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This is a continuation of the problem : Composition of smooth maps.

At the moment, I am on the same problem.

I am not quite sure of the continuation of the comment '' The point here is another. Are you sure that $F$ maps $U$ into $V$? This is not guaranteed. Indeed you should maybe restrict $U$ to get this condition satisfied.''

I thought I could proceed using $Y$, but it is not necessarily an open set of $\mathbb{R}^m$ or for the image of f(i.e., $f(U)$), so that $f(U)$ is an open set, it would have to the inverse of $f$, denoted $f^{-1}$, as a continuous function. They do not give informations about it.

What should I do to complete this problem?

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In general, if you just take arbitrary open sets $U$ and $V$, you are right that we might not get $F(U)\subset V$. But this isn't an issue, since we can simply replace $U$ by $U' =U\cap F^{-1}(V)$.

Since $V$ open and $F$ smooth, this is open, and since both $U$ and $F^{-1}(V)$ contain $x\in X$ (remember $f(x) = F(x) =y\in V$, we see that it is an open set containing $x\in X$.

Now with our new open set, there isn't an issue as we get $F(U')\subset V$.

Spencer Leslie
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