I'm trying to do below exercise from this lecture note, i.e.,
Ex 3.1. Given any two smooth atlases $\mathcal A_1, \mathcal A_2$ on a topological space $M$, we have $\mathcal A_1$ is smoothly equivalent to $\mathcal A_2$ if and only if $\mathcal C^\infty(M, \mathcal A_1) = \mathcal C^\infty(M, \mathcal A_2)$.
I have found a proof here that uses an extension theorem. My proof does not use this theorem and is thus simpler.
Could you check if I made some subtle mistake?
My attempt The direction "$\implies$" is straightforward. Let's prove the reverse. Let $\mathcal C^\infty(M, \mathcal A_1) = \mathcal C^\infty(M, \mathcal A_2)$. We will show that $\mathcal A_1$ is smoothly equivalent to $\mathcal A_2$. Fix charts $f:U \to \mathbb R^m$ in $\mathcal A_1$ and $g:V \to \mathbb R^m$ in $\mathcal A_2$. Let $f = (f_1, \ldots, f_m)$ and $N := U \cap V$. Then $N$ is open in $M$. It suffices to prove that $$ f_1 \circ g^{-1} : g(N) \to f(N) $$ is smooth in the usual sense.
Let $\mathcal A_1' := \{h |_{N} : h \in \mathcal A_1\}$ and $\mathcal A_2' := \{h |_{N} : h \in \mathcal A_2\}$. Then $\mathcal A_1', \mathcal A_2'$ are two smooth atlases on the topological space $N$ (endowed with subspace topology). It follows from $\mathcal A_1$ is smoothly equivalent to $\mathcal A_2$ that $\mathcal A'_1$ is smoothly equivalent to $\mathcal A'_2$. By direction "$\implies$", we get $\mathcal C^\infty(N, \mathcal A'_1) = \mathcal C^\infty(N, \mathcal A'_2)$.
Clearly, $f_1 |_N \in \mathcal C^\infty(N, \mathcal A'_1)$. So $f_1 |_N \in \mathcal C^\infty(N, \mathcal A'_2)$. On the other hand, $g|_N \in \mathcal A'_2$. It follows that $f_1 \circ g^{-1} |_{g(N)}$ is smooth. This completes the proof.
