Suppose $(Y,\tau')$ is Hausdorff and a function $f : (X,\tau)\to (Y,\tau')$ Suppose $(Y,\tau')$ is Hausdorff and a function $f:(X,\tau)\to (Y,\tau')$ is a bijection such that $f^{-1}$ is continuous.
Can you show that $(X, \tau)$ is Hausdorff?
 A: Recall that:


*

*$f$ is continuous if $f^{-1}(U)\in\tau$ for all $U\in\tau'$; 

*and that $(X,\tau)$ is Hausdorff if and only if for every $a,b\in X$ there are $U,V\in\tau$ such that $a\in U, b\in V$ and $U\cap V=\varnothing$.


Hint:
Now, take $a,b\in X$. Since $f$ is a bijection $f(a)\neq f(b)$. In $Y$ find two disjoint open sets one which contains $f(a)$ and the other containing $f(b)$, and show that their preimage have to be disjoint, and open by continuity.

As stated with $f^{-1}$ continuous rather than $f$ continuous this is false.
Consider $X=Y=\mathbb N$, $\tau=\{\varnothing, X\}$ and $\tau'=\mathcal P(X)$, and take $f(x)=x$ the identity function. Certainly $f$ is a bijection, and certainly $f^{-1}$ is continuous, because there are only two open sets in $\tau$ and both are open in $\tau'$. 
However $(X,\tau)$ is as far from Hausdorff as possible, and $(Y,\tau)$ is as Hausdorff as it gets.
A: Corrected: Let’s see what happens if you follow your nose and use the definitions. To show that $\langle X,\tau\rangle$ is Hausdorff, you need to show that if $p,q\in X$ and $p\ne q$, then there are disjoint sets $U_p,U_q\in\tau$ such that $p\in U_p$ and $q\in U_q$. The only possible source of information about $X$ is $Y$, so look at $f(p)$ and $f(q)$; $f$ is a bijection, so $f(p)\ne f(q)$, and $\langle Y,\tau'\rangle$ is Hausdorff, so there are disjoint $V_p,V_q\in\tau'$ such that $f(p)\in V_p$ and $f(q)\in V_q$. Now you’d like to use $V_p$ and $V_q$ to get $U_p$ and $U_q$. Unfortunately, the only obvious way to try to do this is to set $U_p=f^{-1}[V_p]$ and $U_q=f^{-1}[V_q]$, and it doesn’t work: we know that $f^{-1}$ is continuous, but we have no reason to think that it is an open map, meaning one that takes open sets to open sets.
Of course it’s conceivable that the result is true even though the obvious, natural approach doesn’t work, but in fact it’s easy to come up with counterexamples. The simplest one that I can think of is to let $X=Y=\{0,1\}$, $\tau=\{\varnothing,X\}$, and $\tau'=\{\varnothing,\{0\},\{1\},X\}$, and to let $f$ be the identity map: $f(0)=0$ and $f(1)=1$. You can easily check that all of the hypotheses are satisfied, but that $\langle X,\tau\rangle$ is definitely not Hausdorff.
If you replace the condition that $f^{-1}$ is continuous with the condition that $f$ is continuous, or with the equivalent condition that $f^{-1}$ is open, the result is true. (Note that these conditions are equivalent only under the assumption that $f$ is a bijection, not in general.)
