Exercise 19.5

Proof. As in Theorem 19.6 suppose that $f:A\to {\prod <!--\; FUNCTION\; APPLICATION\; -->}_{\alpha \in J}{X}_{\alpha }$ be given by

where $f_{\alpha }:A\to X_{\alpha }$ for each $\alpha \in J$. Here $\prod <!--\; FUNCTION\; APPLICATION\; -->X_{\alpha }$ has the box topology. Suppose that $f$ is continuous and consider any $\beta \in J$. We show that $f_{\beta }$ is continuous, which of course shows the desired result.

So let $V$ be any open set of $X_{\beta }$ and define

Then, since each $B_{\alpha }$ is clearly open in $X_{\alpha }$, we have that $B=\prod <!--\; FUNCTION\; APPLICATION\; -->B_{\alpha }$ is a basis element of the box topology by definition and is therefore open. Hence $U=f^{-1}(B)$ is open in $A$ since $f$ is continuous. We claim that $U={f_{\beta }}^{-1}(V)$, which shows that $f_{\beta }$ is continuous since $U$ is open in $A$ and $V$ was an arbitrary open set of $X_{\beta }$.

$\text{(⊂)}$ If $x\in U=f^{-1}(B)$ then of course $f(x)\in B$ so that each $f_{\alpha }(x)\in B_{\alpha }$ since $f(x)={(f_{\alpha }(x))}_{\alpha \in J}$ and $B=\prod <!--\; FUNCTION\; APPLICATION\; -->B_{\alpha }$. In particular $f_{\beta }(x)\in B_{\beta }=V$ so that $x\in {f_{\beta }}^{-1}(V)$. Hence $U\subset {f_{\beta }}^{-1}(V)$ since $x$ was arbitrary.

$\text{(⊃)}$ If $x\in {f_{\beta }}^{-1}(V)$ then $f_{\beta }(x)\in V=B_{\beta }$. Since of course every other $f_{\alpha }(x)\in X_{\alpha }=B_{\alpha }$ we have that $f(x)\in \prod <!--\; FUNCTION\; APPLICATION\; -->B_{\alpha }=B$. Hence $x\in f^{-1}(B)=U$ so that ${f_{\beta }}^{-1}(V)\subset U$ since $x$ was arbitrary. □