What is the characteristic property of surjective submersions?

Here's what I had in mind:

Theorem: Suppose $M$ and $N$ are smooth manifolds and $\pi:M\to N$ is a surjective smooth submersion. Then the given topology and smooth structure on $N$ are the only ones that satisfy the characteristic property.

(That's what Problem 4-7 asks you to prove.)

The reverse implication, as it is, is not true, for quite an obvious reason (though I think a local version of it should be true).

Start by any smoothly final map $f_0:X_0\rightarrow Y$ (e.g. any surjective submersion), and a smooth map $f_1:X_1 \rightarrow Y$ which is not a submersion. Then, the disjoint union $f:=f_0\sqcup f_1: X_0\sqcup X_1 \rightarrow Y$ is not a submersion, nevertheless it is still smoothly final (indeed, for any smooth manifold $Z$ and any map $g:Y\rightarrow Z$, if $g\circ (f_0\sqcup f_1)=(g\circ f_0)\sqcup (g\circ f_1) $ is smooth, so is $g\circ f_0$, hence $g$ because $f_0$ is smoothly final).

It is true that a smoothly final map $f:X\rightarrow Y$ is necessarily surjective (note e.g. that the above construction $f_0\sqcup f_1$ was surjective). In fact, for any $y\in Y$ there exists a map $g:Y\rightarrow\mathbb{R}$ differentiable in $Y\setminus\{y\}$ and not in $y$ (e.g., a map supported in the domain of a local chart at $y$, that in a local chart is $\|\cdot\|$ near $0$). Then, clearly, if $f:X\rightarrow Y$ is not surjective, say because there is $y\in Y\setminus f(X)$, then $g\circ f$ is smooth though $g$ is not, so $f$ is not smoothly final.