When and why do products preserve pushouts?

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. – Zhen Lin Nov 19 '12 at 14:26
Interesting; can you give a counterexample? I'd like to see why/how this fails. – Julian Kniephoff Nov 19 '12 at 16:49
Any non-distributive lattice (e.g. the lattice of vector subspaces of a non-trivial vector space) will give a counterexample. – Zhen Lin Nov 19 '12 at 17:50

The body question doesn't seem to quite match the title (don't you want to take the product with $E$?) so I am answering the title question. If $A$ is empty then a pushout is just a coproduct, so a special case of your question (in categories with an initial object) is asking whether products distribute over coproducts. Categories with this property are called distributive. As Zhen notes in the comments, distributive categories include cartesian closed categories, where taking the product with a fixed element has a right adjoint and therefore preserves arbitrary colimits (including pushouts).
However, many familiar categories fail to be distributive; take, for example, $\text{Ab}$. This is straightforward to verify for finite abelian groups.
It is also known that $\text{Top}$ is not cartesian closed. However, it is in some sense not far from being cartesian closed; the full subcategory of compactly generated Hausdorff spaces (and it seems one can weaken the second condition even further) is cartesian closed.
Thanks. You're right, I confused $D$ with $E$. It should be correct now. I also changed the title. – Julian Kniephoff Nov 20 '12 at 21:07