My question is: Does the forgetful functor F:(Mfd) $\to$ (Top) preserve pullbacks?

Detailed explanation is following.

A pullback is defined as a manifold/topological space satisfying a universal property: a manifold/topological space $Z$ (more precisely a diagram $Y \overset{g}{\leftarrow}Z\overset{g'}{\rightarrow}Y'$) is said to be a pullback if for any diagram $Y \overset{h}{\leftarrow}W\overset{h'}{\rightarrow}Y'$, there is a unique smooth/continuous map $u:W \to Z$ such that $h=g \circ u$ and $h'=g' \circ u$.

Now suppose that we have a manifold $Z$ which is a pullback of the diagram $Y \overset{f}{\rightarrow}X\overset{f'}{\leftarrow}Y'$ in (Mfd). Then I wonder whether the manifold $Z$ is a pullback in (Top) or not.


  • (Top) is the category of topological spaces and continuous maps.
  • (Mfd) is the category of smooth manifolds and smooth maps.
  • It is well known that a pullback is given by a submanifold $$Y \times_X Y' = \lbrace (y,y') \in Y \times Y'\mid f(y)=f'(y') \rbrace$$ of $Y \times Y'$ if either $f$ or $f'$ is a submersion. But we do not assume that a pullback is of this form. (If a pullback is always given by this form in (Mfd), then this implies that the forgetful functor preserves pullbacks.)
  • I am NOT asking about an example of a pullback in (Top) which is not a manifold. Sorry for my confusing question.
  • $\begingroup$ If $Y\times_X Y'$ is a submanifold of $Y\times Y'$, then it is the pullback in Mfd. So your question is really: is it possible that the pullback exists in Mfd even if the topological pullback $Y\times_X Y'$ is not a manifold. $\endgroup$ – Eric Wofsey Mar 12 '13 at 13:20

Here's a counterexample. Let $X=Y=\mathbb{R}$, and let $Y'$ be a point. Let $f:\mathbb{R}\to\mathbb{R}$ be a smooth map such that $f^{-1}(\{0\})=\{1,1/2,1/3,\dots\}\cup\{0\}$, and let $f'$ map $Y'$ to $0$. Then the topological pullback is just the space $W=\{1,1/2,1/3,\dots\}\cup\{0\}$. But since manifolds are locally connected, any map from a manifold to $W$ factors through $Z$, a space with the same points as $W$ but with the discrete topology. Thus $Z$ is the pullback in manifolds, but the canonical map $Z\to W$ is not a homeomorphism.

An interesting question is whether this is the only way to get a counterexample. That is, can you prove the following: A pair of smooth maps has a pullback in manifolds iff every path component of the topological pullback is a manifold.

EDIT: I believe you can get a path-connected counterexample by taking a pair of maps for which the topological pullback is a topologist's sine curve which has been connected up so that if you "unglued" the two halves of it, it would be homeomorphic to $\mathbb{R}$. That is, it should look kind of like a letter P, except that as the loop of the P approaches the vertical line, it wobbles up and down infinitely like a topologist's sine curve. Then any map from a manifold in should factor through $\mathbb{R}$, since manifolds are locally path-connected.

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  • $\begingroup$ Can you give me an explicit description of $Z$? $\endgroup$ – H. Shindoh Mar 12 '13 at 13:57
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    $\begingroup$ $Z$ is just an infinite discrete space. $\endgroup$ – Eric Wofsey Mar 12 '13 at 14:00
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    $\begingroup$ I love this counterexample! It's so simple. $\endgroup$ – Piotr Pstrągowski Mar 12 '13 at 14:11
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    $\begingroup$ I see. Do you have any explicit example of your $f$? $\endgroup$ – H. Shindoh Mar 12 '13 at 14:20
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    $\begingroup$ Something like $f(x)=e^{-1/x^2}\sin(1/x)$ should work. $\endgroup$ – Eric Wofsey Mar 12 '13 at 14:27

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