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I want to prove the following:

A continuous function $f:M\to N$ is smooth if and only if $g\circ f:f^{-1}(V)\to\Bbb R$ are smooth for all $g\in C^\infty(N)$, where $V\subset N$ is open.

In our class, we defined differentiability as follows:

A continuous function $f:M\to N$ is smooth (or infinitely differentiable $\mathcal C^\infty$) in $p\in M$ if there exist charts $(U,\varphi)$ of $M$ (with $x_0\in U$) and $(V,\psi)$ of $N$ with $f(U)\subset V$ such that $\psi\circ f\circ \varphi^{-1}$ is differentiable $(\mathcal C^\infty)$ in $\varphi(x_0)$. $f$ is differentiable if it is in all $p\in M$.

I did the following, not sure if it is correct or not:

$(\Leftarrow)$ Suppose that $g\circ f$ are differentiable for all $g\in C^\infty(N)$. We can define the chart of $N$ $(V,g)$ for each $g$, and define a chart $(U,\varphi)$ from $M$. Since $\varphi$ is a homeomorphism, it is differentiable, so we can compose it to our preceding function and get that $g\circ f\circ \varphi^{-1}$ is differentiable. From the definition, $f$ is differentiable.

$(\Rightarrow)$ Suppose now that $f$ is differentiable. Then, by definition, take charts $(U,\varphi)$ from $M$ and $(V,\psi)$ from $N$ such that $\psi\circ f\circ\varphi^{-1}$ is differentiable. Now we see that $\varphi$ is invertible, since it is a homeomorphism. So $\varphi^{-1}$ exists and it is differentiable. Hence, $(\psi\circ f \circ \varphi^{-1})\circ\varphi = \psi\circ f$ is differentiable. Rename $\psi = g$, and we are done.

Is this correct? If not, how can I improve it?

Daniele Tampieri
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GreekCorpse
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    Your argument is difficult to follow, and doesn't make sense at points (e.g. homeomorphisms are not necessarily differentiable). Unless you're invoking the definition of a smooth map $M\to N$, or an element of $C^\infty M$, you should only be referring to the smoothness of maps $\mathbb{R}^a\to\mathbb{R}^b$. – Kajelad Sep 27 '21 at 01:04
  • I've looked up to see what smoothness is in relationship to functions, and that's basically what differentiable is for us (differentiable of class $\mathcal C^\infty$). So when we suppose $f$ is differentiable, we suppose $f$ is smooth. Sorry for the lack of info earlier. I'll edit the question (IDK if it would change anything in my proof). – GreekCorpse Sep 27 '21 at 19:26
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    both arguments are not right, you need to explain why the fact that $g\circ f\circ \varphi^{-1}$ is smooth for any $g$ imply that $f$ is smooth (hint: an smotth function on an Euclidean space is smooth if and only if each canonical projection is smooth). Now, in the second part you cannot set $g=\psi $ because there are smooth fucntions that are not diffeomorphisms (moreover, if $\dim N\neq 1$ then any smooth function $g\in C^{\infty }(N)$ cannot be a diffeomorphism) – Masacroso Sep 27 '21 at 19:32
  • Wait, I just noticed that $g$ is also smooth. So the composition of two smooth functions is also smooth. Is this alright? – GreekCorpse Sep 27 '21 at 19:52
  • @GreekCorpse It's not clear what you're asking. The composition of smooth maps is smooth (you should be able to prove this), but adding that detail doesn't fix either of your arguments. – Kajelad Sep 28 '21 at 03:58
  • It's not to add anything to my proof, but to prove the implication to the right altogether (since we suppose that $f$ and $g$ are smooth). I'll try and write what i've thought of the other implication in an answer. – GreekCorpse Sep 28 '21 at 12:36

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$(\Rightarrow)$ This one is clear, since both $f$ and $g$ smooth functions imply that $g\circ f$ is smooth.

$(\Leftarrow)$ I've looked at the answer of this question, but couldn't we just pick a chart $(N,\psi)$ of $N$, and since $\psi\in\mathcal C^\infty(N)$, $\psi\circ f$ is smooth? So if we take a chart $(U,\varphi)$ of $M$, we can deduce that $\varphi^{-1}$ is smooth and the composition $\psi\circ f\circ\varphi$ is smooth, so $f$ is smooth as well.

GreekCorpse
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