$H_0(S^2,A)$ where $A$ is a finite set of points.
Now, my first guess was that homology group would be $\mathbb{Z}$ because $S^2 - A$ has one path component. Unfortunately, this is not the correct way to think about this. Can somebody offer some insight please?
The portion of the L.E.S. that is helpful is:
$0 \rightarrow H_1(S^2,A) \rightarrow H_0(A) \rightarrow H_0(S^2) \rightarrow H_0(S^2,A) \rightarrow 0$
Let X be path connected and A any nonempty subspace of X.
The sequence
$$...\rightarrow H_{i}(A) \rightarrow H_{i}(X) \rightarrow H_{i}(X,A) \rightarrow H_{i-1}(A) \rightarrow ...$$ is then exact. Recall that $C_{i}(Y)$ is defined to be zero when $i<0$, then obviously $H_{i}(Y)$ is also zero so we get that $$...\rightarrow H_{0}(A) \rightarrow H_{0}(X) \rightarrow^{i_*} H_{0}(X,A) \rightarrow 0$$ is exact which means that $im (i_*)=H_{0}(X,A)$.
Let $c$ be a $0$-cycle, then $c$ is homologous to a $0$-cycle $z$ in $C_0(A)$ since $X$ is path-connected. Thus $i_*([c]) = (z+C_0(A))+im(\partial_1) = C_0(A)+im(\partial_1) = 0$
where we take $\partial_*$ to be the boundary operator in $C_*(X,A)$ and $[c]$ to be the homology class of $c$. Thus $i_* = 0$ and $H_0 (X,A)=im(i_*)=0$