*(f)=f \circ F \in C(M)$ is smooth because it's the composition of smooth maps $f$ and $F$. Since $f \in C^{\infty}(N)$ was arbitrary, we get $F^*(C^{\infty}(N)) \subseteq C^{\infty}(M).$

But I have a question concerning the other direction, as I'm still trying to understand charts: Suppose $F^*(C^{\infty}(N)) \subseteq C^{\infty}(M).$ For any $p \in M,$ we can find a smooth chart $(U, \phi)$ for M with $p \in U$ simply because $M$ is a smooth manifold. Now since $F(p) \in N$, we can find a chart $(V, \psi)$ containing $F(p),$ but since I'm trying to show $F$ is smooth, I feel that my only option is to $choose$ a chart $(V, \psi)$ for $N$ such that $F(U) \subseteq V,$ and then go on to (try to) show that $\psi \circ F \circ \phi^{-1}$ is smooth.

$\textbf{Question}:$ What allows me to just select a chart $(V, \psi)$ on $N$ satisfying $F(U) \subseteq V$? Phrased differently, what guarantees that such a chart exists? Could I have also selected a chart $(V, \psi)$ such that $F(U)=V$? A related question: can I select any diffeomorphism defined on the set $V$ to be the diffeomorphism in my chart? The essence of my question is: does $any$ open set $V \subseteq N$ and $any$ diffeomorphism $\psi :V \rightarrow \mathbb{R^n}$ constitute a chart? If not, what allows me to choose such charts? For the n-sphere, for example, it seems like the standard charts are defined for specific open sets and specific diffeomorphisms, and that any open set (with diffeomorphism) does not constitute a smooth chart, so I'm slightly confused as to how we seem to be able to choose charts that satisfy conditions we need. Sorry for writing like 6 questions but they're all in essence the same; I just wanted to be clear as to exactly where my confusion lies. Any relevant insight is very much appreciated.

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  • $\begingroup$ If $(V,\psi)$ is a chart, then for any open subset $W\subseteq V$, you have that $(W,\psi|_W)$ is also a chart. Smoothness is a local property, so restricting to small open sets doesn't affect the compatibility of charts. So you're free to consider $U\cap F^{-1}(V)$ on $M$ as your neighborhood. Note also that since $\psi$ is only locally defined, it doesn't meet your hypothesis for $\psi\circ F$ to be smooth. However a bump function should fix that. $\endgroup$ – Matt Sep 27 '18 at 8:57
  • $\begingroup$ Thanks for the reply. You've definitely provided some clarity. Although helpful, I don't see why we are able to choose the chart $(V,\psi)$ for $N$ such that $F(U) \subseteq V$ based on the fact that, given a chart, we can restrict the chart's map to any open subset and get another chart. I don't see how we know that the image of $U$ will land in some domain $V$ of a chart $(V,\psi)$ for $N$. $\endgroup$ – innerproduct Sep 27 '18 at 22:34
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Suppose $F^*(C^\infty(N))\subseteq C^\infty(M)$, that is, for any $g\in C^\infty(N)$, we have that $g\circ F\in C^\infty(M)$.

Fix $p\in M$, and let $(\psi:W\subseteq N\to\tilde{W}\subseteq\mathbb{R}^n)$ be a coordinate chart about $F(p)$ in $N$. Let $(\phi:U\subseteq M\to\tilde{U}\subseteq\mathbb{R}^m)$ be a coordinate chart about $p$ in $M$. Let $V\subset W$ be some compactly contained neighborhood of $p$ in $W$, and let $\tilde{V}=\psi(V)$. Let $\theta$ be a bump function on $\overline{V}$ supported in $W$, and let $\tilde{\psi}=\theta\psi\in C^\infty(N)$.

Then $F^{-1}(V)$ is an open neighborhood of $p$ by continuity, and so $U\cap F^{-1}(V)$ is an open neighborhood of $p$ and is a subset of $U$. Hence $\phi|_{F^{-1}(V)\cap U}:F^{-1}(V)\cap U\to\phi(F^{-1}(V)\cap U)$ is still a chart map. So we need only show the following map is smooth $$\psi\circ F\circ\left(\left.\phi\right|_{F^{-1}(V)\cap U}\right)^{-1}:\phi(F^{-1}(V)\cap U)\to\tilde{V}.$$

However, since $\psi\circ F=\tilde{\psi}\circ F$ on $V$, and $\tilde{\psi}\circ F$ is smooth by assumption, the result follows.

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