# Proof of the composition of smooth functions is smooth.

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 $G \circ F$.

• 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, 2016 at 0:02

## 2 Answers

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 $g$.

• 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, 2016 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, 2016 at 16:44
• Oh, and also I think smooth in the context of Milnor refers to $C^1$.
– Dair
Sep 23, 2016 at 16:47

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.