Hausdorff of $X$ implies Hausdorff of $Y$ under some strange condition 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
 A: 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.
A: (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.
A: (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.
