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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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1 Answer 1

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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. –  Kyle Schlitt Oct 28 '12 at 0:05
    
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? –  Kyle Schlitt Oct 29 '12 at 2:50
1  
I've added some details. –  levap Oct 29 '12 at 10:21
    
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? –  Kyle Schlitt Oct 29 '12 at 20:13
1  
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). –  levap Oct 30 '12 at 16:35

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