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Let $f: X \rightarrow Y$ be a closed surjective continuous map such that $f^{-1}(\{y\})$ is compact for each $y \in Y$. I want to show that if $X$ is regular (Hausdorff) then $Y$ is regular. I tried to do this using the definition of regularity but got no where. Can you please help? |
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The following steps lead to a solution: (1) Let $y\in Y$ and let $U$ be an open neighborhood of $y$. We wish to find a neighborhood $V$ of $y$ such that $y\in \overline{V}\subseteq U$. (Let us recall that $\overline{V}$ is the closure of $V$.) (2) Note that $f^{-1}(y)\subseteq f^{-1}(U)$. Prove that if $Z$ is a regular (Hausdorff) space and if $C$ is a compact subset of $Z$ contained in an open subset $W$ of $Z$, then there exists an open neighborhood $N$ of $C$ such that $C\subseteq \overline{N}\subseteq W$. Deduce that there is an open neighborhood $N$ of $f^{-1}(y)$ such that $f^{-1}(y)\subseteq \overline{N}\subseteq f^{-1}(U)$. (3) Prove that there is a neighborhood $V$ of $y$ in $Y$ such that $f^{-1}(y)\subseteq f^{-1}(V)\subseteq N$. (Hint: use the fact that $f$ is a closed mapping.) (4) Prove that if $g:A\to B$ is any surjective map and if $S\subseteq B$, then $f(f^{-1}(S))=S$. (5) Finally, conclude that $V$ is a neighborhood of $y$ and $y\in V\subseteq \overline{V}\subseteq U$. I hope this helps! Please feel free to ask if you have any questions regarding the above steps. |
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