Local compactness is preserved under continuous open onto mappings

Question

If $f$ is a continuous open mapping of a locally compact space $(X,\tau)$ onto a topological space $(Y,\tau_1)$, show that $(Y,\tau_1)$ is locally compact.

The definition of locally compact is that all points have compact neighborhood. I know that without the "open mapping" condition, that statement is false, but I can't show how this condition make it different.

Answer 1

so take $y \in Y$ since $f$ is surjective there is $x \in X$ such that $f(x)=y$, so there is a compact neighborhood $U$ of $x$ , $f(U)$ is a compact subset of $Y$ and since $U$ is a neighborhood of $x$ so there is a open subset of $U$, $V$ such that $x \in V$, $f$ is a open mapping so $f(V)$ is open in $Y$ and $f(V) \subset f(U)$ such that $y \in f(V)$ so $f(U)$ is a compact neighborhood of $y$ as requested!

Answer 2

Take a point $y$ in $Y$. Since $f$ is onto, there exists a point $x$ in $X$ such that $f(x)=y$.

Now, take a neighborhood $U$ of $x$ such that $K=\overline{U} $ is compact.

Since $f$ is an open mapping, $f(U) $ is an open neighborhood of $y$.

Since $f$ is continuous, $f(K)$ is compact. Now, can you show that $f(K)$ contains $f(U)$, and that $\overline {f(U)}$ is contained in $f(K)$?

Answer 3

Below, I'll take "neighborhood of $x$" to mean "any set whose interior contains $x$", which appears to be the sense used in the question.

This is essentially voldemort's answer, but slightly reconstituted for the case where $X$ is not Hausdorff (so that compact sets may not be closed). For example, if one considers the topology $\mathcal{T} = \{ ( x , +\infty ) : x \in \mathbb R \}$ on $\mathbb R$, then the space is locally compact in the sense defined in the question: for each $x \in \mathbb R$, $[ x-1 , + \infty )$ is a compact neighbourhood of $x$. However, it is easy to check that $\overline{ U } = \mathbb R$ for every non-empty open $U \subseteq \mathbb{R}$, and since this topology is not compact, it follows that no point has an (open) neighborhood with compact closure.

---

Given $y \in Y$, pick any $x \in f^{-1} ( y )$. As $X$ is locally compact, $x$ has a compact neighborhood $K$. By continuity of $f$, it follows that $f (K)$ is a compact subset of $Y$, and by openness of $f$ we have that $f ( \operatorname{Int} (K) )$ is an open subset of $Y$. Clearly $y = f(x) \in f ( \operatorname{Int} (K) ) \subseteq \operatorname{Int} ( f(K) ) \subseteq f (K)$.