# Commuting smooth maps

Suppose $f:A\to B$ and $g:C\to D$ are smooth embeddings, $h:B\to D$ is a smooth map, and $i:A\to C$ is a continuous map such that $g(i(x))=h(f(x))$.

Then, how to show that $i$ is smooth?

An embedding f is a smooth map such that the inclusion $X\to f(X)$ is a diffeomorphism, and f(X) is a submanifold.

A n-submanifold is a subset S of M, such that for every p in S, there is a chart $x:U\to V$, with p in U, such that $x(U\cap S)=V\cap R^n\times 0$

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Hints, comments, ideas? – TROLLKILLER Mar 27 '12 at 4:03
When a map is an embedding, you can, as a corollary to the local inversion theorem, find suitable charts in which the map has an easy to understand expression. Say $f:M\rightarrow N$ is smooth and an immersion at $p\in M$, then there are (centered) charts $X$ of $M$ in the neighbourhood of $p$ and $Y$ chart of $N$ in the neighborhood of $f(p)$ such that locally $Y\circ f\circ X^{-1}(x_1,\dots,x_m)=(x_1,\dots,x_m,0,\dots,0)$. Work in such a chart and you'll almost be done I think. – Olivier Bégassat Mar 27 '12 at 4:27