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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?

Thanks in advance!

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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.

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  • $\begingroup$ 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
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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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