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I'm looking at Exercise 2.6b (p.58), which is to prove that for smooth manifolds $M$ and $N$ and a continuous map $F:M\rightarrow N$, we have that $F$ is smooth iff $F^*(C^\infty(N))\subseteq C^\infty(M)$, where $F^*(f)=f\circ F$ is the pullback by $F$. The forwards direction is obvious, but I'm stuck on the backwards direction (i.e. that $F^*(C^\infty(N))\subseteq C^\infty(M)$ implies $F$ is smooth). It seems clear that the right approach should something like "the component functions of each chart on $N$ are smooth functions to $\mathbb{R}$, hence so are their pullbacks, and a function to $\mathbb{R}^n$ is smooth iff its component functions are smooth." But I'm having trouble making that work, because the component functions of a chart $\psi:V\rightarrow\mathbb{R}^n$ of $N$ are, of course, functions from $V$ to $\mathbb{R}$, and I don't see a way of showing that $F^*(C^\infty(N))\subseteq C^\infty(M)$ implies $(F|_{F^{-1}(V)})^*(C^\infty(V))\subseteq C^\infty(F^{-1}(V))$ for an arbitrary open $V\subset N$. If I could get that to work, then because $F$ being smooth is a property that can be checked locally (i.e., $F$ is smooth iff for any $p\in M$, there is an open $U\subset M$, $p\in U$ such that $F|_U$ is smooth) I think I would be done.

Any hints in the right direction?

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I'm confused, $C^\infty(F^{-1}(V))$ are functions on a subset of $M$, so you can't pull these back via $F$. Maybe you want to show $C^\infty(F^{-1}(V)) \subset F^* C^\infty(V)$? –  Eric O. Korman Mar 30 '11 at 3:00
    
Whoops, corrected. Thanks! –  Zev Chonoles Mar 30 '11 at 3:01
    
By using a bump function, you can extend any smooth function defined on only a subset to a global smooth function. I think this may help. –  Eric O. Korman Mar 30 '11 at 3:03
    
I had considered that, but the theorems Lee proves about bump functions only seem to let us extend functions from closed subsets of the manifold. –  Zev Chonoles Mar 30 '11 at 3:06
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But you're working locally so take your coordinate neighborhood. Then take say the compact ball. Then you have a bump function which is identically 1 on the compact ball. But now look at the interior of the ball which is also homeomorphic to $\mathbb R^n$ and so gives you another chart in your atlas. So you can always take a chart contained in a compact set and then extend any function on that chart using a bump function. –  Eric O. Korman Mar 30 '11 at 3:11
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up vote 3 down vote accepted

Just to add closure to the question and a little more detail...

You have the right idea. Let $p\in U \subset M, F(p) \in V \subset N$ be coordinate neighborhoods with $F(U) \subset V$ and coordinate functions $\phi: U \to \mathbb R^n, \psi:V \to \mathbb R^m$. Then we need to show that $\psi \circ F \circ \phi^{-1}$ is smooth at $\phi(p)$. As you say, we want to look at the component functions so we want to show $\psi^j \circ F \circ \phi^{-1}$ is a smooth real valued function. Now we can find $V' \subset K \subset V$ with $K$ compact and $V'$ open and containing $F(p)$. Let $f$ be a bump function supported in $V$ and identically 1 on $K$. Then $f\psi$ extends by zero to a smooth function on all of $N$. Thus $f\psi \circ F$ is a smooth function on $M$ by assumption. Since $f\psi$ agrees with $\psi$ on a neighborhood of $F(p)$ and smoothness is a local property, we get that $\psi \circ F \circ \phi^{-1}$ is smooth at $\phi(p)$.

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