On an exercise from Hatcher What is the homology group of $S^{1}\times (S^{1}\vee S^{1})$? $\vee$ denotes wedge sum. Problem 9 sec 2.2.I was trying to use cellular homology, but not able to understand the CW complex structure of this space and maps d_{n}? If someone could give a hint how to proceed, it can help me in learning the application of cellular homology. Thanks in advance.
 A: Hints:


*

*The homology of a wedge sum is the direct sum of homologies.

*Generically, the Kunneth formula is used to compute homology of Cartesian products.

*$S^1 \vee S^1$ is isotopic to "$8$".

*$S^1 \times (S^1 \vee S^1)$ is   
and a cutaway to see the "8" (although rotated to an "$\infty$" in this embedding in $\Bbb{R}^3$).  

*as a cellular decomposition, this object has a $0$-cell, three $1$-cells and two $2$-cells.  The $2$-cells are assigned different colors in the images.

*if you label the identifications needed in the cutaway to produce the completed image, you will have specified enough information to determine the labels and orientations of the four $1$-cells on the boundary of each $2$-cell.

A: So the cellular structure isn't too complicated but perhaps it is good to first get a mental image of what the space looks like.
If $S^1 \times S^1$ is a torus and $S^1 \vee S^1$ is a figure-8, then $S^1 \times (S^1 \vee S^1)$ would be a torus constructed out of a figure-8. That looks like two toruses stacked on top of each other and glued along a common circle. The key word here is "glued." That suggests you use Meyer-Vietoris.
You can also use the Künneth theorem which describes the homology of a product of two spaces, but I don't remember exactly where Hatcher covers that. But I think he covers Meyer-Vietoris early on. You can probably figure out the homology without Meyer-Vietoris. For instance, for $H_1$, each torus has two independent loops but you identify two of these loops together when you glue the toruses. So $H_1 \cong \mathbb{Z}^3$.
So going back to Meyer-Vietoris, you have two toruses, $T_1, T_2$ glued along a common circle $S = T_1 \cap T_2$. Recall (in reduced homology):
$$ 0 \to H_2(T_1) \oplus H_2(T_2) \to H_2(T_1 \cup T_2) \to H_1(T_1 \cap T_2) \to H_1(T_1) \oplus H_1(T_2) \to H_1(T_1 \cup T_2) \to 0 $$
Hopefully you already know what $H_i(T_1 \cap T_2)$ is for $i \ne 1,2$. So we just need to focus on the other maps. The key is this: if $\alpha_i, \beta_i$ are the two generators of $H_1(T_i)$ then the gluing has $S = \alpha_1 = \alpha_2$.
The key map here is $H_1(S) \to H_1(T_1) \oplus H_1(T_2)$. Recall that this takes an element $x$ of $H_1(S)$ to its images inside of $T_1$ and $T_2$ respectively. Since $H_1(S)$ is generated by $[S]$ and the image of $[S]$ inside of $H_1(T_i)$ is $\alpha_i$, the map $H_1(S) \to H_1(T_1) \oplus H_2(T_2)$ is given by $[S] \mapsto (\alpha_1, \alpha_2)$. For the next part, I will write this ordered pair as a sum $\alpha_1 + \alpha_2$.
Using the fact that this map is injective, you obtain $H_2(T_1 \cup T_2)$.
Using what you know about the image, you obtain $$H_1(T_1 \cup T_2) \cong \frac{H_1(T_1) \oplus H_1(T_2)}{{\rm im}(H_1(S) \to H_1(T_1) \oplus H_1(T_2))} \cong \frac{\mathbb{Z} \cdot \{\alpha_1,\alpha_2,\beta_1,\beta_2\}}{\mathbb{Z} \cdot(\alpha_1 + \alpha_2)}.$$

If you want to consider the cellular structure of this space, take the cell structure of each torus $T_1, T_2$. Let's say the simplest one where you have one edge for $\alpha_i$, one for $\beta_i$ intersecting at a common point and having just one face. Then you combine these two cell structures by gluing the $\alpha_1$ edge to the $\alpha_2$ edge.
That gives you two faces, three edges and one vertex. These will generate $H_2, H_1, H_0$ respectively.
