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The category of topological spaces has a forgetful functor to set which commutes with both small limits and colimits (it has both a left and a right adjoint). Moreover Set is a Grothendieck topos and in any Grothendieck topos colimits are stable under pull-backs in the sense that $$ colim_\alpha (X_\alpha \times_Y Z) \cong (colim_\alpha X_\alpha) \times_Y Z.$$

I'm not an expert, but I don't think the topological spaces form a Grothendieck topos. I'd be happy to learn the contrary. But this raises the question: Does this formula hold in topological spaces? Notice that because this equation holds in Set this is just a question about two topologies on the same set.

We could also ask some related questions:

  1. If this doesn't hold generally, then does it hold if we assume the map $Z \to Y$ is open?
  2. What if $Z \to Y$ and all the maps $X_\alpha \to Y$ are open?
  3. What if we assume something nice about our spaces? For example suppose they are sober? or compactly generated? or Hausdorff?
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Computing quotients in Top does not commute with cartesian products, in general. –  Mariano Suárez-Alvarez Aug 6 '10 at 21:37
    
Good point. I guess this is pretty basic/standard, but what is the best example where $(X/\sim) \times Y \neq (X \times Y/\sim)$? –  Chris Schommer-Pries Aug 6 '10 at 23:13
    
@Chris. Sufficient conditions for this to be true can be found in Bredon's "Topology and Geometry", proposition 13.19, page 43. You need Y to be a locally compact Hausdorff space. In fact, you can see the result here: books.google.es/… . It's funny because you need this result in order to prove something as trivial as the fact that the cone of every topological space is contractile! –  a.r. Aug 7 '10 at 7:31
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Another sufficient condition, this one involving $(X,~)$ rather than $Y$, is: $X\to X/~$ is closed and each of its fibers is compact. (The compactness of the fibers guarantees that the map is closed after forming the product with any $Y$, and more generally after forming any fiber product. –  Tom Goodwillie Aug 7 '10 at 15:20
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If you ask whether a quotient map $q:X\to Q$ yields a quotient map $q^{-1}Y\to Y\subset Q$ from a subspace of $X$ to a subspace of $Q$, the answer is no in general. The first counterexamples that I come up are such that $Q$ is not Hausdorff. The answer is easily seen to be yes if $Y\subset Q$ is either open or closed. –  Tom Goodwillie Aug 7 '10 at 15:34

1 Answer 1

up vote 8 down vote accepted

In compactly generated weak hausdorf spaces (CGWH), proclusions are compatible with base change. (A proclusion $f:X\to Y$ is a map of topological spaces which is homeomorphic to a quotient map.) That is, if $f:X\to Y$ is a proclusion between CGWH-spaces, and $B\to Y$ any map between CGWH-spaces, then $A=X\times_Y B\to B$ is also a proclusion, where the fiber product is taken in CGWH. (Note that $A$ is set theoretically the fiber product, with the topology obtained by replacing the product topology with the closest compactly generated one.)

So in CGWH, quotients are always compatible with base change.

In Top, a colimit is always be computed as a quotient of a coproduct. Unfortunately, this is not true in CGWH; the colimit in Top of a diagram of CGWH-spaces will be compactly generated, but not necessarily weak hausdorf.

There are some results which will tell you that a certain colimit in Top of CGWH-spaces is also the CGWH-colimit. For instance, given $f:X\to Y$ and $g:X\to Z$ maps of CGWH-spaces, if $f$ is the inclusion of a closed subspace, then the Top-pushout of $f$ along $g$ is actually in CGWH. Since the pushout is a quotient of the coproduct of $Y$ and $Z$, you can see that this particular pushout is compatible with fiber products, by the proclusion result I mentioned above.

If you're looking for a counterexample in CGWH, I'd look for a Top-pushout of CGWH-spaces which is not weak hausdorf. I don't have a particular one handy, though I'm sure there are many.

Most or all of the above comes from Gaunce Lewis's thesis, though I don't have it at hand to give precise references.

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Thanks Charles, This is great! So am I right in understanding this: If you have a diagram of CGWH spaces and its colimit in TOP is weak Hausdorff (hence CGWH), then it automatically compatible with base change in CGWH spaces? –  Chris Schommer-Pries Aug 8 '10 at 15:54

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