Dtyjlui

Let $X$ and $Y$ be topological spaces such that $Y$ is compact, and let $f colon X to Y$ be a closed, surjective, and continuous map such that, for each $y in Y$, the inverse image $f^{-1} ( {y } )$ is compact.

Then how to show that $X$ is compact also?

My effort:
(A Mere Write-Up / Presentation Of The Proof Based On The Answers Below)

Let $mathscr{A}$ be an open covering of $X$. Then, for each $y in Y$, the collection $mathscr{A}$ is a covering of $f^{-1} ( {y} )$ by sets open in $X$, and since $f^{-1} ( {y} )$ is compact, some finite subcollection of $mathscr{A}$, say $mathscr{A}_y$, also covers $f^{-1} ( {y} )$. That is, for each $y in Y$, there exists a finite subcollection $mathscr{A}_y$ of $mathscr{A}$ such that
$$ f^{-1} ( { y } ) subset bigcup_{A in mathscr{A}_y } A. tag{0} $$
For each $y in Y$, let us put
$$ U_y colon= bigcup_{A in mathscr{A}_y } A, qquad mbox{ and } qquad C_y colon= X setminus U_y. tag{A} $$
Then $U_y$ is an open set in $X$ and $C_y$ is a closed set; moreover we also have
$$ f^{-1} ( { y } ) subset U_y, tag{1} $$
and so
$$ f^{-1} ( { y } ) cap C_y = emptyset. tag{2} $$
Now as $f colon X to Y$ is a closed map and as $C_y$ is a closed set in $X$, so the image set $f left( C_y right)$ is a closed set in $Y$. Let us put
$$ W_y colon= Y setminus f left( C_y right). tag{B} $$
Then $W_y$ is an open set in $Y$.

Now if $x in f^{-1}left( W_y right)$, then by definition $x in X$ is such that $f(x) in W_y$, which implies that $f(x) notin f left(C_y right)$, [Refer to (B) above.] and so $x notin C_y$, which in turn implies that $x in U_y$. [Refer to (A) above.] Therefore we have
$$ f^{-1} left( W_y right) subset U_y. tag{3} $$

As $f$ is surjective, so, for each $y in Y$, the set $f^{-}( { y } )$ is non-empty, that is, there exists at least one $x_y in X$ for which
$$ y = f left( x_y right). tag{4} $$
Any such $x_y in f^{-1}( { y } )$ of course, and so any such $x_y notin C_y$ because of (2) above.

So if $y in f left( C_y right)$, then $y = f(v)$ for some element $v in C_y$, which would imply that $v in f^{-1}( { y } ) cap C_y$, which contradicts (2) above. Thus $y notin f left( C_y right)$, and therefore
$ y in W_y$. [Refer to (B) above.]

Thus we have shown that, for each $y in Y$, there exists an open set $U_y$ in $X$ and there exists an open set $W_y$ in $Y$ such that
$$ y in W_y qquad mbox{ and } qquad f^{-1} left( W_y right) subset U_y; tag{5} $$
[Refer to (3) above.]
moreover the set $U_y$ is the union of some finite subcollection $mathscr{A}_y$ of the open covering $mathscr{A}$ of $X$. [Refer to (0) above.]

In this way, we obtain an open covering of $Y$, which is as follows:
$$ left{ W_y colon y in Y right}. $$
And, since $Y$ is compact, this open covering has a finite subcollection that also covers $Y$; let
$$ left{ W_{y_1}, ldots, W_{y_n} right} $$
be this finite subcollection. Then we have
$$ Y = bigcup_{j=1}^n W_{y_j}. tag{6} $$

Now we show that
$$ X = bigcup_{j=1}^n U_{y_j}. tag{7} $$
Let $x$ be an arbitrary point of $X$. Let $y colon= f(x)$. Then $y in Y$, and so by (6) above we have $y in W_{y_k}$ for at least one $k = 1, ldots, n$; that is, $f(x) in W_{y_k}$; this implies that $x in f^{-1} left( W_{y_k} right)$ and hence $x in U_{y_k}$, because of (3) above; therefore
$$ x in bigcup_{j=1}^n U_{y_j}. $$
Thus one of the inclusions required for (7) above to hold holds; the other inclusion follows from the fact that, by our construction of these sets, each set $U_{y_j}$, for $j = 1, ldots, n$, being a union of subsets of $X$, is itself a subset of $X$. [Refer to (0) and (1) above.]

Finally, as $X$ is a union of the collection $left{ U_{y_1}, ldots, U_{y_n} right}$ consisting of finitely many sets and as each set $U_{y_j}$, for $j = 1, ldots, n$, in this collection is itself a union of a finite subcollection $mathscr{A}_{y_j}$ of our original open covering $mathscr{A}$ of $X$ [Refer to (A) above.], so we can conclude that $X$ is eventually the union of the finite subcollection $mathscr{A}_{y_1} cup ldots cup mathscr{A}_{y_n}$ of $mathscr{A}$.

