# Is every representable map a submersion?

Recall that a morphism $f:C \to D$ in a category $\mathscr{C}$ is representable if for all maps $g:E \to D$ in $\mathscr{C},$ the pullback $C \times_{D} E$ exists.

Let now $\mathscr{C}$ be the category of smooth manifolds. Then any submersion is representable. Is the converse true? I have heard from various people that the converse is true, but the only reference I have found is David Metzler's preprint on the arXiv:

Topological and Smooth Stacks

However, the proof he gives there is not complete, for it assumes implicitly that if $M \times_N L$ is a pullback of manifolds, then the induced map $$M \times_N L \to M \times L$$ is a a smooth embedding. I do not see how this is automatic.

I do have a sketch of a proof that this map must be a topological embedding (using diffeological spaces) but is it necessarily an immersion? I would like to argue this using curves, however, this is difficult without knowledge of how to differentiate them in the pullback.

Does anyone have either have a proof or a counterexample for this statement?

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I hadn't heard that usage of "representable" before. Where does it come from? I note that it clashes with the standard usage of "representable" in the case of Set-valued morphisms in CAT. – Tom Leinster Aug 3 '11 at 1:22
@Tom: I am sort of making up the usage. More correctly, I am borrowing it by considering the morphism as actually being a morphism of presheaves via Yoneda. – David Carchedi Aug 3 '11 at 4:44
Ah, I see where you're coming from. Thanks. – Tom Leinster Aug 3 '11 at 9:40
Indeed, I do not see why the proof of Lemma 71 in Metzler's preprint works. Let's look at an example: suppose $M=N=L=\mathbb{R}$ and the maps M->N and L->N are x^2 and y^3, respectively. Isn't the pullback just a line? More explicitly: suppose f and g are differentiable functions such that f^2=g^3. Clearly, f/g is continuous. But actually, it seems that it is differentiable. If so, the pullback does exist, but the induced map $M\times_N L\to M\times L$ is not a smooth embedding. – t3suji Aug 3 '11 at 14:30
1+, nice question. It asks for a categorical characterization of submersions in the category of smooth manifolds. – Martin Brandenburg Aug 5 '11 at 9:20

## 1 Answer

Consider $f(x)=x^3$ on the real line, $C=D=\mathbb R$. Then for any smooth $g:E\to \mathbb R$ the pullback $\mathbb R\times_{f,\mathbb R,g}E$ is a smooth manifold diffeomorphic to the graph of $g$, but is is not a submanifold of $\mathbb R\times E$ in general.

So this is not really a counterexample.

Edit:

The following shows, that not every pullback is embedded into the product.

Consider the topological pullback $N = \lbrace(x,y)\in \mathbb R^2: x^2=y^3\rbrace$ of the two smooth mappings $x^2, x^3: \mathbb R\to \mathbb R$ which is Neill's parabola, and consider the manifold $P=\mathbb R$ with the two mappings $x^3, x^2:\mathbb R\to \mathbb R$ which give a topological homeomorphism $P\to N$: $$\begin{array}{ccccc} P=\mathbb R & \xrightarrow{(x^3,x^2)} & N & \rightarrow & \mathbb R \newline & & \downarrow & & \downarrow x^2 \newline & & \mathbb R & \xrightarrow{x^3} & \mathbb R \end{array}$$ Claim: The triple $(P,x^3,x^2)$ has the universal property of a pullback.

Namely, let $M$ be a smooth manifold and let $f,g:M\to \mathbb R$ be smooth mappings with $f^2= g^3$. Note that then $g\ge 0$. I claim that $f_1:=f^{1/3}:M\to \mathbb R$ is a smooth mapping which gives a smooth factorization $f_1:M\to P$.

Indeed, by convenient calculus (see 1) it is sufficient to show, that $f_1\circ c: \mathbb R\to \mathbb R$ is smooth for each smooth curve $c:\mathbb R\to M$. But $(f_1\circ c)^2=g\circ c$ is smooth and $(f_1\circ c)^3 = f\circ c$ is smooth, so by the theorem of Joris (http://mathoverflow.net/questions/127724), $f_1$ is smooth. QED

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