Perfect map preserves paracompactness property This is an interesting problem about perfect map and paracompactness but it stucked me for a long time. Hope some one can help me solve this.


Let $p: X \to Y$ be a perfect map.
    $(a)$ Show that if $Y$ is paracompact , so is $X$.
    $(b)$ Show that if $X$ is paracompact Hausdorff space, then so is $Y$.


Thanks so much
 A: In both cases we come to a point where we wish to obtain a cover of $Y$ by means of a cover of $X.$
For a set $A$ of $X$ define $$g(A)=\{y\in Y\mid f^{-1}(y)\subseteq A\}=Y-f[X-A]$$
If $A$ is open, then so is $g(A)$ by closedness of $f.$ Also if $\mathcal U=\{U_i\mid i\in I\}$ is an open cover of $X,$ then since fibers are compact, $$\left\{g\left(\bigcup_{k=1}^n U_{i_k}\right)\mid n\in\Bbb N, i_k\in I\right\}$$ is an open cover of $Y.$
Here's a solution for ($a$): Let $Y$ be paracompact and $\mathcal U=\{U_i\mid i\in I\}$ an open cover of $X.$ Then the above is an open cover of $Y,$ which has a locally finite open refinement $\mathcal V=\{V_j\mid j\in J\}.$ The preimages $f^{-1}(V_j)$ form a locally finite open cover of $X.$ Still, we dont have $f^{-1}(V_j)\subset U_i$, but rather $f^{-1}(V_j)\subseteq\bigcup_{k=1}^n U_{i_k},$ so in order to make it a refinement, we split each preimage by intersecting it with $U_{i_1},...,U_{i_n}.$ Since each $f^{-1}(V_j)$ is split into a finite number of open sets, the local finiteness is still satisfied. $\square$
Next, to show that $X$ Hausdorff implies $Y$ Hausdorff, let $x$ and $y$ be distinct point in $Y.$ The compact preimages $f^{-1}(x)$ and $f^{-1}(y)$ can be separated by disjoint open sets $U$ and $V$ since compact sets behave like points in a Hausdorff space. Then $g(U)$ contains $x$ and $g(V)$ contains $y$ and both are open and disjoint.
