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I have seen a couple questions involving the composition of smooth maps is smooth, but the proof I came up with does not look quite like the proofs given and I want to know if there is anything wrong with my proof.

Let $X \subset R^k$, $Y \subset R^l$, and $Z \subset R^m$. Claim: If $f : X \to Y$ and $g : Y \to Z$ are smooth, then their composition $g \circ f$ is smooth.

Proof:

We have that $f$ is smooth. Therefore, for any $x \in X$, $\exists U \subset R^k$ that contains $x$ and there exists $F : U \to R^l$ that coincides with $f$ in $U \cap X$. Notice that $f(x) \in Y$. As $f(x) \in Y$ and $g$ is smooth we know that for any $x \in X$, $\exists W \subset R^l$ containing $f(x)$ and $\exists G : W \to R^m$ such that $G$ coincides with $g$ in $W \cap Y$. As a result, we can guarantee $G \circ F : U \to R^m$, therefore, $g \circ f$ must be smooth as there exists an appropriate open set $U$ and corresponding function $G \circ F$.

Dair
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    Your proof seems mostly correct with a few notes: You should likely mention that $F$ and $G$ are themselves smooth and that $U$ and $W$ are open (or at least neighborhoods of $x$ and $f(x)$, respectively. – Tom Sep 23 '16 at 00:02

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Your alleged proof does nothing. You just have replaced the given open sets $X\subset{\mathbb R}^k$ and $Y\subset{\mathbb R}^l$ by smaller sets, with no apparent reason or purpose.

You have not made clear what you mean by "smooth". If you just mean $C^1$ the proof is just a reference to the chain rule, see the answer in the linked question. If you mean $C^\infty$ you have to set up an induction proof showing that all partial derivatives of $g\circ f$ of all orders exist and are continuous if this is the case for $f$ and $g$.

Christian Blatter
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  • So, if I understand you correctly. First of all, I should mention that $F$ and $G$ are guaranteed too be smooth. Then, the part that I messed up was not showing $G \circ F$ is smooth as well. But by the chain rule, $d_{x_i}(G \circ F)$ is simply $dG_{f(x_i)} \circ dF_{x_i}$. And from this I can conclude $f \circ g$ is smooth. – Dair Sep 23 '16 at 16:42
  • Also, the definition of a smooth map is given in Milnor by: "More generally let $X \subset R^k$ and $Y \subset R^l$ be arbitrary subsets of euclidean spaces. A map $f : X \to Y$ is called smooth if for each $x \in X$ there exists an open set $U \subset R^k$ containing $x$ and a smooth function $F : U \to R^l$ that coincides with $f$ throughout $U \cap X$. – Dair Sep 23 '16 at 16:44
  • Oh, and also I think smooth in the context of Milnor refers to $C^1$. – Dair Sep 23 '16 at 16:47
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I think the answer to your question is already given in the answers to the couple questions you mention, it just doesn't appear integrated into a single answer, so I think it's convenient to write it here.

It is convenient to say that this problem corresponds to Exercise 1.3 (with a star) of Guillemin and Pollack's book, Differential Topology.

Let us therefore remember the definitions that Guillemin and Pollack offer in that book:

First we write the definition of smoothness in the classical sense, i.e., on open sets of $ \mathbb R ^ n $​ (Guillemin and Pollack do not use the term "classical" but it suits us to handle it like this for our explanation):

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

Then, we write the "extended" version (on arbitrary sets):

A map $f:X\to\mathbb R^m$ defined on an arbitrary subset $X$ in $\mathbb R^n$ 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 $U\subset\mathbb R^n$ and a smooth map [in the classical sense] $F:U\to\mathbb R^m$ such that $F$ equals $f$ on $U\cap X$​​. (Pages 1 and 2)

Let then $ X \subset \mathbb R ^ n $, $ Y \subset \mathbb R ^ m $, $ Z \subset R ^ k $ be arbitrary subsets and let $ f: X \to Y $ y $ g: Y \to Z $ smooth maps (extended sense). We want to show that the composition $ h = g \circ f: X \to Z $ is smooth (extended sense).

In the case of $ X $​, $ Y $​ and $ Z $​ open this is a typical calculus result that is explained here.

In the general case we proceed as follows:

Let $ x \in X $. Since $ f $ is smooth (extended sense), there exists an open $ U \subset \mathbb R ^ m $ and a function $ F: U \to \ R ^ n $​ smooth (classical sense) such that $$ x\in U \qquad \text y \qquad F|_{U\cap X}=f. $$ Let $ y = f (x) \in Y $. Since $ g $ is smooth (extended sense), there is an open $ V \subset \mathbb R ^ n $ and a function $ G: V \to \mathbb R ^ k $ smooth (classical sense) such that $$ y=f(x)\in V \qquad \text y \qquad G|_{V\cap Y}=g. $$ Of course we would like to take the composition $ G \circ F $​ as the smooth function (classical sense) that smoothly extends the function $ h $​, but it is not possible to do this for the reasons explained here.

So if we set $$ \widehat F=F|_{U\cap F^{-1}(V)}:U\cap F^{-1}(V)\to \mathbb R^n $$ then $ \widehat F $ is smooth (classical sense) and since $ U \cap X \cap F ^ {- 1} (V) \subset U \cap X $, it follows that $$ \widehat F|_{U\cap X}=F|_{U\cap X}=f. $$ We set $$ H=G\circ\widehat F:U\cap F^{-1}(V)\to\mathbb R^k. $$ Then $ H $ is smooth (classical sense) and if $ \xi \in U \cap X $, $f(\xi)\in V\cap Y$, and therefore $$ H(\xi)=G(\widehat F(\xi))=G(f(\xi))=g(f(\xi))=h(\xi). $$ That is $H|_{U\cap X}=h$ and $h$​ is smooth (extended sense).

Many other solutions to starred exercises from Guillemin and Pollack's book can be found here.

G al Cubo
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