Idiot moment. Of course it's true.
and are both continuous. So, of course any topological property invariant under continuous mappings transfers from to and
Sorry for the hiccup
I came up with the following question, and would appreciate if I could get some validation/invalidation.
If is connected Hausdorff and continuous then is connected? ( is the graph of ). I think the answer is yes.
Proof: We first need a lemma
Lemma: Let be connected Hausdorff, then is connected ( is the diagonal).
Proof: Clearly the map given by is continuous. But, it is not hard to see that . But, connectedness is invariant under continuous mapping. So, the conclusion follows
Now, it is easy to prove that the product of two continuous maps is continuous. In particular, since and the identity map are continuous we have that is continuous. I now claim that .
To see this, let then for some . Clearly then and so
Conversely, let . Then, for some but that means that and so .
Recalling the lemma and that connectedness is invariant under continuous mappings finishes the argument.
Is that right? I feel as though I am making a stupid mistake...especially because I realized that I didn't use Hausdorffness anywhere.
What I'm talking about is that usually one writes a function with and the correspondence relation with (not trying to impose, but it really bugs me out...I really need to relax a bit with notation...)