12
$\begingroup$

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 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?

$\endgroup$
7
  • 3
    $\begingroup$ 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. $\endgroup$ Aug 3, 2011 at 1:22
  • $\begingroup$ @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. $\endgroup$ Aug 3, 2011 at 4:44
  • $\begingroup$ Ah, I see where you're coming from. Thanks. $\endgroup$ Aug 3, 2011 at 9:40
  • 1
    $\begingroup$ 1+, nice question. It asks for a categorical characterization of submersions in the category of smooth manifolds. $\endgroup$ Aug 5, 2011 at 9:20
  • 1
    $\begingroup$ I have seen the term "carrable" for what is here called "representable", in Paul Taylor's work. I have no idea where that comes from. At least it is unique! $\endgroup$ Jun 4, 2013 at 23:08

1 Answer 1

4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.