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 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?
 A: 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 
