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From section 1, problem 3 of Differential Topology by Guillemin and Pollack:

Let $X \subset R^N, Y \subset R^M, Z \subset R^L$ be arbitrary subsets, and let $f : X \to Y, g : Y \to Z$ be smooth maps. Then [show that] the composite $g \circ f : X \to Z$ is smooth.

I don't really know where to go with this. The definition of smooth I'm working with is:

A mapping $f$ of an open set $U \subset R^n$ into $R^m$ is called smooth if it has continuous partial derivatives of all orders. (1)

A map $f : X \to R^m$ defined on an arbitrary subset X in $R^m$ is called smooth if it may be locally extended to a smooth map on open sets; that is, if around each point $x \in X$ there is an open set $U \subset R^n$ and a smooth map $F : U \to R^m$ such that $F$ equals $f$ on $U \cap X$. (2)

Using definition (2), take $F,G$ to be smooth local extensions of $f,g$ around arbitrary points $x \in X, y \in Y$ defined on open sets $U \subset R^N, V \subset R^M$ (respectively). Let $h = g \circ f$. I need to show that for arbitrary $x \in X$, there is a smooth (definition (1)) local extension $H$ of $h$ on an open set around $x$.

Take $H = G \circ F : U \to Z$. Given that $F,G$ are smooth (definition (1)), I now need to show that $H$ is smooth by definition (1), i.e., that it has "continuous partial derivatives of all orders". I'm not quite sure how to go about that.

Red
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    $H$ is defined on an open set as are $G,F$. Since $G,F$ are smooth, then so is $H$ (the chain rule shows that compositions are as smooth as their components). – copper.hat Feb 25 '13 at 07:57

2 Answers2

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First off, you will probably want to use a total differential as opposed to partial derivatives. Than, from the chain rule, we can actually find the derivative at any order:

Let $h = g\circ f$. Then $h'=f'g'(f)$. Differentiate this to get an idea for what $h^{(2)}$ is. Using induction to complete this constructive proof should be easy enough.

cderwin
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$H$ is smooth because the composition of differentiable functions is differentiable, and if the differential are continuous also the differential of the composition is continuous.

The point here is another. Are you sure that $F$ maps $U$ into $V$? This is not guaranteed. Indeed you should maybe restrict $U$ to get this condition satisfied.

Emanuele Paolini
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