Proposition 2.40

Ian_mi
First, suppose that $$f$$ is continuous and let $$G$$ be an open subset of $$ Y$$. Let $$x \in f^{-1}(G)$$. Then $$f(x) \in G$$ so there exists an open set $$U_x \subseteq X$$ containing $$x$$ such that $$f(U_x) \subseteq G$$. Therefore for all $$y \in U_x$$, $$f(y) \in G$$ so $$y \in f^{-1}(G)$$. Hence $$U_x \subseteq f^{-1}(G)$$ so $$\bigcup_{x \in f^{-1}(G)} U_x \subseteq f^{-1}(G)$$. In addition, $$x \in U_x$$ for all $$x \in f^{-1}(G)$$ so $$f^{-1}(G) \subseteq \bigcup_{x \in f^{-1}(G)} U_x$$. Thus $$f^{-1}(G) = \bigcup_{x \in f^{-1}(G)} U_x$$ which is the union of open sets and therefore open.

Suppose conversely that the inverse image under $$f$$ of any open set in $$ Y$$ is open in $$X$$. Let $$x \in X$$ and let $$G$$ be an open set containing $$f(x)$$. Then $$f^{-1}(G)$$ is open, $$x \in f^{-1}(G)$$, and $$f(f^{-1}(G)) \subseteq G$$ so $$f$$ is continuous.

Ian mi 06:29, March 24, 2011 (UTC)Ian_mi