$X$ is Hausdorff then $Y$ is Hausdorff

Question

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

Let $a\neq b\in Y$. Fix $x\in X$ such that $p(x)=a$. Let $y_i\in X$ such that $p(y_i)=b$

As $X$ is Hasudorff there exists $U_i$ containing $x$ and $V_i$ containing $y_i$ such that $U_i\cap V_i=\emptyset$

See that $\{V_i\}$ is an open cover for $p^{-1}(\{b\})$. As this is finite there exists a finite subcover $\{V_i\}_{i=1}^n$

Fix $y\in X$ such that $p(y)=b$.. Let $x_i\in X$ such that $p(x_i)=a$

As $X$ is Hasudorff there exists $M_j$ containing $x_j$ and $N_j$ containing $y$ such that $M_j\cap N_j=\emptyset$

See that $\{M_j\}$ is an open cover for $p^{-1}(\{a\})$

As this is finite there exists a finite subcover $\{M_j\}_{i=1}^m$

Consider $\bigcup_{i=1}^n V_i$ and $\bigcup _{j=1}^m M_j$ open sets in $X$

Their complement $\bigcap_{i=1}^n V_i^c$ and $\bigcap _{j=1}^m M_j^c$ are closed in $X$

As $p$ is a closed map we have $p\left(\bigcap_{i=1}^n V_i^c\right)$ and $p\left(\bigcap_{j=1}^m M_j^c\right)$ are closed

So, $\left(p\left(\bigcap_{i=1}^n V_i^c\right)\right)^c$ and $\left(p\left(\bigcap_{j=1}^m M_j^c\right)\right)^c$ are open

I expect to have $a\in\left(p\left(\bigcap_{j=1}^m M_j^c\right)\right)^c$ and $b\in \left(p\left(\bigcap_{i=1}^n V_i^c\right)\right)^c$ but could not prove

Suppose $a\in p\left(\bigcap_{j=1}^m M_j^c\right)$ i.e., $a=p(x)$ for some $x\in \bigcap_{j=1}^m M_j^c$ i.e., $x\notin M_j$ for all $j$

We have $p^{-1}(a)\subset \bigcup_{j=1}^n M_j$ in particular, we must have $x\in \bigcup_{j=1}^n M_j$ but then $x\notin M_j$ for any $j$, contradiction

Similarly for $b$ also we have same thing

Is this justification sufficient enough?

EDIT : I do not see why should $\bigcup_{i=1}^n V_i$ and $\bigcup _{j=1}^m M_j$ in $X$.. Accepted answer below assume that they are disjoint.. It is my mistake.. User asked me to check if it is disjoint and i said yes..

Answer 1

Given $y_1, y_2 \in Y$, the preimages are compact, hence may be separated by open sets $A_1$ and $A_2$. (Proof below). Then $A_1^c$ and $A_2^c$ are closed, and so their images are closed. We claim that $p(A_1^c)^c$ and $p(A_2^c)^c$ separate $y_1$ and $y_2$. First, $y \in p(A_i^c)^c$ if and only if $p^{-1}(y) \subset A_i$. Therefore $y_1 \in p(A_1^c)^c$ and $y_2 \in p(A_2^c)^c$. In particular, $y \in p(A_1^c)^c \cap p(A_2^c)^c$ if and only if $p^{-1}(y) \subset A_1 \cap A_2 = \emptyset$, hence $p(A_1^c)^c \cap p(A_2^c)^c = \emptyset$.

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EDIT: I did not include the proof that the compact sets can be separated because it was linked in the comments. But for completeness I include the argument here.

Step 1: For each $z \in p^{-1}(y_2)$, we will separate $z$ from $p^{-1}(y_1)$. That is, we will construct an open neighborhood $B(z)$ of $z$ and an open set $A(z)$ containing $p^{-1}(y_1)$, such that $A(z)$ and $B(z)$ are disjoint. Here is the construction:

For each pair of points $x \in p^{-1}(y_1)$, $z \in p^{-1}(y_2)$, we can choose open disjoint neighborhoods $A_{xz}$ and $B_{xz}$, containing $x$ and $z$, respectively. For fixed $z$, $A_{xz}$ is a cover of $p^{-1}(y_1)$, and we may pass to a finite subcover $A_{x_iz}$. Now define $$A(z) = \bigcup_i A_{x_iz} \ \ \ \ \ B(z) = \bigcap_i B_{x_iz}.$$ Then $A(z)$ is an open set containing $p^{-1}(y_1)$ and $B(z)$ is an open neighborhood of $z$. Moreover, if $u \in B(z)$, then $u \in B_{x_iz}$ for all $i$, hence $u \notin A_{x_iz}$ for any $i$, hence $u \notin A(z)$. Thus we have completed step 1.

