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Suppose $f:X\longrightarrow{Y}$ is closed and continuous with X locally compact,then if $f^{-1}(\{y\})$ is compact in $X$ for all $y\in{Y}$,then $Y$ is locally compact.

Let us see why:

Let $y\in{Y}$ and $U$ be a neighborhood of $y$ in $Y$, since $f^{-1}(U)$ is a neighborhood of each element of $f^{-1}(\{y\})$, then for each $a\in{f^{-1}(\{y\})}$ there is a compact neighborhood $K_a$ of $a$ with $K_a\subseteq{f^{-1}(U)}$, it follows that $f^{-1}(\{y\})\subseteq\bigcup_{a\in{f^{-1}(\{y\})}}K_a^{o}$, but since $f^{-1}(\{y\})$ is compact, there exist $a_1,...,a_n\in{f^{-1}(\{y\})}$ such that $f^{-1}(\{y\})\subseteq\bigcup_{i=1}^nK_{a_i}^{o}$. Since each $K_{a_i}$ is compact in $X$, $\bigcup_{i=1}^nK_{a_i}$ is compact in $X$, then $f(\bigcup_{i=1}^nK_{a_i})$ is compact. But $f^{-1}(\{y\})\subseteq\bigcup_{i=1}^nK_{a_i}^{o}$, in consequence $y\notin{f(X-\bigcup_{i=1}^nK_{a_i}^{o}})$, but $f(X-\bigcup_{i=1}^nK_{a_i}^{o})$ is closed , thus $Y-f(X-\bigcup_{i=1}^nK_{a_i}^{o})$ is an open set containing $y$, but since $f$ is onto we have $Y-f(X-\bigcup_{i=1}^nK_{a_i}^{o})\subseteq{f(\bigcup_{i=1}^nK_{a_i}^o)}\subseteq{f(\bigcup_{i=1}^nK_{a_i})}$, thus $f(\bigcup_{i=1}^nK_{a_i})$ is a compact neighborhood of $y$ with $f(\bigcup_{i=1}^nK_{a_i})\subseteq{U}$, again since $f$ is onto. This proves $Y$ is locally compact.

However the converse to this property is not true; a counterexample is gotten identifying $[1,\infty)$ in $\mathbb{R}$.

I just want to know if my reasoning is correct.

Thanks

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If your definition of local compactness is that each point has a local base of compact nbhds, then the argument is correct (apart from the fact that you never actually stated the hypothesis that $f$ is a surjection). I don’t understand the assertion about the converse, however, since $(1,\infty]$ isn’t a subset of $\Bbb R$. – Brian M. Scott Oct 26 '12 at 20:51
Thanks for noticing, I changed $(1,\infty]$ for $[1,\infty)$, and I did use the fact that $f$ is surjective, notice that I wrote many times "since $f$ is onto" – Camilo Arosemena Oct 26 '12 at 21:36
Yes, you mentioned several times in your proof that $f$ is onto; my objection is that you did not state this as a hypothesis when you stated the theorem. The statement of the theorem should include all of the essential hypotheses, and surjectivity of $f$ is certainly one of them. – Brian M. Scott Oct 26 '12 at 21:40
Could you please be a bit more explicit about your counterexample and what you mean by "converse"? – commenter Oct 26 '12 at 21:40
@commenter: It took me a while, but it appears that Camilo is taking $X=\Bbb R$ and the quotient of $\Bbb R$ obtained by identifying $[1,\to)$ to a point as $Y$. The quotient map $f$ is closed and continuous, and $X$ and $Y$ are locally compact, but $f$ does not have all compact fibres. It’s the converse that you get if you state the original theorem as follows. Let $f:X\to Y$ be closed and continuous, and let $X$ be locally compact. If $f$ has compact fibres, then $Y$ is locally compact. One could get other converses by taking other parts of the hypothesis as background, so ... – Brian M. Scott Oct 26 '12 at 21:44
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1 Answer

up vote 0 down vote accepted

Let $\mathfrak{D}=\{\{x\}|x\in{\mathbb{R}-[1,\infty)}\}\cup\{[1,\infty)\}$.

Let $P:\mathfrak{D}\longrightarrow{\mathbb{R}}$ be the quotient map, the topology on $\mathfrak{D}$ is the initial topology from the family $\{P\}$. Let us see $\mathfrak{D}$ is locally compact. By our choice of $\mathfrak{D}$, $P$ is clearly closed.

Let $B\in{\mathfrak{D}}$.

  • If $B=\{x\}$ for some $x\in{(-\infty,1)}$,let $U$ be a neighborhood of $B$, choose $\epsilon>0$ such that $x+\epsilon<1$ and $P([x-\epsilon,x+\epsilon])\subseteq{U}$ then $P^{-1}(P(x-\epsilon,x+\epsilon))=(x-\epsilon,x+\epsilon)$, which implies $P(x-\epsilon,x+\epsilon)$ is an open set containing $B$ in $\mathfrak{D}$, but $P([x-\epsilon,x+\epsilon])$ is compact, this proves all $B\in{\mathfrak{D}}$ have a fundamental family of compact neighborhoods.
  • If $B=[1,\infty)$, let $U$ be a neighborhood of $B$, pick $\epsilon>0$ such that $P[1-\epsilon,\infty)\subseteq{U}$. But $P[1-\epsilon,\infty)$ is a compact subset of $\mathfrak{D}$ since $P[1-\epsilon,\infty)=P([1-\epsilon,1])$.

Therefore $\mathfrak{D}$ is locally compact, but $P^{-1}((1,\infty])$ is not compact in $\mathbb{R}$.

share|improve this answer
In the last line you probably want $P^{-1}([1,\infty))$. As Brian pointed out in a comment, there are more possible interpretations of "converse" that might be interesting to think about: Is it possible to find a continuous and onto function $f\colon X \to Y$ between locally compact spaces such that $f^{-1}(y)$ is compact for all $y \in Y$ but such that $f$ isn't closed? (Hint: examples can be gotten by finding two distinct locally compact topologies on the same set). – commenter Oct 27 '12 at 11:46

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