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Jan 27, 2011 at 17:11 comment added Johannes Ebert Thats a good habit, but the universal coefficient theorem does not work for all sorts of coefficients. In my example, if you take $ZG$ as coefficients, then the map won't induce an isomorphism, because $H_* (EG \times_G K(V;n); ZG) \cong H_* (K(V;n)) = V$; and my map induces $0$ on $V$.
Jan 27, 2011 at 16:38 comment added Oscar Randal-Williams Ah, my mistake. I have a habit of understanding "with coefficients" as "with local coefficients".
Jan 27, 2011 at 15:33 comment added Johannes Ebert There is a theorem as follows: $f:X \to Y$ a $\pi_1$-isomorphism and an isomorphism for homology with all local coefficient systems, then $f$ is a weak homotopy equivalence. This follows from the natural isomorphism H_p(X,Z\pi_1 (X))\cong H_p (\tilde{X}).
Jan 27, 2011 at 15:11 comment added Oscar Randal-Williams The homology with coefficients will not be the same: one can see this already in your example, taking the coefficient system given by one of the representations.
Jan 27, 2011 at 13:55 history edited Johannes Ebert CC BY-SA 2.5
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Jan 26, 2011 at 21:31 history answered Johannes Ebert CC BY-SA 2.5