Thread: Topological Groups

Can anybody help me with this question

Given topological groups X and Y and homomorphism f:X->Y, verify the following are equivalent:
(a) f is an open map
(b) for each neighborhood N of e(x), f(N) is a neighborhood of e(y)
(c) for each open neighborhood N of e(x), f(N) is a neighborhood of e(y)
(d) for each open neighborhood N of e(x), f(N) is an open neighborhood of e(y)

I already know that f is continuous which implies f is continuous at the identity e(x) of x

Originally Posted by Turloughmack

Can anybody help me with this question

Given topological groups X and Y and homomorphism f:X->Y, verify the following are equivalent:
(a) f is an open map
(b) for each neighborhood N of e(x), f(N) is a neighborhood of e(y)
(c) for each open neighborhood N of e(x), f(N) is a neighborhood of e(y)
(d) for each open neighborhood N of e(x), f(N) is an open neighborhood of e(y)

I already know that f is continuous which implies f is continuous at the identity e(x) of x

I helped you with the other one. Any ideas for this one?

(a) Well I know that to show f is an open map I need something like

For all G a subset of X , f(G) is open in Y.

(b) the identity e(x) is an element of G which is a subset of N. So f(e(x)) = e(y) which is an element of open f(G). This is however a subset of f(N) so it makes that f(N) is a neighborhood of e(y).

(c) I think this is implied from the previous answer but not sure how I would do it, maybe by finding an open neighborhood of e(x)?

(d) To solve this I'd have to prove that f(N) is an open neighborhood which would be implied form the previous answer.

Am I going about the right way to solve these 4 questions?