Perfect Map $p:\ X\to Y$, $Y$ compact implies $X$ compact

I was assigned the following homework problem for an introductory course in topology:

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

My question is, are the assumptions that $p$ be continuous and surjective even necessary for the result to hold? I've done the following:

Let $U$ be an open set containing $p^{-1}(\{y\})$. Since $X-U$ is closed, $p(X-U)$ is closed in $Y$. Then $W=Y-p(X-U)$ is an open set that contains $y$. Since $(X-U)\cap p^{-1}(W)=\varnothing$, $p^{-1}(\{y\})\subset p^{-1}(W) \subset U$.

Let $\{U_\alpha\}$ be an open cover of $X$. Then $p^{-1}(\{y\})\subset \bigcup_{k=1}^{N}U_k$. For each $y\in Y$, define $W_{y}$ as above; then $\{W_y\}$ is an open cover of $Y$, so can be covered by $W_{y_1},...,W_{y_m}$, and $p^{-1}(\{y_j\})\subset p^{-1}(W_{y_j})\subset \bigcup_{k=1}^{N}U_k$. Since $X=p^{-1}(Y)$, $X=\bigcup_{j=1}^{n} p^{-1}(W_{y_j})$. Since each $p^{-1}(W_{y_j})$ is covered by finitely many $U_k$, so is $X$, so it is compact.

If we dispense with surjectivity of $p$, the only thing that happens is that $p^{-1}(\{y\})$ may be empty, but then $p^{-1}(\{y\})\subset p^{-1}(W)$ holds vacuously, and the empty set is still compact.

If $p$ is continuous, then $p^{-1}(W)$ is open, but I don't use the fact that it is open at all in my proof.

I know that all 3 assumptions make a map "perfect" so that may be why they were mentioned, but can we dispense with continuity and surjectivity?

Thanks.

Solution 1:

but can we dispense with continuity and surjectivity?

Yes. Your - correct - proof uses only the closedness of $p$ to show that every $y\in Y$ has an open neighbourhood $W_y$ such that $p^{-1}(W_y)$ is covered by finitely many of the $U_\alpha$, and since $Y$ is covered by finitely many $W_y$, it then follows that $X$ is covered by finitely many $U_\alpha$.

If one can prove something without using some assumptions in the statement of the theorem, these assumptions are then unnecessary to reach the conclusion.