Question

I was recently trying to learn a little bit about group cohomology, but one point has been confusing me. According to wikipedia (http://en.wikipedia.org/wiki/Group_cohomology and some other sources on the internet), given a (topological) group $G$, we have that the group cohomology $H^n(G)$ is the same as the singular cohomology $H^n(BG)$ (with coefficients in a trivial $G$-module $M$). Moreover, it says that given any group $G$, if we don't care about its topology, we can always give it the discrete topology and look at the cohomology of $K(G,1)$. This seems to suggest that when $G$ has topology that we do care about, we can just look at $BG$ with whatever topology $G$ is supposed to have. The relevant citation in this section is a reference to a book called Cohomology of Finite Groups, but I was wondering if this result would work with groups such as $U(1)$ which are not finite? Moreover, it would seem then that there is some sort of natural way to define group cohomology to detect the topology of the group; for example maybe look at continuous $G$-modules and continuous cochains. However, I heard that when doing this, one has to be careful because in general, the category of continuous $G$-modules might not have enough injectives. Also, I found this article by Stasheff (http://projecteuclid.org/DPubS/Repository/1.0/Disseminate?view=body&id=pdf_1&handle=euclid.bams/1183540920) which seems to suggest that for continuous cohomology, the we might not have the equality $H^n(BG)=H^n(G)$ between singular and group cohomology. I was wondering if someone could explain these connections to me (including what "continuous cohomology" is) and/or clarify what is happening? It would also be great if someone could tell me how one might compute something like $H^n(U(1);M)$ where $U(1)$ carries the discrete topology. Thanks.