In Wayne Patty's book titled Foundations of Topology, local compactness is defined as follows: $X$ is locally compact at a point $p$ in $X$ provided that there is an open set $U$ and a compact set $K$ such that $p \in U$ and $U \subseteq K$. The proof of the fact that the image of a locally compact space under a continuous open map is locally compact goes like this: Suppose $X$ is locally compact, $Y$ any topological space and $f:X \to Y$ is onto. Let $y \in Y$ and let $x \in {f^{ - 1}}(y)$. Then there is an open set $V$ and a compact subset $C$ such that $x \in V \subseteq C$. Take $U = f(\operatorname{int} C)$ and $f(C) = K$. Then $U$ is open, $K$ is compact and $y \in U \subseteq K$.

Why does one take $U = f(\operatorname{int} C)$ here? Is $U = f(V)$ not enough?

  • $\begingroup$ You're right. Since $f$ is an open map, $f(V)$ is open. And $y\in f(V)\subset f(C)$. $\endgroup$ – DanielWainfleet Apr 16 '17 at 6:45

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