TITLE: Is the cartesian product of homeomorphisms again a homeomorphism?
QUESTION [6 upvotes]: If we have two homeomorphisms $f:A\to X$ and $g:B\to Y$, then is it true that $f\times g:A\times B\to X\times Y$ defined by $(f\times g)(a,b)=(f(a),g(b))$ is again a homeomorphism?
I think the answer is yes; 
It's clearly a bijection. Intuitively it seems to be continuous but I don't know how to show it. If, however, this is not true, can you give me a counterexample?

REPLY [5 votes]: A categorical argument would be like this: Consider a category $C$ with products. (In this case, $\mathbf{Top}$, the category with objects the topological spaces and arrows the continuous functions, has products: the product topology has the required universal property). Let $f_i:a_i\rightarrow b_i$ be an iso for each $i\in I$, where $I$ is a discrete category (a set). Consider the following commutative diagram
$$\require{AMScd}
\begin{CD}
b_i @<q_i<< \prod b_i\\
@Vf_i^{-1}VV @VV\prod f_i^{-1}V\\
a_i @<p_i<< \prod a_i\\
@Vf_iVV @VV\prod f_iV\\
b_i @<<q_i< \prod b_i.
\end{CD}$$ Consider the "same" diagram with $f$ and $f^{-1}$ interchanged. Then, by the universal property of products, the arrow $\prod f_i:\prod a_i\rightarrow\prod b_i$ is an iso, with inverse $\prod f_i^{-1}$. In $\mathbf{Top}$, isos are homeomorphisms. Everything is done by the universal property, so the same argument works for $\mathbf{Set}$, $\mathbf{Grp}$, etc.