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.


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

Thanks in advance!


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.

  • $\begingroup$ I read your answer, but I cannot understand what is wrong with my proof. Could you please explain? $\endgroup$ – Pedro Gomes Nov 3 '18 at 16:54
  • $\begingroup$ 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. $\endgroup$ – José Carlos Santos Nov 3 '18 at 16:57
  • $\begingroup$ 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! $\endgroup$ – Pedro Gomes Nov 3 '18 at 17:49
  • $\begingroup$ 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. $\endgroup$ – 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.


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