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Let $A,B,C$ topological spaces and then $D$ the pushout of a diagram

$$B\stackrel{b}{\leftarrow}A\stackrel{c}{\rightarrow}C.$$

It seems logical to me that for a fifth topological space $E$ the pushout of

$$B\times E\stackrel{b\times\operatorname{id}_{E}}{\leftarrow}A\times E\stackrel{c\times\operatorname{id}_{E}}{\rightarrow}C\times E$$

is homeomorphic to the product $D\times E$.

However, several tries to come up with a simple proof ended at some point where there was no obvious next step, so to say. This makes me think that this result doesn't hold in a general setting like the above.

On the other hand, I don't see, why something like this should not work even in more general settings like other categories.

Can somebody point me in the right direction here? In which (preferably very general) situation does the above hold and how would you go about giving a simple, abstract proof of it?

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    You are asking when the functor $(-) \times D$ preserves pushouts. One situation where this happens is when $(-) \times D$ has a right adjoint: this happens, for example, in $\textbf{Set}$, or more generally in any cartesian closed category. Unfortunately, $\textbf{Top}$ is not such a category.2012-11-19
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    Interesting; can you give a counterexample? I'd like to see why/how this fails.2012-11-19
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    Any non-distributive lattice (e.g. the lattice of vector subspaces of a non-trivial vector space) will give a counterexample.2012-11-19

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