homology of 3.

We need for this problem knowledge of the homology groups of S2 and 2. We will simply assume the former:

Hk(S2;) ={k=0,20else

Now, for 2, we can argue without Mayer-Vietoris. X=2 is connected, so H0(X;)=. X is non-orientable, so H2(X;) is 0. Last, H1(X;) is the abelianizationMathworldPlanetmath of the already abelianMathworldPlanetmath fundamental groupMathworldPlanetmathPlanetmath π1(X)=/2, so we have:

Hk(2;) ={k=0/2k=10k2

Now that we have the homology of 2, we can compute the homology of 3 from Mayer-Vietoris. Let X=3, V=3\{pt}2 (by vieweing 3 as a CW-complexMathworldPlanetmath), UD3{pt}, and UVS2, where denotes equivalence through a deformation retractMathworldPlanetmath. Then the Mayer-Vietoris sequence gives