Let $X$ be a set.
Let $\mathscr{A}$ and $\mathscr{B}$ be $C^k$-atlases on $X$. Let $\tau_1,\tau_2$ be the canonical topologies on $X$ determined by $\mathscr{A}$ and $\mathscr{B}$ respectively.
I'm curious on the relation between [equivalence of $C^k$-atlases] $\leftrightarrow$ [canonical topologies determined by $C^k$-atlases]
It is true that if $\mathscr{A}$ and $\mathscr{B}$ are equivalent, then $\tau_1=\tau_2$.
What about the converse?
Since topology does not say anything about differentiability, the converse would be false. But what is a countetexample for this?
How about this? Take $X = \mathbb R$. My first atlas has a single chart $\varphi_1 : X \to \mathbb R$ with $\varphi_1(t) = t$. My second atlas has a single chart $\varphi_2: X \to \mathbb R$ with $\varphi_2(t) = t^3$. The topologies induced by both atlases are the same, since $\varphi_1 \circ \varphi_2^{-1}$ is a homeomorphism, which means that $\varphi_1(V)$ is open iff $\varphi_2(V)$ is open for any $V \subset X$. But the two atlases are not compatible, since $\varphi_1 \circ \varphi_2^{-1}$ is not smooth at the origin.