continuous image of a locally compact space is locally compact

Is continuous image of a locally compact space is locally compact?

Let $X$ be locally compact(l.c.). Let $f: X\to Y$ be continuous and surjective.

A space $X$ is locally compact if for each $x\in X$ such that $\exists $ a compact set $V$ such that $x\in U\subset V$ for some open set $U$ containing $x$.

To show that $Y$ is locally compact

My try: Let $y=f(x)\in Y$ and let $V\subset Y$ be an open set containing $y$.then $x\in f^{-1}(V)\subset X$ .As $f$ is continuous then $f^{-1}(V)$ is open in $X$ which is locally compact so there exists a compact set $C\subset X$ such that $x\in f^{-1}(V)\subset C\subset X\implies f(x)\subset V\subset f(C)\subset Y$ Now $f(C)$ being continuous image of a compact set is compact.

Thus proved. But the problem is it has been given that only continuity will not do the map has to be open also.

Please find mistakes in the proof if it exists.

Solution 1:

I have found an answer ;

Let $y=f(x)\in Y $ then since $X$ is locally compact $x\in X$ has a compact neighborhood i.e there exists an open set $U $ contained in a compact set $V$ i.e. $x\in U\subset V$ then $f(x)\in f(U)\subset f(V)$

Now $f$ being open $f(U)$ is open in $Y$ and continuous image of a compact set being compact implies $f(V)$ is compact

Solution 2:

The open set $f^{-1}(V)$ need not be contained in a compact set, so what you have to do is that using locally compactness of $X$ there is a open set $x \in U$ such that there is a compact set $C$ and $U \subset C$. Now take the intersection $U \cap f^{-1}(V)$ and take the image $f( U \cap f^{-1}(V)) \subset f(C)$ now $f$ being open the image is open and $f$ being continuous $f(C)$ is compact. Thus we get locally compactness. The openness condition is required.