- #71

jbergman

- 406

- 164

Ahh, I finally get it!fresh_42 said:You were done when you showed that ##m'## is continuous. This means that ##m'^{-1}(U)## is open. It is the definition of a continuous function.

To show that ##\pi## is open, take an open set ##V## and calculate ##\pi(V)=VK=\ldots .##

$$\pi^{-1}(\pi(V)) = \pi^{-1}(VK) \bigcup_{k_i \in K} k_iV$$

Which is just the union of open sets, again using left multplication is a homeomorphism.

Well, my argument to show that ##m'## is continuous is essentially the same for the inverse map ##i'## and then from 5.1 we know that is equivalent to showing that ##\phi## is continuous.fresh_42 said:We have defined a topological group by continuity of ##(g,h)\mapsto gh^{-1}##. So you have to show that ##\varphi ## as defined in post #63 is continuous.

Briefly,

##\pi(i(g)) = i'(\pi(g))## and again the left is continuous so the right side must be, and since ##\pi## is surjective and continous, ##i'## must also be.