Is the cartesian product of homeomorphisms again a homeomorphism? 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?
 A: In general, the Cartesian product of two continuous functions is again continuous. Use the fact that the basic open sets in $X\times Y$ are of the form $U\times V$ where $U$ is open in $X$ and $V$ is open in $Y,$ then show that $$(f\times g)^{-1}[U\times V]=f^{-1}[U]\times g^{-1}[V],$$ which is open in $A\times B$ by continuity of $f$ and $g$ and the definition of the product topology.
By the same sort of reasoning, the basic open sets of $A\times B$ are of the form $U\times V$ where $U$ is open in $A$ and $V$ is open in $B,$ so we show that $$(f\times g)[U\times V]=f[U]\times g[V],$$ which is open since $f$ and $g$ are open maps.
Just to clear up my notation (in case it's new to you): Given a function $h:C\to Z,$ $E\subseteq C,$ $F\subseteq Z,$ I denote $$h^{-1}[F]:=\{x\in C:h(x)\in F\}$$ and $$h[E]:=\{f(x):x\in E\}.$$
A: 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.
A: To check that it is continuous, you just have to check the individual coordinates, since $f\times g$ is continuous iff $p_X\circ (f\times g)$ and $p_Y\circ (f\times g)$ are. But these are just the mappings $(a,b)\mapsto f(a)$ and $(a,b)\mapsto g(b)$. Can you express them as compositions of continuous functions?
More generally


*

*$f\times g$ is continuous if  $f$ and $g$ are continuous.

*$f\times g$ is injective if  $f$ and $g$ are injective.

*$f\times g$ is surjective if  $f$ and $g$ are surjective.

*$f\times g$ is open if $f$ and $g$ are open.


The above holds for arbitrary many maps, not just two. If all spaces are non-empty, then the converse implications hold as well.
But:
If $f$ and $g$ are quotient maps, $f\times g$ need not be a quotient map.