Since $mathscr{A}$ is an arbitrary open covering of $X$, we can conclude that $X$ is compact.

Is my write-up correct? If so, then is it clear and easy enough to understand? If not, then where are the problems?

Continuing from your effort: Let $C(y)$ be the closed set of $X$ that is not covered by $mathscr A(y)$. We know that $f(C(y))subseteq Y$ is closed, which means that the complement $U(y)subseteq Y$ of that again is open.

$mathscr A(y)$ cover the fiber of $y$, so no point of $f^{-1}({y})$ is in $C(y)$, and therefore $y in U(y)$, so the $U(y)$ cover $Y$. Since $Y$ is compact, it can be covered by finitely many such $U(y)$, let's say $U(y_i), 1 leq i leq n$ for some $n$. I claim that the corresponding $mathscr A(y_i)$ for all $i$ cover $X$.

To prove it, take an $x_0 in X$. The point $f(x_0)$ is in some $U(y_j)$. That means that $f(x_0) notin f(C(y_j))$, which again means that $x_0 notin C(y_j)$, which again means that $x_0$ is inside one of the open sets in $mathscr A(y_j)$.

Let ${ U_i }_{i in I}$ be an open cover of $X$. Then this is also an open cover of $f^{-1}({y})$ for each $y in Y$. Since $f^{-1}({y})$ is compact, there exists a finite subset $I_y subset I$, such that
$$ f^{-1}({y}) subset bigcup_{i in I_y} U_i =: U_y ; $$
Since $U_y$ is open, $X backslash U_y$ is closed subset of $X$, and since $f$ is a closed map, $f(X backslash U_y)$ is a closed subset of $Y$. Note that $y notin f(X backslash U_y)$. We define $W_y := Y backslash f(X backslash U_y)$. Now we see that $W_y$ is open in $Y$ and $y in W_y$, so $W_y$ is an open neighbourhood of $y$. This means that ${ W_y }_{y in Y}$ is an open cover of $Y$, and since $Y$ is compact, there exists a finite subset ${ y_1, ldots, y_m } subset Y$, such that
$$ Y = bigcup_{j=1}^m W_{y_j} ; .$$
We note that
$$ f^{-1}(W_{y_j}) = f^{-1}( Y backslash f(X backslash U_{y_j})) subset X backslash f^{-1}(f(X backslash U_{y_j})) subset X backslash (X backslash U_{y_j}) = U_{y_j}$$
for each $j in {1, ldots, m}$, and from that it follows that
$$ X = f^{-1}(Y) = bigcup_{j=1}^m f^{-1}(W_{y_j}) subset bigcup_{j=1}^m U_{y_j} = bigcup_{j=1}^m bigcup_{i in I_{y_j}} U_i ; , $$
so we have found a finite subcover of ${ U_i }_{i in I}$, which means, that $X$ is compact.

Please check all these steps carefully, I'm not completely sure, if everything is working.

Let ${U_alpha}_{alphain Gamma}$ be an open cover of $X$. Then for each $yin Y$, $f^{-1}(y)subset cup_{alpha in Gamma}U_{alpha}$. Since $f^{-1}(y)$ is compact, for each $yin Y$, there exists a finite subcollection ${U^y_{j}}_{j=1}^n$ of ${U_alpha}_{alphain Gamma}$ such that $f^{-1}(y)subsetcup_{j=1}^nU^y_{j}$. So, $cap_{j=1}^n(X-U^y_j)subset X-f^{-1}(y)$. Notice that the left hand side of the previous expression is closed, so applying $f$ to both sides, the image remains closed. So using the fact that $f$ is surjective, we get, $f(cap_{j=1}^n(X-U^y_j))subset Y-y$. Therefore, $yin Y-f(cap_{j=1}^n(X-U^y_j))$, and $Y-f(cap_{j=1}^n(X-U^y_j))$ is open. Now, taking union over $yin Y$, we get, $Y=cup_{yin Y}(Y-f(cap_{j=1}^n(X-U^y_j)))$. Now, since $Y$ is compact, there exists a finite collection of $yin Y$ such that $Y=cup_{k=1}^l(Y-f(cap_{j=1}^n(X-U^{y_k}_j)))$. This implies
$$
begin{align}
X = f^{-1}(Y)
&=f^{-1}(cup_{k=1}^l(Y-f(cap_{j=1}^n(X-U^{y_k}_j)))) \
&= cup_{k=1}^l(f^{-1}(Y-f(cap_{j=1}^n(X-U^{y_k}_j)))) \
&= cup_{k=1}^l(X-f^{-1}(f(cap_{j=1}^n(X-U^{y_k}_j)))) \
& subset cup_{k=1}^l(X-cap_{j=1}^n(X-U^{y_k}_j)) \
&=cup_{k=1}^lcup_{j=1}^nU^{y_k}_j.
end{align}
$$
So, we've found a finite subcover, so $X$ is compact.