I'm preparing for an exam. I want to solve this past exam question:
Define three-dimensional torus by $T^3 = I^3/[(x,y,0)\sim (x,y,1), (x,0,z)\sim (x,1,z), (0,y,z)\sim (1,y,z)]$ and then other three-dimensional manifolds by $K^3 = I^3/[(x,y,0)\sim (1-x,y,1), (x,0,z)\sim (x,1,z), (0,y,z)\sim (1,y,z)]$.
I want to compute the homology groups of $T^3$ and $K^3$. For $T^3$, I think Hatcher does the the proof in page 142 and 143 of his book. Is the proof applicable for this question? More specifically, am I not supposed to use Mayers-Vitoris for this ? Also, on the same page 143, the homology groups are computed for $K\times S^1$, is this same as $K^3$? I think I may be confusing a bunch of stuff with each other.
I'd really appreciate if someone could write a solution for this especially for $K^3$
Note that: $$T^{3} \cong S^1 \times S^1 \times S^1,$$ and $$K^3 \cong K \times S^1 $$ (as you had correctly noted.) To prove the second fact note that the image of the plane $\{y=0\}$ is a Klein bottle. Then, one may check that the map $K^3 \rightarrow K \times S^1$ given by $[(x,y,z)] \mapsto ([x,0,z],[y])$ is a diffeomorphism.
(but I think that in reality it is more instructive to draw a picture of a cube and see that the subsets $U_{K} = \{[(x,k,y)] : x,z \in [0,1] \}$ are Klein bottles, and indeed when the Klein-bottles are glued together there is no "twisting" in the $y$ direction i.e. we are just taking $K \times \mathbb{R}$ and just quotienting by the map $(p,t) \mapsto (p,t +1)$, so it is natural that we get a direct product with the circle).
Now the computation of the homology manifolds follows from the Kunneth Formula (i.e. the product formula for homology). The computations of the homology of the circle and the Klein bottle are easy to find.