Ok, so you're using the theorem that if is Hausdorff, compact and continuous then is closed, I assume? But then, I don't understand the problem. You just proved that and since Hausdorffness is preserved under homeomorphism (in fact, it is preserved under open surjections with closed kernels) you know that is Hasudorff, no?b) is connected and compact, so also will be connected and compact. We must show at this point that is (Hausdorff). That is to prove that the set
is a closed .
where are the vertices of the square.
is closed because is . , are closed because they are closed subsets of a space and images of the compact of through continuous applications. is closed because it is a finite set of points in space . Since , then also will be closed, this means that is . We conclude that is closed.
Good, assuming your readership will understand that you are using the fact that since is connected there exists no nontrivial clopen sets.c) Then we know that if is an open set in (in our case) is said to be an open application if will be open. Then we take where as a general case assuming that none of the parties are empty. Similarly . At this point we compute and then and since and are closed the application can not be open. In fact, the application will be closed (if is closed).
Good job overall!