# Prove that a function is an open mapping.

Let A be a set;let {$X_\alpha$}$_\alpha{_\in{_J}}$ be an indexed family of spaces and let {$f_\alpha$}$_\alpha{_\in{_J}}$ be an indexed family of functions $f_\alpha$:A $\rightarrow X{_\alpha}$. Let $\mathfrak J$ be the coarsest topology on A relative to which each of the functions $f_\alpha$ is continuous. Define $f$ by the equation $f$(a)=($f_\alpha$(a))$_\alpha{_\in{_J}}$;let Z denote the subspace $f$(A) of the product space $\prod X{_\alpha}$. Show that the image under $f$ of each element of $\mathfrak J$ is an open set of Z.

This is an exercise from Munkre's Topology. In the exercise, it proved that $S$=$\bigcup S_\beta$ where $S_\beta$={$f^{-1}_{\beta}({U_\beta})$|$U_\beta$ is open in $X_\beta$} is a subbasis for $\mathfrak J$. It suffices to prove that the image of every basis element under $f$ is an open set of Z. Suppose $\bigcap _{i=1}^{n}f^{-1}_{\alpha i}{(U_{\alpha i})}$ is an arbitrary basis element of $\mathfrak J$, how to prove that $f$($\bigcap _{i=1}^{n}f^{-1}_{\alpha i}{(U_{\alpha i})}$) is an open set of Z?

## 1 Answer

Observe that $f(a) \in f\left( \cap_{i\le n} f^{-1}_{\alpha_i}\left( U_{\alpha_i} \right)\right)$ if, and only if, $(f_{\alpha_i}(a) \in U_{\alpha_i}$ for all $i = 1,2,...,n$.

But, each $U_{\alpha_i}$ is open. Then, for each $i$ there exists $V_{\alpha_i} \subset U_{\alpha_i}$ containing $f_{\alpha_i}(a)$. Thus the set $V = \prod_{\alpha \in J} V_{\alpha}$ such that $V_{\alpha} = X_{\alpha}$ if, and only if, $\alpha \notin \{ \alpha_1,..., \alpha_n\}$ is open in $Z$.

It is clear that $V \subset f\left( \cap_{i\le n} f^{-1}_{\alpha_i}\left( U_{\alpha_i} \right)\right)$ which implies $f\left( \cap_{i\le n} f^{-1}_{\alpha_i}\left( U_{\alpha_i} \right)\right)$ is open.