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>.

Solution

Past exam of the mathematics course of the University of Cambridge / 2017 / ib / Paper 3 / 3E / a / Solution

= Open basis
{synonym}

A basis of a topology is a family of open sets such that every open set is a union of members of that family. Equivalently, every point of every open set lies in a basis member contained in it. The rectangles $U\times V$ form a basis for the <product topology> of two spaces. This topological meaning differs from a linear-algebra <basis>.

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 Incoming links: Basis of a topology