# Does pullback in the category of smooth manifolds always exists?

I am looking for an example where $f:Y\to X$ and $f':Y'\to X$, are both smooth maps of smooth manifolds, but the pullback does not exist.

Remarks:

1) A pullback in a certain category is defined as a space satisfying a universal property, not as a fiber product. (Which is just the usual form of many pullbacks...). See a clarification at the bottom.

2) I know there are examples where the pullback in the smooth category exists, but is different from the fiber product $Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$ (which is always the pullback in the topological category).

This is not what I am looking for, since in this example the smooth pullback (as the space satisfying the required universal property) exists.

3) In any case where the fiber product is a (smooth) submanifold, it is the pullback. Therefore, any possible example must be one whose fiber product is not a (smooth) submanifold. (In particular $f,f'$ can't be transverse to each other)

4) In this answer there is a possible way of poving some limits does not exist. Maybe it's possible to use this method here also, but so far I didn't find an example

Clarification of the definition of pullback:

A space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$ which complete $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}Y'$ into a commutative square) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$ (which commutes with $Y\overset{f}{\rightarrow}X\overset{f'}{\leftarrow}X$), there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$. (see Wikipedia).

• If, for you, the categorical fiber product is not the same as "pullback", can you please define what you mean by "pullback"? – Jason Starr Oct 7 '15 at 10:21
• @JasonStarr : I have edited the question to clarify this. – Asaf Shachar Oct 7 '15 at 10:40
• Your definition of "pullback" is the definition of categorical fiber product. – Jason Starr Oct 7 '15 at 12:18
• On the nLab, there is I think a growing consensus to apply the phrase "fiber product" to the limit of the cospan, and to apply the word "pullback" to a functor $f^\ast: E/X \to E/Y$ which sends (in the notation of OP's clarification) $f': Y' \to X$ to $g: Z \to Y$ arising in the fiber product, as in "pulling back along $f$". – Todd Trimble Oct 7 '15 at 13:19
• @JasonStarr It's certainly related. One way of thinking about a cleavage of a fibration is that it is given by a pseudofunctor $F: C^{op} \to Cat$ which takes $f: X \to Y$ to a suitable $f^\ast: F(Y) \to F(X)$. A cleavage of the codomain fibration $cod: C^\mathbf{2} \to C$ would then be given by a choice of pullback $f^\ast: C/Y \to C/X$ for each $f: X \to Y$. So having a choice of fiber product for all cospans would give you such a cleavage, although I was really just referring to a single $f$ in my comment. – Todd Trimble Oct 7 '15 at 15:20

So take for example the pullback of $\{0\}$ along the map $(x,y) \rightarrow xy$. If this pullback existed it would be a smooth structure on the union of the horizontal and the vertical line in $\mathbb{R^2}$ such that a map into it is smooth if and only if it is smooth when seen as a map into $\mathbb{R^2}$.
• Thanks for your answer. I have two questions: 1) I see why the pullback would have to contain as a set all the points in the fiber product. How can I convince myself that it does not contain any other points? 2) Why such a smooth structure on the union of lines does not exist? (The topology on the pullback can be different from the subspace top' of the product $Y \times Y'$ (like in the weird example in the link...), so in general we 'have freedom' in the topology as well not just in the smooth structure – Asaf Shachar Oct 7 '15 at 10:55
• 1) a point of a manifold $M$ is a morphism from the one point manifold to $M$, as such they obey the universal properties of the pullback and hence are the points of the fiber product. 2) Once you know what the smooth maps to fiber product are, the atlas has to be given by smooth injective maps from open subsets of $\mathbb{R^n}$ to you manifold. In our case, the only such maps are map from $\mathbb{R}$, so our manifold would be one dimensional and it is not possible in a one dimensioncal manifold to have two non constant path that cut into only one point like you have in $0$ here. – Simon Henry Oct 7 '15 at 11:02