Let $p:X\to Y$ be continuous surjective closed mapping s.t. $p^{-1}(y)$ is compact $\forall y\in Y$, prove that:
(a) If $X$ is Hausdorff, then $Y$ is Hausdorff
(b) If $Y$ is compact, then $X$ is compact.
I have no idea but a messy mind, could you please give me some hints? Thank you
On
Such functions are called proper. In general, they have the property that preimages of compact sets are compact.
For the first question, note that a space $A$ is Hausdorff if and only if for any $a_1,a_2$ we have closed sets $F_1,F_2$ such that $a_i\notin F_i$ and $F_1\cup F_2=A$, and that in a Hausdorff space, you can separate any two disjoint compact sets.
For the second one, ping me later if you still have trouble and I will add the hints.
On
(a)
Let $y_{1},y_{2}$ be distinct elements of $Y$.
The fibers of $p$ are compact and $X$ is Hausdorff, so open disjoint sets $U_{i}\subseteq X$ can be constructed with $p^{-1}\left(y_{i}\right)\subseteq U_{i}$ for $i=1,2$.
$p$ is a closed map so the $p\left(U_{i}^{c}\right)$ are closed subsets of $Y$.
This with $p\left(U_{1}^{c}\right)\cup p\left(U_{2}^{c}\right)=p\left(U_{1}^{c}\cup U_{2}^{c}\right)=p\left(X\right)=Y$ where the last equality follows from surjectivity of $p$.
The sets $p\left(U_{i}^{c}\right)^{c}$ are open and above it has been shown now that they are disjoint.
From $p^{-1}\left(y_{i}\right)\subseteq U_{i}$ it follows that $y_{i}\notin p\left(U_{i}^{c}\right)$ or equivalently $y_{i}\in p\left(U_{i}^{c}\right)^{c}$
Proved is now that $Y$ is Hausdorff.
(b)
First something that will simplify the proof:
Let it be that for every $y\in Y$ we have $p^{-1}\left(y\right)\subseteq U_{y}$ with $U_{y}$ open.
Then $y\in p\left(U_{y}^{c}\right)^{c}$ so that $Y=\bigcup_{y\in Y}p\left(U_{y}^{c}\right)^{c}$.
The sets $p\left(U_{y}^{c}\right)^{c}$ are open and $Y$ is compact, so $Y=\bigcup_{i=1}^{n}p\left(U_{y_{i}}^{c}\right)^{c}$ for a finite subset $\left\{ y_{1},\dots,y_{n}\right\} \subseteq Y$.
We claim that $X=\bigcup_{i=1}^{n}U_{y_{i}}$.
If not then $x\in\bigcap_{i=1}^{n}U_{y_{i}}^{c}$ for some $x\in X$ and consequently $p\left(x\right)\in\bigcap_{i=1}^{n}p\left(U_{y_{i}}^{c}\right)$ contradicting that $Y=\bigcup_{i=1}^{n}p\left(U_{y_{i}}^{c}\right)^{c}$.
Now we will prove that $X$ is compact.
Let it be that $\left(U_{\lambda}\right)_{\lambda\in\Lambda}$ is an open cover of $X$.
For every $y\in Y$ there is a finite set $\Lambda_{y}$ such that $p^{-1}\left(y\right)\subseteq\bigcup_{\lambda\in\Lambda_{y}}U_{\lambda}$.
Defining $U_{y}:=\bigcup_{\lambda\in\Lambda_{y}}U_{\lambda}$ we are in the situation described above.
So there is a finite subset $\left\{ y_{1},\dots,y_{n}\right\} \subseteq Y$ such that: $$X=\bigcup_{i=1}^{n}U_{y_{i}}=\bigcup_{i=1}^{n}\bigcup_{\lambda\in\Lambda_{y}}U_{\lambda}=\bigcup_{\lambda\in\Lambda_{f}}U_{\lambda}$$
where $\Lambda_{f}:=\bigcup_{i=1}^{n}\Lambda_{y_{i}}\subseteq\Lambda$ is a finite union of finite sets, hence is a finite set.
(a) Note that p being a closed continous surjective mapping implies that it is a quotient map, ie. U is open in Y iff $p^{-1} (U)$ is open in X.
Let y1, y2 be any two disjoint points in Y. Let $y\in p^{-1}(y2)$. Since X is Hausdorff, $p^{-1}(y1)$ being compact implies that there exist open sets $U_y$ and $V_y$ such that
$y\in V_y$, $p^{-1}(y1)\subseteq U_y$ and $V_y\cap U_y = \phi$ $\forall y\in X$
Now, $(V_y)_{y\in Y}$ gives an open cover for $p^{-1}(y2)$, hence $\exists$ z1, z2,..,zn such that $(V_{z_i})^n_{i=1}$ gives an open cover for $p^{-1}(y2)$. Hence, $V = \cup (V_{z_i})^n_{i=1}$ and $U = \cap (U_{z_i})^n_{i=1}$ are disjoint open sets covering $p^{-1}(y2)$ and $p^{-1}(y1)$ respectively.
Now, since U is open in X, $U^c$ is closed in X. p is a closed map, so $p(U^c)$ is closed in Y, and hence $Y\setminus p(U^c)$ is open in Y. Since $y1\in Y\setminus p(U^c)$, $\exists W\subseteq Y\setminus p(U^c)$, W open such that $y1\in W$. Using set inclusions, we obtain $p^{-1}(W)\subseteq U$. Similarly, we get an open set Z in Y such that $y2\in Z$ and $p^{-1}(Z)\subseteq V$.
If $W\cap Z \ne \phi$, then $p^{-1}(W)\cap p^{-1}(Z) \ne \phi$, which would contradict $U\cap V \ne \phi$. Hence we obtain disjoint open sets W and Z such that $y1\in W$ and $y2\in Z$. Thus Y is Hausdorff.
(b) Let $(U_{\alpha})_{\alpha \in A}$ be an open cover of X. Then for every x in $U_{\alpha}$, $\exists$ an open set $W_x$ in Y containing p(x) such that $p^{-1}(W)\subseteq U_{\alpha}$ (as constructed in (a)). Since p is a surjective map, $(W_x)_{x\in X}$ gives an open cover of Y. Since Y is compact, $\exists$ x1, x2, ...,xn such that $(W_{xi})^n_{i=1}$ is an open cover of Y. As $p^{-1}(Y) = X$, therefore $(p^{-1}(W_{xi}))^n_{i=1}$ gives an open cover of X. This is contained in finitely many $\alpha_i$s, which give a finite open cover for X. So every open cover of X has a finite subcover, and we are done.