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.


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).

  • $\begingroup$ If, for you, the categorical fiber product is not the same as "pullback", can you please define what you mean by "pullback"? $\endgroup$ Oct 7, 2015 at 10:21
  • $\begingroup$ @JasonStarr : I have edited the question to clarify this. $\endgroup$ Oct 7, 2015 at 10:40
  • $\begingroup$ Your definition of "pullback" is the definition of categorical fiber product. $\endgroup$ Oct 7, 2015 at 12:18
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    $\begingroup$ 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 $f$". $\endgroup$
    – Todd Trimble
    Oct 7, 2015 at 13:19
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    $\begingroup$ @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. $\endgroup$
    – Todd Trimble
    Oct 7, 2015 at 15:20

1 Answer 1


If a pullback exists in the category of smooth manifold then, its underlying set of points has to be what you described simply by looking at morphism from the point. Moreover a map into the pullback is smooth if and only if the map to the product is smooth (because it is smooth if and only if each component is smooth by the universal properties of the pullback).

So the only things surprising things that can happen if the categorical pullback exists, is that the fiber product has a smooth structure which is not induced by the differentiable structure on the product, but still have the same smooth map into it, like the example you gave a link to... But this situation is rather weird, and it is a lot more common to simply have no pullback:

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 $\mathbb{R^2}$.

  • $\begingroup$ 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 $\endgroup$ Oct 7, 2015 at 10:55
  • $\begingroup$ Also, a minor technical point: You wrote 'Moreover a map into the fiber product is smooth if and only if the map to the product is smooth'. I think You should replace fiber product with the pullback object, since they are not necessarily equal (at least as topological spaces, although as sets you claim they are...This is of course the subtle point lurking here, showing the fiber product is not a submanifold is not enough to conclude a pullback does not exists). $\endgroup$ Oct 7, 2015 at 11:02
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    $\begingroup$ 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. $\endgroup$ Oct 7, 2015 at 11:02
  • $\begingroup$ For your second comment you are absolutely right, I wanted to emphasize the fact that the pullback and the fiber product are the same as sets, but I will correct this. $\endgroup$ Oct 7, 2015 at 11:03

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