# Continuous mapping into a Hausdorff space

[=>]: Let $$y_1, y_2 \in Y, y_1 \neq y_2$$.
Since $$f$$ is surjective, $$y_1 = f(x_1), y_2 = f(x_2)$$ for some $$x_1,x_2 \in X$$.
Since $$Y$$ is Hausdorff, there exist in $$Y$$ open neighborhoods $$N(y_1), N(y_2)$$ of $$y_1,y_2$$ respectively, such that $$N(y_1) \cap N(y_2) = \emptyset$$.
Since $$f$$ is continuous, $$f^{-1}(N(y_1)) \text{ and } f^{-1}(N(y_2))$$ are open in $$X$$, necessarily disjointed and contain $$x_1, x_2$$, respectively.

It's here that I got stuck. My idea is to use what I know about $$f^{-1}(N(y_1)) \text{ and } f^{-1}(N(y_2))$$ to show that $$(X \times X)\backslash R$$ is open, hence $$R$$ is closed. But I don't know how to proceed.

[<=]: I might need some hints here also.

• Is this supposed to be some extension of the fact that any topological space $X$ is Hausdorff iff the diagonal is closed in $X \times X$ ? See here
– user636532
Commented Apr 25, 2019 at 10:48

You have to show that if $$Y$$ is Hausdorff, a certain subset of $$X\times X$$ is closed. So you are not going to begin by taking two distinct elements of $$Y$$.

Take $$(x_1, x_2)$$ in $$(X\times X)\setminus R$$, then by definition $$f(x_1)\neq f(x_2)$$, so you have two open neighborhoods $$U, V$$ of $$f(x_1)$$ resp. $$f(x_2)$$ in $$Y$$, such that $$U\cap V=\emptyset$$. Then, $$U^\prime:=f^{-1}(U)$$ and $$V^\prime:=f^{-1}(V)$$ are open neighborhoods of $$x_1$$ resp $$x_2$$, and you have that $$f(U^\prime)\cap f(V^\prime)=\emptyset$$. So, $$U^\prime\times V^\prime$$ is an open neighborhood of $$(x_1, x_2)$$ in $$(X\times X)\setminus R$$.

I let you try the second part.

• Is this supposed to be some extension of the fact that any topological space $X$ is Hausdorff iff the diagonal is closed in $X \times X$ ? See here
– user636532
Commented Apr 25, 2019 at 10:51
• Well, it is, with $f=1_X$ Commented Apr 25, 2019 at 11:08
• Oh is that what the hint means?
– user636532
Commented Apr 25, 2019 at 11:13
• @elidiot: thanks for the answer. Since we had not got to the part about product topology in our class yet, I was a bit hesitated with the claim that "$U' \times V'$ is an open neighborhood in $(X \times X) \backslash R$", although I had the hunch that this is the case. Could you briefly say which property is used here? Commented Apr 26, 2019 at 21:23
• Also, even if it's not stated in your answer, I think the conclusion is that "since every point of $(X \times X) \backslash R$ admits such an open neighborhood, $(X \times X) \backslash R$ is open"? But how do I know that $(U' \times V') \subset (X \times X) \backslash R$? Apologize if these questions seem too basic. Commented Apr 26, 2019 at 21:28

If $$Y$$ is Hausdorff, to show that $$R$$ is closed, we can show that $$R'=X \times X \setminus R$$ is open. Let $$(x_1, x_2)$$ be an arbitrary point in $$R'$$. Then by definition, this means that $$y_1=f(x_1) \not= f(x_2)=y_2$$. Because $$Y$$ is Hausdorff, this means there exist disjoint open sets $$U$$ and $$V$$ in $$Y$$ such that $$y_1 \in U$$ and $$y_2\in V$$. By continuity of $$f$$, we have that $$f^{-1}(U)$$ and $$f^{-1}(V)$$ are open in $$X$$ and they are disjoint so that $$f^{-1}(U) \times f^{-1}(V)$$ is open in $$R'$$ (Why is it open in $$R'$$? Because first $$f^{-1}(U) \times f^{-1}(V)$$ is open in $$X \times X$$ and being contained in $$R'$$ gives us that it is open in the subspace $$R'$$ of $$X \times X$$, which has open sets of the form $$E \times F \cap R'$$, where $$E \times F$$ is open in $$X \times X$$. We can also write $$f^{-1}(U) \times f^{-1}(V) \cap R'= f^{-1}(U) \times f^{-1}(V)$$ to see that it is open in $$R'$$.) and $$(x_1, x_2) \in f^{-1}(U) \times f^{-1}(V)$$. This means that $$R'$$ is the union of open sets hence open itself.

Conversely, assume that $$R$$ is closed in $$X \times X$$. Let $$f(x_1) = y_1, f(x_2)=y_2$$ be two distinct points in $$Y$$ (here we're using surjectivity of $$f$$). We can then say that $$(x_1, x_2) \in A \times B \subseteq R'$$, where $$A$$, $$B$$ are open in $$X$$. Then $$f(A)$$ and $$f(B)$$ are open sets in $$Y$$ (because $$f$$ is open) containing $$y_1$$ and $$y_2$$, respectively. We show that $$f(A) \cap f(B)$$ is disjoint. Suppose $$y \in f(A) \cap f(B)$$. Then $$y=f(a)=f(b)$$ for $$a \in A$$ and $$b\in B$$. Hence $$(a, b) \in R$$, which is a contradiction because $$A \times B \cap R = \emptyset$$.