Showing that a space is Hausdorff

Question

Suppose $X$ is compact and Hausdorff and that $f:X \to Y$ is continuous, closed, and surjective. How can I show that $Y$ is Hausdorff?

Answer 1

Since $X$ is compact Hausdorff, $X$ it is normal. Next, since $f$ is closed, the single-element subsets of $Y$ are closed (as $\{y\} = f''\{x\}$ for some $x \in X$.) It follows that for each pair of distinct $y_0, y_1 \in Y$, the sets $X_0 = f^{-1}(y_0)$ and $X_1 = f^{-1}(y_1)$ are disjoint closed subsets of $X$. Noting that $X$ is normal, you can separate $X_0$ and $X_1$ using open-sets with disjoint closures. The remaining bit is to turn the open-sets separating $X_0$ and $X_1$ into open subsets of $Y$ separating $y_0$ and $y_1$.

Answer 2

Let $y_1 \neq y_2$ be distinct points of $Y$. By surjectivity we have $x_1 ,x_2 \in X$ such that $f(x_1) = y_1 , f(x_2) = y_2$. In particular, $f[\{x_i\}] = y_i$ for both $i$ and as in $X$ singletons are closed (Hausdorff implies $T_1$) and $f$ is a closed map (first time we use it), we have that both $\{y_i\}$ are closed in $Y$.

Now, define $F_i = f^{-1}[\{y_i\}] \subseteq X$, $i=1,2$. $F_1$ and $F_2$ are disjoint closed (by continuity of $f$) subsets of $X$, which is compact and Hausdorff and thus $T_4$ or normal. So in $X$ we can find open disjoint subsets $U_1, U_2$ such that $F_1 \subseteq U_1$ and $F_2 \subseteq U_2$.

Now define $O_i = Y\setminus f[X\setminus U_i]$ for $i=1,2$.

These sets are open as $f$ is a closed map and complements of open sets are closed and complements of closed sets are open.

For each $i$ $y_i \in O_i$: suppose $y_i \notin O_i$ then by definition $y_i \in f[X\setminus U_i]$ so $y_i = f(p)$ for some $p \in X \setminus U_i$. But by definition $p \in F_i$ as it maps to $y_i$ and so should be in $U_i$ by $F_i \subseteq U_i$, a contradiction with $p \in X \setminus U_i$. This contradiction shows that $y_i \in O_i$.

$O_1 \cap O_2 = \emptyset$. Suppose that $y$ lies in both. Then for some $x \in X$, $f(x) = y$. As $U_1$ and $U_2$ are disjoint, $x$ must lie in $X\setminus U_1$ or in $X \setminus U_2$ (or both); suppose it lies in $X \setminus U_j$ for some $j \in \{1,2\}$. But then $y = f(x) \in f[X\setminus U_j]$ which means $y \notin O_j$, which is our contradiction.

So $O_1$ and $O_2$ are the required disjoint open neighbourhoods of $y_1$ resp. $y_2$.

Sanity check: we used the closedness of $f$ and its Hausdorffness too, (also to get normality from the combination with compactness), and the continuity of $f$ and its surjectivity. The proof would also have worked if $X$ would have been merely $T_4$ (normal and $T_1$) and $f$ the same (continuous closed surjection).

Answer 3

Let $y_1 \neq y_2$ in Y. Pick $x_1 ,x_2$ such that $f(x_1)=y_1$ and $f(x_2)=y_2$ Claim: there exist neighborhoods U and V of $x_1$ and $x_2$ such that $f(U) \cap f(V)$ is empty. [ This is where continuity is used]. If not, for any neighborhoods U and V there exist points $t_{UV}$, $s_{UV}$ in U and V such that $f(t_{UV})=f(s_{UV})$ Order the pairs (U,V) by $(U,V) \geq (S,T)$ if U is a subset of S and V is a subset of T. We get nets $\{t_{UV}\}$ and $\{t_{UV}\}$ converging to $x_1$ and $x_2$ respectively. Since f is continuous and $f(t_{UV})=f(s_{UV})$ we get $y_1=f(x_1)=f(x_2)=y_2$ which is a contradiction. This proves the claim. Next pick open sets A, B containing $x_1$ and $x_2$ such that their closures are contained in U and V respectively. Now the complements of the images of the closures of A and B are disjoint neighborhoods of $y_1$ and $y_2$.