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.

  • $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.