What is the homology group of the sphere with an annular ring? I'm trying to compute the homology groups of $\mathbb S^2$ with an annular ring whose inner circle is a great circle of the $\mathbb S^2$.
space X
Calling this space $X$, the $H_0(X)$ is easy, because this space is path-connected then it's connected, thus $H_0(X)=\mathbb Z$
When we triangulate this space, it's easy to see that $H_2(X)=\mathbb Z$.
But I've found the $H_1(X)$ difficult to discover, I don't know the fundamental group of it, then I can't use the Hurowicz Theorem. I'm trying to find this using  the triangulation of it, but there are so many calculations.
I have the following questions:
1- How we can use Mayer-Vietoris theorem in this case? 
2-What is the fundamental group of this space? 
3- I know the homology groups of the sphere and the annulus, this can help in this case? 
I need help, please
Thanks a lot.
 A: You can deformation retract the annulus to the great circle, so $X$ is homotopy equivalent to the sphere and in particular has the same fundamental group and homology groups. This answers $(2)$ and $(3)$.
Mayer-Vietoris doesn't help you much here. Assuming you split $X = A \cup B$ where $A = S^2$ and $B$ is the annulus and $A \cap B = S^1$, the relevant terms are
$$H_1(A \cap B) \to H_1(A) \oplus H_1(B) \to H_1(X) \to H_0(A \cap B).$$
All of these groups are isomorphic to $\mathbb Z$, so we need more information. This information is again that $S^1$ is deformation retract of the annulus and therefore the first map is actually bijection and so $H_1(X) = 0$.
Alternatively, suppose that $H_1(X) \neq 0$. Then there is some $\eta \in H_1(X)$ such that $\langle \eta \rangle = \mathbb Z$ and which maps nontrivially into $H_0(A \cap B)$. But this would then imply that the image of $H_0(A \cap B)$ in $H_0(A) \oplus H_0(B)$ is killed and consequently $H_0(X) = \mathbb Z^2$, a contradiction.
A: If I understand your space correctly, it seems you could do a deformation retraction onto $S^2$, and hence $H_1(X)=H_1(S^2)=0$.
