Recall that a space $X$ is locally compact if for any point $x$ in $X$, and any neighborhood $U$ of $x$ in $X$,there is a neighborhood $V$ of $x$ in $X$,and a compact subspace $C$ of $X$ such that $x \in V \subset C \subset U.$ Let $X$ be a locally compact space, and let $f : X \rightarrow Y$ be a continuous, closed surjection. Suppose that $f^{-1}(y)$ is compact for each $y \in Y$ . Show that $Y$ is locally compact.


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Take any $y_0\in Y$, and let $U_0$ be an open neighbourhood of $y_0$. We are looking for an open set $V$ and a compact set $C$ such that $y_0\in V\subset C\subset U_0$.

By assumption, $f^{-1}(\{ y_0\})$ is a nonempty compact subset of $X$, and $O=f^{-1}(U_0)$ is an open set, with $f^{-1}(\{ y_0\})\subset O$. Since $X$ is locally compact, one can find for any point $x\in f^{-1}(\{ y_0\})$ an open neighbourhood $W_x$ of $x$ and a compact set $E_x$ such that ${W_x}\subset E_x\subset O$. By compactness, one can cover $f^{-1}(\{ y_0\})$ by finitely many $W_x$'s, say $W_{x_1},\dots ,W_{x_N}$. Then $E=E_{x_1}\cup\dots \cup E_{x_N}$ is compact, $W=W_{x_1}\cup\dots \cup W_{x_N}$ is open and we have $f^{-1}(\{ y_0\})\subset W\subset E\subset f^{-1}(U_0)$.

Since $f$ is continuous, $C=f(E)$ is compact, and $C\subset U_0$. Since $f$ is closed, the set $V=Y\setminus f(X\setminus W)$ is open; and since $f$ is onto, we also have $V\subset f(W)$, so $V\subset C$. Finally, since $f^{-1}(\{ y_0\})\subset W$, we have $y_0\not\in f(X\setminus W)$, i.e. $y_0\in V$. So the sets $C$ and $V$ have the required properties.


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