# Homeomorphism between spaces equipped with cofinite topologies

I was given this question on my midterm. Currently I am studying for finals and am still unsure how to properly solve this question.

Let X and Y be two sets and f be a map from X to Y be a bijection. Prove that, when we consider X and Y with their respective T1-topologies (cofinite topology), this map is a homeomorphism.

I know that I must show that f is continuous in order to show X and Y are homeomorphic. On the midterm I tried showing that open sets in Y have preimage that are open in X and closed sets in Y have preimage that are closed in X. However I'm unsure how to approach the sets in X,Y that are neither open nor closed.

So my question is, would it be easier to show continuity by first showing local continuity at every point in X and using that to prove the continuity of f? Or is this the wrong way to approach this problem?

• What is the $T_1$-topology? Commented Apr 22, 2014 at 17:34
• Ya, the cofinite topologies. Sorry for forgetting to mention that, I'll add that in. Commented Apr 22, 2014 at 17:35
• Then you just need to show that the preimage of a finite set is finite. Commented Apr 22, 2014 at 17:35
• Isn't that the same as showing the preimage of closed sets in Y are closed sets in X? Because my professor that marked this, wrote a comment talking about the sets that are neither open or closed in Y. So it suffices to say that I am required to talk about them as well? Commented Apr 22, 2014 at 17:37
• @StefanHamcke In the eyes of my professor it does not suffice to show that the preimage of a finite set is finite proves continuity. Commented Apr 22, 2014 at 17:53

In general, given topological spaces $(X, \tau)$ and $(Y, \sigma)$, a bijection $f\colon X\to Y$ between the underlying sets is a homeomorphism $(X, \tau)\to(Y, \sigma)$ if and only if it is continuous and closed (i.e. it sends closed subsets of the domain into closed subsets of the codomain). This is clear, since $f$ being closed is equivalent to $f^{-1}$ being continuous.
Let us then apply this observation to our concrete case. Here $\tau$ is the cofinite topology on $X$ and $\sigma$ is the cofinite topology on $Y$. Therefore closed subsets in both cases are just finite subsets. If $B\subseteq Y$ is a finite set, then also $f^{-1}(B)$ needs to be so, because $f$ is (bijective hence) injective. So $f^{-1}(B)$ is closed, thus $f$ is continuous. In addition, if $A\subseteq X$ is a finite subset of $X$, $f(A)\subseteq Y$ is certainly finite (note that this is true for each function $f\colon X\to Y$, even if it is not a bijection). Thence, $f$ is closed as well and we are done.
• @tamefoxes I don't see the problem. Given topological spaces $(X, \tau)$ and $(Y,\ \sigma)$ a map $f\colon X\to Y$ is continuous, by definition, if and only if the preimage (under $f$) of every open subsets in $(Y, \sigma)$ is open in $(X,\ \tau)$. This is equivalent to say that the preimage of every closed subset in $(Y, \sigma)$ is closed in $(X,\ \tau)$, by set theory. Therefore, showing that a map is continuous can be done either by proving the continuity property on open subsets or on closed subsets. That's it, nothing else matters... Commented Apr 22, 2014 at 18:09