Prob. 7 (c), Sec. 31, in Munkres' TOPOLOGY, 2nd ed: The image of a locally compact space under a perfect map is also a locally compact space

Question

Here is Prob. 7 (c), Sec. 31, in the book Topology by James R. Munkres, 2nd edition:

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

My Attempt:

Let $y$ be any given point of $Y$. We show that $Y$ is locally compact at $Y$.

Since $p \colon X \rightarrow Y$ is a surjective mapping and since $y \in Y$, so the inverse image set $p^{-1}\big( \{ y \} \big)$ is a non-empty subset of $X$, that is, there exists at least one point $x \in X$ such that $y = p(x)$.

Moreover, since $X$ is locally compact, therefore $X$ is locally compact at $x$ for any point $x \in p^{-1}\big( \{ y \} \big)$, and so we can find an open set $U_x$ in $X$ and a compact subspace $C_x$ of $X$ such that $$ x \in U_x \subset C_x. \tag{0} $$

Thus the collection $$ \left\{ \, U_x \, \colon \, x \in X, p(x) = y \, \right\} \tag{1} $$ is a cover of $p^{-1}\big( \{ y \} \big)$ by sets open in $X$, and since $p^{-1}\big( \{ y \} \big)$ is a compact subspace of $X$, we can find a finite subcollection of the collection in (1) above that also covers $p^{-1} \big( \{ y \} \big)$; let this finite subcollection be $$ \left\{ \, U_{x_1}, \ldots, U_{x_n} \, \right\}, \tag{2} $$ where $x_1, \ldots, x_n \in X$ satisfy $$ p \left( x_1 \right) = \cdots = p \left( x_n \right) = y. $$ Let us put $$ U \colon= \bigcup_{i = 1}^n U_{x_i} \qquad \mbox{ and } \qquad C \colon= \bigcup_{i = 1}^n C_{x_i}. \tag{Definition 0} $$ Then the set $U$ is an open set of $X$ containing $p^{-1}\big( \{ y \} \big)$, and the set $C$ is a compact subspace of $X$, by Prob. 3, Sec. 26, in Munkres.

Thus by (0), (1), (2), and (Definition 0) above, we have $$ p^{-1} \big( \{ y \} \big) \subset U \subset C. \tag{3} $$

From (3) as $$ p^{-1} \big( \{ y \} \big) \subset U, $$ so we have $$ X \setminus p^{-1} \big( \{ y \} \big) \supset X \setminus U, $$ which implies $$ p \left( X \setminus p^{-1} \big( \{ y \} \big) \right) \supset p \big( X \setminus U \big), $$ and hence $$ Y \setminus p \left( X \setminus p^{-1} \big( \{ y \} \big) \right) \subset Y \setminus p \big( X \setminus U \big), $$ that is, $$ \{ y \} \subset Y \setminus p \big( X \setminus U \big). \tag{4} $$

Here and here are two Math Stack Exchange posts of mine where I've shown that $$ Y \setminus p \left( X \setminus p^{-1} \big( \{ y \} \big) \right) = \{ y \}. $$

As $U$ is an open set in $X$ [Refer to (1), (2), and (Definition 0) above.], so $X \setminus U$ is closed in $X$, and as the map $p \colon X \rightarrow Y$ is a closed map, so the set $p \big( X \setminus U \big)$ is a closed set in $Y$, which implies that $Y \setminus p \big( X \setminus U \big)$ is an open set of $Y$.

Let us put $$ V \colon= Y \setminus p \big( X \setminus U \big). \tag{Definition 1} $$ Then $V$ is an open set of $Y$, and (4) gives $y \in V$.

Moreover, if $x$ is any point of $p^{-1}(V)$, then we have $x \in X$ such that $p(x) \in V$, and so $p(x) \in Y$ such that $p(x) \not\in p \big( X \setminus U \big)$, which implies that $x \not\in X \setminus U$, and hence $x \in U$. Thus it follows that $$ p^{-1}(V) \subset U, $$ and since $p \colon X \rightarrow Y$ is a surjective mapping, therefore we have $$ V = p \left( p^{-1} (V) \right) \subset p(U), $$ that is, $$ y \in V \subset p(U). \tag{5} $$

And from (3) above and the surjectivity of the map $p \colon X \rightarrow Y$ we obtain $$ \{ y \} = p \left( p^{-1} \big( \{ y \} \big) \right) \subset p(U) \subset p(C), $$ that is, $$ \{ y \} \subset p(U) \subset p(C). \tag{6} $$

Thus from (5) and (6) together we obtain $$ y \in V \subset p(U) \subset p(C), $$ and so $$ y \in V \subset p(C). \tag{7} $$

Moreover, as $C$ is a compact subspace of $X$ [Refer to the lines immediately following (Definition 0) above.] and as the map $p \colon X \rightarrow Y$ is continuous, so the image set $p(C)$ is a compact subspace of $Y$, by Theorem 26.5 in Munkres. Let us put $$ V_y \colon= V \qquad \mbox{ and } \qquad C_y \colon= p(C). \tag{Definition 2} $$

Thus we have shown that, for any point $y \in Y$, there exists an open set $V_y$ in $Y$ and a compact subspace $C_y$ of $Y$ such that $$ y \in V_y \subset C_y. $$ [Refer to (7) and (Definition 2) above.] Therefore $Y$ is locally compact at any point $y \in Y$.

Hence $Y$ is locally compact.

Is this proof correct and accurate in each and every detail? If so, is each and every step of my presentation easy enough to follow? If not, then where are the issues, if any, of correctness or clarity?

Answer 1

As I said before, re-use the closed map lemma in fibres for better exposition.

If $y \in Y$, $F_y:=p^{-1}[\{y\}]$ is compact and so in the locally compact $X$ it has a compact neighbourhood $C$ (just use the union of the finite subcover of the cover of compact neighbourhoods of all points of $F_y$).

So we have by the lemma on fibres then there is an open subset $U$ of $y$ such that $p^{-1}[U] \subseteq C$. But then clearly $p[C]$ is compact (the only use we use continuity of $p$) and contains $U$ and so is the required neighbourhood of $y$.