# What is the characteristic property of surjective submersions?

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

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Why are people trying to close this? –  David Carchedi Jan 24 at 19:11
Since SS (surjective submersion) is a strictly stronger property compared with SF (smoothly final), it remains the curiosity of giving a characterization of both. Note that a SS map $f:X\to Y$ produces by restriction, on any open set $U\subset X$, an SS (thus SF) $f_{|U}:U\to f(U)$ onto an open subset of $Y$. Therefore it is not only SF, but also "locally SF" in the above sense. I'm not sure if this stronger property is enough to characterize SS maps. On the other hand, it would be interesting a characterization of SF, e.g. in the category of $C^1$ manifolds and maps. –  Pietro Majer Jan 25 at 11:09

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.)

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I wasn't expecting the author of the text to turn up! Thanks, what you had in mind wasn't what I was expecting, but I see now I should have done, its exactly analogous to final maps in Top. –  Mozibur Ullah Jan 25 at 1:49
@Mozibur Ullah, you're welcome! –  Jack Lee Jan 25 at 5:23

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 $y\in Y$.

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The OP asked for the other implication: If f is smoothly final, then it is a surjective submersion. –  Dylan Wilson Jan 24 at 16:51
Hmm, got mixed up with which implication was in question. –  Igor Khavkine Jan 24 at 17:23
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.
More generally: for smooth maps $h:U\to X$ and $f:X\to Y$, if $fh$ is smoothly final, then $f$ is smoothly final. –  Pietro Majer Jan 24 at 23:18