Question about quotient of a compact Hausdorff space

I am reading the book 'Algebraic Topology' by Tammo Tom Dieck. On page 12 in the proposition 1.4.4 he states that :

Let $X$ be a compact Hausdorff space and $f : X \rightarrow Y$ be a quotient map. Then the following assertions are equivalent : (1) Y is a Hausdorff space, (2) $f$ is closed, (3) $R=\{ (x_1,x_2)|f(x_1)=f(x_2)\}$ is closed in $X \times X$.

I am able to prove that (1) implies (2) and that (1) implies (3) but not able to prove the other implications. I will appreciate any help.

Thinking about this question, a related issue comes up. We all know that compact subsets of Hausdorff spaces are closed. Is it true that if all compact subspaces of a space are closed then the space is Hausdorff ?

• See this for your last question. Commented Jan 6, 2015 at 19:22
• You can also look at my related question here. Another related question here. Commented Jan 6, 2015 at 21:09

As noted in the comments, this answer shows that $f$ is closed if $R$ is closed in $X\times X$, i.e., that (3) implies (2). Now suppose that $f$ is closed. Note first that since $X$ is Hausdorff, $\{x\}$ is closed for each $x\in X$, and therefore $\{f(x)\}$ is closed for each $x\in X$. And $f$ is a surjection, so $\{y\}$ is closed for each $y\in Y$.

Now let $y_0$ and $y_1$ be distinct points of $Y$, and let $F_i=f^{-1}[\{y_i\}]$ for $i=0,1$; $F_0$ and $F_1$ are disjoint closed sets in $X$. $X$, being compact Hausdorff, is normal, so there are disjoint open sets $V_i$ for $i=0,1$ such that $F_i\subseteq V_i$. For $i=0,1$ let $K_i=X\setminus V_i$, and let $W_i=X\setminus f^{-1}[f[K_i]]$; $f$ is closed and continuous, so $W_i$ is open. It’s easy to check that

$$F_i\subseteq W_i=f^{-1}[f[W_i]]\subseteq V_i$$

for $i=0,1$ and hence that $f[W_0]$ and $f[W_1]$ are disjoint open nbhds of $y_0$ and $y_1$, respectively. Thus, $Y$ is Hausdorff, and (2) implies (1). Since you’ve already shown that (1) implies (2) and (3), the proof that all three are equivalent is complete.

• why is $F_i \subseteq W_i$ and $f^{-1}(f(W_i))\subseteq V_i$ ?
– user228168
Commented Sep 5, 2016 at 11:15
• @SaunDev: If $x\in W_i$, then $f(x)\notin f[K_i]$, so $x\notin K_i$, and therefore $x\in V_i$. If $x\in F_i\setminus W_i$, then $f(x)\in f[K_i]$, so there is an $x'\in X\setminus V_i$ such that $f(x')=f(x)=y_i$. But there is no such $x'$, since $F_i\subseteq V_i$. Commented Sep 5, 2016 at 17:40
• @BrianM.Scott Why $f[W_{i}]$ is open? Thank you. Commented Jan 26, 2017 at 16:00
• @Mr.Lilly It's because $W_i = f^{-1}[f[W_i]]$ is open in $X$. Since $f : X \to Y$ is a quotient map, a subset $S$ of $Y$ is open if and only if $f^{-1}[S]$ is an open subset of $X$. Commented Jan 29 at 21:40