# Proving a space is Hausdorff if $f$ is continuous and bijective

Exercise: Let $$(X,\tau_1)$$ and $$(Y,\tau_2)$$ be topological spaces and $$f:(X,\tau_1)\to (Y,\tau_2)$$ a continuous map. If $$f$$ is one-to-one, prove that $$(Y,\tau_2)$$ Hausdorff implies $$(X,\tau_1)$$ Hausdorff.

Attempted proof:

As $$f$$ is continuous and bijective $$\exists \ x_1,x_2$$ such that $$x_1\neq x_2$$ then $$f(x_1)=y_1$$ and $$f(x_2)=y_2$$ and $$y_1\neq y_2$$.

As $$(Y,\tau_2)$$ is Hausdorff $$\exists \ U,V\in\tau_2$$ such that $$y_1\in U$$ and $$y_2\in V$$ and $$U\cap V=\emptyset$$. As $$f$$ is continuous and bijective $$f^{-1}(y_1)=x_1\in f^{-1}(U)\in\tau_1$$ and $$f^{-1}(y_2)=x_2\in f^{-1}(V)\in\tau_1$$.

By the fact that $$(Y,\tau_2)$$ is Hausdorff and $$\emptyset$$ is open.

$$\emptyset=f^{-1}(U\cap V)=f^{-1}(U)\cap f^{-1}(V)=$$ such that $$y_1\in f^{-1}(U)$$ and $$y_2\in f^{-1}(V)$$. Then $$(X,\tau_1)$$ is Hausdorff.

Questions:

Is my proof right? If not. Why? What could be alternative proofs?

This is wrong from the start. What you are supposed to prove is that for any two distinct elements $$x_1,x_2\in X$$, there are open sets $$U_1$$ and $$U_2$$ such that $$x_1\in U_1$$, $$x_2\in U_2$$, and $$U_1\cap U_2=\emptyset$$. Take open sets $$V_1,V_2$$ in $$Y$$ such that $$f(x_1)\in V_1$$, $$f(x_2)\in V_2$$ and $$V_1\cap V_2=\emptyset$$; such sets exist, since $$Y$$ is Hausdorff and $$f$$ is injective. Now, let $$U_1=f^{-1}(V_1)$$ and let $$U_2=f^{-1}(V_2)$$. Then $$x_1\in U_1$$, $$x_2\in U_2$$ and, since $$f$$ is continuous, $$U_1$$ and $$U_2$$ are open. And, since $$V_1\cap V_2=\emptyset$$, $$U_1\cap U_2=\emptyset$$ too.

Note that the fact that $$f$$ is surjective is irrelevant.

• I read your answer, but I cannot understand what is wrong with my proof. Could you please explain? – Pedro Gomes Nov 3 '18 at 16:54
• Right in the first sentence you write “$\exists x_1,x_1$ such that…”, but the goal is not to prove that such elements $x_1$ and $x_2$ exist. It is to prove that whenever you have two distinct elements of $X$, there are open sets containing them that don't intersect. – José Carlos Santos Nov 3 '18 at 16:57
• After analysing your answer and mine I understood the point in your comment. I would like to know if there was any other error. Thanks in advance! – Pedro Gomes Nov 3 '18 at 17:49
• There is another error, but a minor one. You wrote “By the fact that $(Y,\tau_2)$ is Hausdorff and $\emptyset$ is open…”, but then you used none of these facts. – José Carlos Santos Nov 3 '18 at 18:42

I think your ideas are correct but imprecise. We ought to prove that each pair of distinct points of $$X$$ have disjoint neighbourhoods.

If $$x_1, x_2 \in X$$ are distinct points, so are $$f(x_1)$$ and $$f(x_2)$$. Since $$Y$$ is Hausdorff, we have open disjoints sets $$U_i \ni f(x_i)$$, and thus $$V_i = f^{-1}(U_i)$$ are open and disjoint, with $$x_i \in V_i$$. Hence $$X$$ is Hausdorff.

Another approach: a space $$Z$$ is Hausdorff if and only if the diagonal

$$\Delta_Z = \{(z,z) : z \in Z \} \subseteq Z \times Z$$

is closed. In your case,

$$\Delta_X = \{(x,x) : x \in X\} = \{(x,x') \in X \times X : f(x) = f(x')\}$$

because $$x = x'$$ if and only $$f(x) = f(x')$$, since $$f$$ is bijective. Thus, $$\Delta_X$$ is the preimage of the (by hypothesis) closed set $$\Delta_Y$$ via the continuous function $$(f \times f) : X \times X \to Y \times Y$$,

$$\Delta_X = (f \times f)^{-1}(\Delta_Y)$$

and so it is closed, which says that $$X$$ is Hausdorff.

As José Carlos Santos notes in his answer, surjectivity is not needed. After all, we have only needed some bijective correspondence between $$X$$ and $$f(X)$$, and since $$f(X) \subseteq Y$$ is Hausdorff as well, we can let go of the surjectivity hypothesis.