# Computing the homology of $\mathbb{S}^2$ via Mayer-Vietoris

I'm trying to compute the homology of the $2$-sphere. I start by decomposing the sphere into a northern hemisphere and southern hemisphere, denoted by $A$ and $B$, respectively, and allow these two to overlap, $A \cap B = \textrm{cylinder} \simeq \mathbb{S}^1$. Now, I feed this into the Mayer-Vietoris sequence to get a long exact sequence $$\cdots\rightarrow H_{n+1}\big( \mathbb{S}^2\big) \rightarrow H_n \big(\mathbb{S}^1\big) \stackrel{\alpha_n}{\rightarrow } H_n\big(\mathbb{D}^2\big) \oplus H_n \big(\mathbb{D}^2\big) \rightarrow H_n\big(\mathbb{S}^2\big) \rightarrow H_{n-1} \big(\mathbb{S}^1\big) \stackrel{\alpha_{n-1}}{\rightarrow}\cdots$$

where $\alpha_n$ is the map generated by the inclusions of $A\cap B$ into $A$ and into $B$, the usual Mayer-Vietoris setup. Now, I know that we have a short exact sequence given by $$0\rightarrow \textrm{coker}\big(\alpha_n\big) \rightarrow H_n\big(\mathbb{S}^2\big) \rightarrow \ker \big(\alpha_{n-1}\big) \rightarrow 0\,,$$ for all $n$. So, one needs to understand the inclusion maps. But these are given by the identity inclusion since the boundaries of A and B are circles and oriented the same as $A\cap B \simeq \mathbb{S}^1$. Hence, the map in homology should be the identity and we see that $\alpha_n$ can be represented by the matrix $$\alpha_n=\left( \begin{array}{c} 1 \\ 1 \end{array} \right) \sim \left( \begin{array}{c} 1 \\ 0 \end{array} \right)$$

So, now I can see the kernel of this map is 0, yet I get $$\textrm{coker}\big(\alpha_n\big) = \Big(H_n \big(\mathbb{D}^2\big)\oplus H_n \big(\mathbb{D}^2\big)\Big) \Big/ H_n \big(\mathbb{S}^1\big)$$ and I don't see how this will be equal to $\mathbb{Z}$ when $n=2$. I'm sure I've messed up somewhere, but can't seem to spot it. Does anyone have any ideas? Thanks.

• Note that $H_{n}D^{2}=0$ for $n>0$. Commented Mar 27, 2013 at 4:56
• Although this is actually another topic, I calculated the homology of the $S^n$ by Mayer-Vietoris here: math.stackexchange.com/questions/1482227/… Commented Oct 19, 2015 at 16:47

I don't understand the part that you said about the inclusion maps and the matrix. So you need to calculate ker $\alpha_{n-1}$ to know $H_nS^2$ from the short exact sequence?
However, Start with the Mayer-Vietoris sequence from $H_2 D^2 \oplus H_2D^2$, you have $0\rightarrow H_2S^2\rightarrow H_1 S^1 \rightarrow 0$, and you know $H_1S^1$ ?