= Solution
The <product topology> has as a <basis of a topology> all sets $U\times V$, where $U$ is open in $X$ and $V$ is open in $Y$. Open sets are arbitrary unions of these rectangles. The <projection maps> are $p(x,y)=x$ and $q(x,y)=y$. For any open $U\subset X$ and $V\subset Y$,
$$
\boxed{p^{-1}(U)=U\times Y,\qquad q^{-1}(V)=X\times V.}
$$
These are open rectangles, including the empty case. Taking inverse images of open sets therefore proves that both <projection maps> are <continuous functions>.
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