Step 2: Here we construct open disjoint sets $A_1$ containing $p^{-1}(y_1)$ and $A_2$ containing $p^{-1}(y_2)$.

The sets $B(z)$ constructed above give a cover of $p^{-1}(y_2)$, so we may pass to a finite subcover $B(z_i)$. Now define $$A_1 = \bigcap_i A(z_i) \ \ \ \ \ A_2 = \bigcup_i B(z_i).$$ Then $A_1$ is an open set containing $p^{-1}(y_1)$ and $A_2$ is an open set containing $p^{-1}(y_2)$. Now if $v \in A_1$, then $v \in A(z_i)$ for all $i$, hence $v \notin B(z_i)$ for any $i$, hence $v \notin A_2$. Therefore $A_1$ and $A_2$ are disjoint.

Answer 2

Let $a\neq b\in Y$, $p^{-1}(a)$ and $p^{-1}(b)$ exist by surjectivity of $p$ and we have by assumption that $p^{-1}(a)$ and $p^{-1}(b)$ are compact.

Note that $p^{-1}(a)\cap p^{-1}(b)=\emptyset$ by $p$ being a function.

For each pair of points $x\in p^{-1}(a)$, and $y\in p^{-1}(b)$, by $X$ Hausdorff, there exist open sets $U_{x,y}$ and $V_{x,y}$ such that $U_{x,y}\cap V_{x,y}=\emptyset$. These form open covers of $p^{-1}(a)$ (by $U_{x,y}$'s), and $p^{-1}(b)$ (by $V_{x,y}$'s). For each $x_0\in p^{-1}(a)$ we have an open cover of $x_0$ by the sets $U_{x_0,y}$ for all $y\in p^{-1}(b)$, hence by compactness, these open covers each have finite subcovers. Let $\mathfrak{U}_0$ denote the intersection of finite cover for $x_0$. Then $\mathfrak{U}_0$ separates $x_0$ from every $y$ because $\mathfrak{U}_0 \cap V_{x_0,y}=\emptyset$ for all $y$. The union $\bigcup_{x_0 \in p^{-1}(a)}\mathfrak{U}_0$ is an open cover of $p^{-1}(a)$. Hence it has a finite subcover, which we denote as $\mathfrak{U}$. An analogous construct can be done for each $y_0\in p^{-1}(b)$, ultimately yielding a finite open subcover of $p^{-1}(b)$ which we denote $\mathfrak{V}$. (Note: must use the exact same $U_{x,y}$'s and $V_{x,y}$'s for this construction)

Now we need to get open sets in $Y$. To do so, we will exploit the last unused hypothesis: $p$ is a closed map. Let $\mathfrak{U}^c$ denote the set of complements of the elements of $\mathfrak{U}$, and likewise for $\mathfrak{V}^c$ and $\mathfrak{V}$.

Consider the images of the sets in these collections through $p$, which will denote as $p(\mathfrak{U}^c)$ and $p(\mathfrak{V}^c)$. These are images of closed sets, hence closed in $Y$. We take their complements which yield open finite collections of open sets in $Y$.

First, since $p^{-1}(a)\cap \bigcap_{P\in \mathfrak{U}^c}P=\emptyset$, $a\notin p(\mathfrak{U}^c)$ and hence $a\in p(\mathfrak{U}^c)^c$. With the analogous result holding for $b$. Finally define $U=\bigcap_{P\in p(\mathfrak{U}^c)^c}$ and likewise $V=\bigcap_{Q\in p(\mathfrak{V}^c)^c}$. (Intersect the open sets we got.)

Now suppose $z \in U \cap V$. Then $p^{-1}(z) \cap U_{x,y}$ and $p^{-1}(z) \cap V_{x,y}$ are nonempty for every $x,y$, but this implies that there exists $x_0\in p^{-1}(a)$ and $y_0\in p^{-1}(b)$ such that intersection $U_{x,y}\cap V_{x,y}$ is nonempty. But this contradicts our earlier construction, hence $U\cap V=\emptyset$. Therefore, $Y$ is Hausdorff.