How can I demonstrate the following result?
(note): I have already searched and found results where, they say that, "the closed continuous image of a locally compact space does not need to be locally compact," is false.
But I don't know how to show you the following:
The closed continuous image of a locally compact space is locally compact, provided the pre-image of each point is compact .
Maybe the question is very obvious but I can't see where to start a demonstration, Any suggestion or contribution is welcome, thanks!
Let $f: X \to Y$ be closed, continuous and onto and have all fibres $f^{-1}[\{y\}]$ compact (a so-called perfect map) and suppose $X$ is locally compact (Hausdorff). Then $Y$ is locally compact too.
In the OP's definition we have to show first that $Y$ is Hausdorff. So let $y_1 \neq y_2$ in $Y$ and note that $f^{-1}[\{y_1\}]$ and $f^{-1}[\{y_2\}]$ are disjoint compact subsets and thus, in a Hausdorff space, have disjoint open neighbourhoods $f^{-1}[\{y_1\}] \subseteq U_1$ and $f^{-1}[\{y_2\}] \subseteq U_2$. But then (by closedness of $f$ and basic set theory) $V_i = Y\setminus f[X\setminus U_i]$ ($i=1,2$) are disjoint open neighbourhoods of $y_1,y_2$ respectively. So $Y$ is indeed Hausdorff.
To finish the proof let $y$ be a point of $Y$ and by local compactness of $X$ (as in the OP's definition) we can find an open subset $W$ of $X$ such that $\overline{W}$ is compact and $f^{-1}[\{y\}] \subseteq W$ (this uses the compactness of the fibre and the fact that a finite union of relatively compact open sets is still relatively compact). As above (for Hausdorff) we define $V = Y\setminus f[X\setminus W]$ which is an open neighbourhood of $y$ and as $$V = Y\setminus f[X\setminus W] \subseteq Y\setminus f[X\setminus \overline{W}] \subseteq f[\overline{W}]$$ and the right hand set is compact, $V$ is a relatively compact neighbourhood of $y$, as we needed.
We do need the fibre condition: if $X = \Bbb R$ and we identify $\Bbb Z$ to a point, we get (in the quotient space) a non-locally compact space $\Bbb R{/}\Bbb Z$ even though $\Bbb R$ is locally compact and the natural quotient map is also closed.