Exercise 16.4

Proof. Suppose that $U$ is an open subset of $X\times Y$. Consider any $x\in {\pi }_1(U)$ so that there is a $y\in Y$ such that $(x,y)\in U$. Then there is a basis element $A\times B$ of the product topology on $X\times Y$ where $(x,y)\in A\times B\subset U$. Then $A$ and $B$ are open sets of $X$ and $Y$, respectively, since $A\times B$ is a basis element of the product topology. Clearly, we have that $x\in A$ since $(x,y)\in A\times B$. Now, for any $x^{\prime }\in A$, we have that $(x^{\prime },y)\in A\times B$ so that $(x^{\prime },y)\in U$. Hence $x^{\prime }={\pi }_1(x^{\prime },y)\in {\pi }_1(U)$, which shows that $A\subset {\pi }_1(U)$ since $x^{\prime }$ was arbitrary. Then, since $A$ is an open subset of $X$, there is a basis element $A^{\prime }$ where $x\in A^{\prime }\subset A\subset {\pi }_1(U)$. This suffices to show that ${\pi }_1(U)$ is an open subset of $X$ since $x$ was arbitrary. An analogous argument shows that ${\pi }_2$ is also an open map. □