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Let $\mathsf{Man}$ be the category of smooth manifolds. Denote by $|~|$ the forgetful functor to $\mathsf{Top}$. If $X \to S$ and $Y \to S$ are morphisms in $\mathsf{Man}$, then $X \times_S Y$ exists and $|X \times_S Y| = |X| \times_{|S|} |Y|$ under suitable assumptions, for example when $X \to S$ and $Y \to S$ are submersions, right? But my actual question is:

Question. What is an explicit example for morphisms $X \to S$ and $Y \to S$ such that $X \times_S Y$ does not exist in $\mathsf{Man}$?

Of course it is not enough (a priori) to prove that $|X| \times_{|S|} |Y|$ is not a manifold. On the other hand, it is easy to see, using morphisms from the point, that the underlying set of $X \times_S Y$ is the fiber product of the underlying sets.

I am also interested in related categories, for example topological manifolds, Banach or Frechet manifolds. Also I would like to add a soft question: The theory of manifolds doesn't really suffer from the failure of the existence of fiber products, right? On the other hand fiber products play an essential role in the theory schemes, which can be seen as "algebraic manifolds". What is a possible reason or explanation for this asymmetry?

Martin Brandenburg
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  • On the last question, I wonder if this is because the smooth category is "soft" and has very good local theory (e.g. there are partitions of unity, the topological spaces are genuinely Hausdorff, etc.) and so the role of a "base space" is not quite as important as when in a "rigid" category like $\textbf{Sch}$... but this is simply my speculation. – Zhen Lin Mar 06 '13 at 13:00
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    Related: http://mathoverflow.net/questions/151/when-do-fibre-products-of-smooth-manifolds-exist – Colin McQuillan Mar 06 '13 at 23:04

2 Answers2

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Here's an argument for the first question. Take $X=Y=S=\mathbb R$ and take $x\mapsto x^2$ for the maps $X,Y\to S$. As you note, the underlying set of $X\times_S Y$ is what you'd expect, $\{(x,y)\mid x=\pm y\}$ (at least up to bijection, and so without loss of generality).

Lemma: Let $Z$ be a subset of $\{(x,y)\mid x=\pm y\}$ such that $Z$ defines a smooth manifold $M$ in the usual topology coming from $X\times Y$. Then the subspace topology induced on $Z$ from $X\times_S Y$ is the usual topology.

Proof: By the universal property of $X\times_S Y$ (and using morphisms from the point), the inclusion $M\to X\times Y$ factorises in $\mathsf{Man}$ into two inclusions: $$M\to X\times_S Y\to X\times Y$$ Restricting to $Z$ we get continuous inclusions which we can denote $M\to M'\to M$. But this implies $M=M'$. $\square$

In particular, applying the lemma three times, with $Z$ equal to $\{(x,y)\mid x=\pm y\}\setminus\{(0,0)\}$ and $\{(x,x)\mid x\in\mathbb R\}$ and $\{(x,-x)\mid x\in\mathbb R\}$, we find that $X\times_S Y$ induces the usual topology on these sets. So $X\times_S Y$ is connected but $X\times_S Y\setminus\{(0,0)\}$ has four connected components. This cannot happen with smooth manifolds: removing a point from a manifold increases the number of connected components by at most $1$.

Colin McQuillan
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  • Why is it the induced topology from $X \times Y$? – Martin Brandenburg Mar 06 '13 at 14:56
  • @MartinBrandenburg: I am not sure what you're asking. The lemma is supposed to establish that $X\times Y$ induces the same topology on $Z$ that $X\times_S Y$ induces on $Z$. – Colin McQuillan Mar 06 '13 at 15:15
  • I don't understand why we can apply the Lemma. We don't know if the manifold has the topology coming from the product. And when you just consider that topology, it doesn't have to come from a manifold. – Martin Brandenburg Mar 06 '13 at 16:07
  • In the last paragraph I apply the lemma three times, using various $Z$. Each of these three $Z$s is a manifold in the usual topology on $\mathbb R^2$. Therefore the lemma applies (the only precondition for the lemma is that $Z$ is a subset of ${(x,y)\mid x=\pm y}$ and is manifold in the usual topology). – Colin McQuillan Mar 06 '13 at 16:23
  • Ok. But I don't understand the proof of the Lemma. Is there anything special about ${x = \pm y}$ here? Why is $M'$ a submanifold? etc – Martin Brandenburg Mar 06 '13 at 18:06
  • The proof of the lemma was nonsense, sorry. I've written a simpler argument. – Colin McQuillan Mar 06 '13 at 22:19
  • Alright, thank you. An alternative argument for the lemma: $M$ satisfies the universal property of the fiber product. And this isn't anything special about ${x = \pm y}$, right? – Martin Brandenburg Mar 07 '13 at 00:02
  • No special properties, just that ${(x,y)\mid x=\pm y}$ is the underlying set of $X\times_S Y$. – Colin McQuillan Mar 07 '13 at 11:20
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Regarding failure of existence of fibre products and the comparison with the case of schemes, the basic point is that the intersection of smooth objects need not be smooth. (The fibre product of $f:X \to S$ and $g: Y \to S$ is the intersection of $\Gamma_f \times Y$ and $X \times \Gamma_g$ in $X \times S \times Y = X \times Y \times S$; here $\Gamma$ denotes graph.)

In algebraic geometry we "remedy" this by allowing non-smooth varieties/schemes, etc. But the possible singularities that can arise by intersecting algebraic varieties are much tamer than those that can arise by intersecting submanifolds of some ambient manifold. So it's not surprising that it's not routine to extend the theory of manifolds to include singular objects. There are such extensions, such as the theory of stratified spaces, which among other things are related to issues of transversality and so on that arise when considering singularities of intersections. But the theory of such objects is certainly not a routine extension of the theory of manifolds.

One reason that people work with O-minimal structures is that these provide a setting which is somewhat close to the topology of manifolds, while being tame enough that one can form pullbacks and fibre products and stay in a world of reasonable objects.

Matt E
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