Computing the homology groups of a quotient space of the sphere I want to solve following question: Let $A$ denote the union of equatorial circle and the north pole on $S^2$. Let $X=S^2 / A$. Compute the homology groups of X. 
I calculated that $H_2(X) = \Bbb Z^2$, and all other homology groups are zero. Is this correct? If not, how do I solve this?
 A: As Stefan Hamcke has indicated in his comments, one can obtain $S^2/A$ step-by-step by first pinching the equator to a point to get $S^2 \vee S^2$ and then pinching the pole to the wedged point to have $S^2 \vee S^2 \vee S^1$. This is homotopy equivalent to our space. 
Hence, $\widetilde{H_n}(S^2/A) \cong H_n(S^2 \vee S^1 \vee S^1)$ which is $\Bbb Z^2$ when $n = 2$ as you have proved and $\Bbb Z$ for $n = 1$, not trivial.

Alternatively, use the long exact sequence at $\widetilde{H_1}(S^2/A)$ : $$\cdots \to \widetilde{H_1}(S^2) \to \widetilde{H_1}(S^2/A) \to \widetilde{H_0}(A) \to \widetilde{H_0}(S^2) \to \cdots$$
$\widetilde{H_1}(S^2) \cong \widetilde{H_0}(S^2) \cong 0$, hence $\widetilde{H_1}(S^2/A) \cong \widetilde{H_0}(A)$. Since $A = S^1 \sqcup pt$, $\widetilde{H_0}(A) \cong \Bbb Z$. Thus, $H_1(S^2/A) \cong \Bbb Z$.
Similarly, we look at the long exact sequence at $\widetilde{H_2}(S^2/A)$ : $$\cdots \to \widetilde{H_2}(A) \to \widetilde{H_2}(S^2) \to \widetilde{H_2}(S^2/A) \to \widetilde{H_1}(A) \to \widetilde{H_1}(S^2) \to \cdots$$
$\widetilde{H_2}(A) \cong \widetilde{H_1}(S^2) \cong 0$, and $\widetilde{H_2}(S^2) \cong \widetilde{H_1}(A) \cong \Bbb Z$, so $\widetilde{H_1}(S^2/A)$ is an extension of $\Bbb Z$ by $\Bbb Z$, and indeed a split extension, which proves $\widetilde{H_2}(S^2/A) \cong \Bbb Z^2$ as you have done. 
