What is the characteristic property of surjective submersions?

Question

In Lee's Introduction to smooth manifolds he states that given smooth manifolds $X,Y$ and a surjective submersion, $f:X \rightarrow Y$, then $f$ is a smoothly final map, that is for any further smooth manifold $Z$, and any map $g:Y\rightarrow Z$, we have $g$ smooth if $g\circ f$ is smooth.

He then says that problem 4.7 shows why this property is 'characteristic'. I can't see why the reverse implication should hold.

Unfortunately, google-books doesn't show that page, nor do I have access to a mathematical library, can some-one enlighten me as to what he means?

One of the answers to this question states a characteristic property, but it doesn't appear on the face of it what Lee has in mind.

Answer 1

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.)

Answer 2

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.

Answer 3

By the implicit function theorem, the submersion property of $f$ tells you that any point $x\in X$ has a neighborhood of the product form $U\times V'$ such that $f$ is constant along each copy of $U$ and such that $f$ induces a diffeomorphism of $V'$ onto $V=f(U\times V')$, which is a neighborhood of $y=f(x)$. Knowing that $g\circ f$ is smooth at $x$ lets you precompose it with the inverse of the above diffeomorphism to get $g$ restricted to $V$. It is then easy to conclude that $g$ is smooth at $y$. Since $f$ is surjective, the same argument can be repeated for every $y\in Y$.