Prove that $A$ is closed. let $f:X\to Y$ continuous, open and onto. Then $ Y $ is Hausdorff if and only if the set $ \{(x, y): f (x) = f (y) \} $ is closed. I already did the one for $ Y$ is Hausdorff so the set is closed. Someone help me to come, I don't know how to do it.
Let $A=\{(x, y): f (x) = f (y) \}$. What I will prove is that the $ A'$ is open. Let $ x, y \in A'$. So, $ f (x) \neq f (y) $, since $ Y $ is $ T_2 $, then we have $ U_ {f (x)} $ and $ V_ {f (y)} $ two open such that $ U_ {f (x)} \cap V_ {f (y)} = \varnothing. $ Now, $$f^{-1}(U_ {f (x)}) \cap f^{-1}(V_ {f (y)})=f^{-1}(U_ {f (x)} \cap V_ {f (y)} )=f^{-1}(\varnothing)=\varnothing $$
From there we will already obtain that $ A '$ is open, so $ A $ is closed. The other implication is missing.
 A: Supposed that $A$ is closed and $u\ne v$, $u,v\in Y$. Since $f$ is onto, then there exist $x,y\in X,$ such that $u=f(x)$ and $v=f(y)$.
Note that the open subsets of $Y\times Y$, in the standard product topology, are exactly the arbitrary unions of products of open subsets of $Y$.
Since $\big(f(x),f(y)\big)\in Y\times Y\setminus A$, which is open, there exist $U,V\subset Y$, open, such that
$$
\big(f(x),f(y)\big)\in U\times V\subset Y\times Y\setminus A.
$$
Clearly $u=f(x)\in U$ and $v=f(y)\in V$.
If $z\in U\cap V$, then $(z,z)\in U\times V$ and hence $(z,z)\in Y\times Y\setminus A$. Contradiction.
Therefore, $Y$ is a Hausdorff space.
A: For a more compact proof, note that $Y$ is Hausdorff if and only if the "complement of diagonal" $E = \{(y_1, y_2) \in Y^2: y_1 \neq y_2\}$ is open in $Y \times Y$.
The continuous map $f$ induces a continuous map $g: X \times X\rightarrow Y \times Y$ such that $g(x_1, x_2)=(f(x_1), f(x_2))$. Since $f$ is open and onto, so is $g$.
Now the set $\{(x_1, x_2) \in X^2: f(x_1) \neq f(x_2)\}$ is nothing but the inverse image $g^{-1}(E)$. Thus the proposition becomes: $E$ is open if and only if $g^{-1}(E)$ is open.
In one direction, if $E$ is open, then $g^{-1}(E)$ is open because $g$ is continuous.
In the other direction, we have $E = g(g^{-1}(E))$ because $g$ is onto. Thus if $g^{-1}(E)$ is open, then $E$ is open because $g$ is open.
A: Suppose $A$ is closed and let $y_1 \neq y_2$ in $Y$. By ontoness there are $x_1, x_2 \in X$ so that $f(x_1)=y_1, f(x_2)= y_2$. By definition $(x_1,x_2) \in X^2\setminus A$ and as $A$ is closed, there is a basic open subset $U \times V$ of $X^2$ so that $$(x_1, x_2) \in U \times V \subseteq X^2\setminus A\tag{1}$$
Now if $z \in f[U] \cap f[V]$ there would be $x_3 \in U, x_4 \in V$ so that $z=f(x_3)=f(x_4)$ but then $(x_3,x_4) \in (U \times V)\cap A$ would contradict the inclusion in $(1)$. So $f[U] \cap f[V]=\emptyset$.
By openness of $f$, $f[U]$ and $f[V]$ are the required disjoint open neighbourhoods of $y_1, y_2$ resp. and $Y$ is $T_2$.
