TITLE: Homology of $S^1 \times (S^1 \vee S^1)$
QUESTION [14 upvotes]: I'm trying to solve question 2.2.9(b) in Hatcher's Algebraic Topology. 
Question: Calculate the homology groups of $X=S^1 \times (S^1 \vee S^1)$.
My attempt: I try to use the Mayer-Vietoris sequence. 
Let $A=S^1 \times (S^1 \vee \text{small bit}) =S^1\times S^1$ and $B=S^1 \times ( \text{small bit} \vee S^1)=S^1 \times S^1$. (Unsure how to express this, but hopefully this is clear enough.) 
Then $A\cap B=S^1\times \{\text{point}\}=S^1$ and $A\cup B=X$.
The Mayer-Vietoris sequence then gives us an exact sequence in reduced homology as follows:
$$0 \to \mathbb{Z}^2 \to \tilde{H}_2(X) \to \mathbb{Z} \to \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$
Is it possible from this exact sequence to determine the homology groups, in some algebraic fashion? Or do I need to determine what the maps are? How do I determine the maps?

REPLY [4 votes]: I have yet another solution, using cellular homology directly. 
Define a CW-structure on $X$ with one 0-cell $e^0$, three 1-cells $a, b, c$, and two 2-cells $U,L$. Attach the 1-cells to $e^0$ so that $X^1 = S^1 \vee S^1 \vee S^1$, and attach the 2-cells via the relations so $\partial U \mapsto aba^{-1}b^{-1}$ and $\partial L \mapsto cac^{-1}a^{-1}$. (In the plane, this looks like a horizontally bisected rectangle with the horizontal edges & bisecting line identified in the same direction, and the opposite vertical edges identified; i.e. two flat tori glued along their top and bottom edges.) Then the nontrivial relative homology groups $H_2(X, X^1)$, $H_1(X^1, e^0)$, and $H_0(e^0)$ form the long exact sequence
$$ 0 \to \mathbb Z^2 \xrightarrow{d_2} \mathbb Z^3 \xrightarrow{d_1}\mathbb Z \to 0.$$ Since there is only one 0-cell, the cellular boundary map $d_1 = 0$ (see Hatcher, paragraph before the Cellular Boundary Formula). Meanwhile one directly computes $d_2(U) = a + b - a - b = 0$ and $d_2(L) = c + a - c - a = 0$, so $d_2 = 0$. Since the sequence is exact, we get $H_2(X) = \mathbb Z^2$, $H_1(X) = \mathbb Z^3$, $H_0(X) = \mathbb Z$, and all other homology groups are $0$.