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 '16 at 0:02

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 '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