# $X$ is locally compact then $Y$ is locally compact

Let $p:X\rightarrow Y$ be a closed continuous surjective map such that $p^{-1}(y)$ is compact for each $y\in Y$. Show that if $X$ is locally compact then $Y$ is locally compact.

Let $y\in Y$ and $x\in p^{-1}(y)$. As $X$ is locally compact there exists a compact subspace $C_x$ and an open set $U_x$ such that $x\in U_x\subset C_x$.

So, we have an open cover $\{U_x\}$ for $p^{-1}(y)$. As this is compact there exists a finite subcover.

Let $\{U_{x_i}\}$ for $1\leq i\leq n$ be an open cover for $p^{-1}\left(\{y\}\right)$. We have $p^{-1}(y)\subset \bigcup_{i=1}^n U_{x_i}\subset\bigcup_{i=1}^n C_{x_i}$.

Goal is to get an open set $U$ and a comactset $C$ in $Y$ such that $y\subset U\subset C$.

Let $M=\bigcup_{i=1}^n U_{x_i}$ and $N=\bigcup_{i=1}^n C_{x_i}$

Continuous image of compact set is compact. So, $p(N)$ seems to be a good choice for this compact set.

As $p$ is not an open map $p(M)$ need not be open. As $p$ is closed, $p(M^c)$ is closed. So, $\left(p\left(M^c\right)\right)^c$ is open.

Claim is that $y\in \left(p(M^c)\right)^c\subset p(N)$.

Suppose $y\notin \left(p(M^c)\right)^c$ i.e., $y\in p(M^c)$ i.e., $y=p(x)$ for some $x\in M^c=\bigcap_{i=1}^nU_{x_i}^c$. So, $x\notin U_{x_i}$ for any $i$.

But, $p(x)=y$ implies $x\in p^{-1}(y)\subset \bigcup_{i=1}^n U_{x_i}$ i.e., we have $x\in U_{x_i}$ for some $i$ a contradiction.

So, we do have $y\in \left(p(M^c)\right)^c$.

To prove $\left(p(M^c)\right)^c\subset p(N)$.

Suppose $m\notin p(N)$ then $m=p(a)$ for some $a\notin N$ i.e., $a\in N^c\subset M^c$.

So, $m\in p(M^c)$. Thus, $m\notin \left(p(M^c)\right)^c$.

So, $y\in \left(p(M^c)\right)^c\subset p(N)$.

Let me know if there are any gaps to fill.

• It’s both correct and complete. Mar 14 '16 at 6:57
• @BrianM.Scott : Please see if math.stackexchange.com/questions/1697564/… is correct.. Excuse me
– user312648
Mar 14 '16 at 22:09