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Let $X$, $Y$ be manifolds, with respective coordinate charts $\{(U_{\alpha}, \varphi_{\alpha})\}_{\alpha\in I}$ and $\{(V_{\beta}, \psi_{\beta})\}_{\beta\in J}$.

I want to show that $f:X\to Y$ is differentiable if and only if for every $C^{\infty}$ map $g:Y\to\mathbb{R}^{n}$, $g\circ f:X\to \mathbb{R}^{n}$ is $C^{\infty}$.

The $(\Rightarrow)$ direction I had no problem with.

For the $(\Leftarrow)$ direction, a hint in the text suggests using a bump function. But I am stuck as to where the bump function will be helpful.

If I suppose that for every $\alpha\in I$, $g\circ f\circ \varphi_{\alpha}$ is $C^{\infty}$ for all $C^{\infty}$ functions $g:Y\to\mathbb{R}^{n}$, then I can observe that for any $\beta\in J$, there is agreement between $g\circ f\circ \varphi_{\alpha}$ and $g\circ \psi_{\beta}\circ \psi_{\beta}^{-1}\circ f\circ \varphi_{\alpha}$ on the domain of the latter map.

I know that $g\circ f\circ \varphi_{\alpha}$ and $g\circ \psi_{\beta}$ are $C^{\infty}$, but this does not directly imply that $\psi_{\beta}^{-1}\circ f\circ \varphi_{\alpha}$ is $C^{\infty}$ which is what is desired.

Can anyone offer advice on how I can use a smooth bump function to fix this proof?

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Take a chart $\psi_\beta : \tilde{V_\beta} \rightarrow V_\beta$, where $\tilde{V}_\beta$ is an open subset of $\mathbb{R}^n$ and $V_\beta$ is an open subset of $Y$. Choose a smaller open set $V \subset Y$ such that $\bar{V}$ is compact and $\bar{V} \subset V_\beta$ and a bump function $\phi$ that equals to $1$ on $\bar{V}$ and has support in $V_\beta$. Consider the map $\phi \psi_{\beta}^{-1} : Y \rightarrow \mathbb{R}^n$. If $p \in V_\beta$, then there is a neighborhood of $p$ which is included in $V_\beta$. Hence, the product $\phi \psi_{\beta}^{-1}$ is well-defined in that neighborhood and smooth in that neighborhood as a product of smooth functions. If $p \notin V_\beta$, then there is a neighborhood of $p$ on which $\phi$ is identically $0$. Hence, the product is a smooth function in a neighborhood of each point and so a smooth function.

Because $\phi \psi_{\beta}^{-1}$ is smooth, you know what $(\phi \psi_{\beta}^{-1}) \circ f : X \rightarrow \mathbb{R}^n$ is smooth. Choose a coordinate system $(U_{\alpha}, \varphi_{\alpha})$ such that $f(U_\alpha) \subset V$. Representing $(\phi \psi_{\beta}^{-1}) \circ f$ in this coordinate system, you see that $$ (\phi \psi_{\beta}^{-1}) \circ f \circ \varphi_{\alpha} : \tilde{U}_{\alpha} \rightarrow \tilde{V}_{\beta}$$ is smooth. But since $\phi = 1$ on $V$, this means that $ \psi_{\beta}^{-1} \circ f \circ \varphi_\alpha $ is smooth.


Answer to the question in the comments:

When someone writes "take a chart" on a smooth manifold $(M, \mathcal{A})$, one means to take a chart from the atlas $\mathcal{A}$ that defines the differentiable structure, not some random homeomorphism from an open subset of $U \subset \mathbb{R}^n$ to $M$. Such chart $\psi : U \rightarrow M$ is neccesarily smooth as map between manifolds, where you consider the open subset $U$ as a smooth manifold with the natural structure induced by the identity chart.

Why? By definition a map is smooth if its representations in local coordinates are smooth. If you represent the map $\psi$ in the coordinates $\mathrm{id}$ on $U$ and $\psi$ on $M$, you get $\psi^{-1} \circ \psi \circ \mathrm{id} = id$ which is a smooth map. So the definition of what it means to be a smooth map between two manifolds makes the charts of the atlas $\mathcal{A}$, a priori only homeomorphisms, smooth maps.

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    Thank you. I think I got it worked out now.2012-10-28
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    I'm sorry, I thought I had it but I don't. I don't see why there is any reason to expect that $\phi\psi_{\beta}^{-1}:Y\to\mathbb{R}^{n}$ would be smooth?2012-10-29
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    I've added some details.2012-10-29
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    Thanks for the details but I still don't see why $\phi\psi_{\beta}^{-1}$ is smooth. You say it is a product of smooth functions but $\psi_{\beta}^{-1}$ is only a homeomorphism which need not be smooth, no?2012-10-29
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    I guess this sounds reasonable, but in the definition of differential manifold, we only insisted that the transition functions be smooth, not the coordinate charts themselves. This seems to me like we're adding an extra feature to the chart that is not necessarily included. But my Professor said basically to use essentially the same argument you gave so I'm not saying I don't believe you. Thanks for the helpful answers and sorry for being so picky.2012-10-30
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    In the definition, you can't insist that the charts $\psi : U \rightarrow M$ are smooth, because you haven't even defined what a smooth map to a manifold means! You first talk about compatible charts (that is, transition functions are smooth $\mathbb{R}^n \rightarrow \mathbb{R}^k$), then you define what it means that a map is smooth, then you get as a consequence of the definition that all those charts you used in the definition of the manifold are smooth maps (diffeomorphisms).2012-10-30
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    OK. That helps a lot. Thank you very much.2012-10-30
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    Actually, I gave this some more thought and it is crystal clear now. I took the image of the coordinate chart and defined it as a manifold of its own (with the identity map as a coordinate chart). Then it was easy to verify that the chart I started with was smooth based on the definition. ( I think this is what you were getting at above. ) This has been bothering me for a long time, now that I see it I can't wipe the smile off my face! Thanks again!2012-10-30