Is a composition of maps as differentiable as a single map that does the same thing?

Question

Let $ U, S, M $ be open sets where U is an open set in $\mathbb {R^2}$ and S and M are open sets in $\mathbb {R^3}$

Define $ \alpha : \mathbb {R} \times \mathbb {R} \to \mathbb {R^3} $

Let $\alpha (U) = S $ where $\alpha$ is a $ C^{ \infty} $ Map/function.

Define $g: S \to M $ to be a diffeomorphism from S to M.

Now i want to get to M from U.

To start Define f to be a function from U to M and note that $g^{-1} $ exists and is $C^{k}$ where $k \geq 1$

I then want to claim the following:

$g^{-1} [f(\alpha ^{-1}$ (S))] = I(S) = $\alpha (U) $

Where I is the identity map; does this imply that the composition of $g^{-1} f \space \alpha ^{-1}$ is also $ C^{ \infty} $ ?

Answer 1

The claim $g^{-1} [f(\alpha ^{-1} (S))] = I(S)$ is not necessarily true, for example when $f$ is a constant function. If $f$ is such that your claim is true (so assuming $g^{-1} f \space \alpha ^{-1}$ is equal to identity), the composition is in $C^{\infty}$. This follows from the definition of the derivative.

Answer 2

By the chain rule for derivatives, the composition of smooth maps is a smooth map. Thus assuming that the maps $g^{-1},f$ and $\alpha^{-1}$ are well-defined and smooth, the map $g^{-1}f \alpha^{-1}$ is smooth.

However, note that the map $\alpha:U \rightarrow S$ cannot be surjective, because $U$ is two-dimensional and $S$ is three-dimensional.

More to the point, suppose that we have a situation \begin{equation*} U \overset{f}{\longrightarrow} V \overset{g}{\longrightarrow} W, \end{equation*} where $g \circ f = h: U \rightarrow W$. Then if $h$ is smooth, then $g \circ f$ is also smooth, however, this does not imply that either $f$ or $g$ is smooth. (It is even possible that neither $f$ nor $g$ is smooth, but their composition is. Exercise: Find an example...)