Q:

Relative homology groups

I have to compute the homology groups $H_{n}(X,A)$ when $X$ is $S^{2}$ or $S^{1}\times S^{1}$ and $A$ is a finite set of points in $X$. So, I write the exact long sequence :

$...\rightarrow H_{n}(S^2)\rightarrow H_{n}(S^2,A)\rightarrow H_{n-1}(A)\rightarrow H_{n-1}(S^2)\rightarrow H_{n-1}(S^2,A)\rightarrow ...$

The sequence becomes :

$...\rightarrow 0\rightarrow H_{n}(S^2,A)\rightarrow H_{n-1}(A)\rightarrow 0\rightarrow H_{n-1}(S^2,A)\rightarrow ...$

And then $H_n(S^2,A)\simeq H_{n-1}(A)\simeq 0$ because $A$ is a finite set or points in $X$. Is that true ? 
Thank you ! 
EDIT : I think there is a problem when $n=2$...

A:

I will do only the case of the sphere (the case of the torus is very similar).  If $n \ge 4$ (and not $3$: you need $H_{n-1}(S^2)$ to vanish) then there's not problem and the long exact sequence gives you $$H_n(S^2, A) \cong H_{n-1}(A) = 0.$$
Let's write down the LES in low dimensions:
$$\require{cancel}
\begin{align}
& \cancel{H_3(S^2)} \to H_3(S^2,A) \to \cancel{H_2(A)} \to \dots \\
& H_2(S^2) \to H_2(S^2, A) \to \cancel{H_1(A)} \to \cancel{H_1(S^2)} \to \dots \\
& H_1(S^2,A) \to H_0(A) \to H_0(S^2) \to H_0(S^2,A) \to 0
\end{align}$$
As you can see
$$H_2(S^2,A) \cong H_2(S^2) = \mathbb{Z}.$$
If $A$ is not empty then $A$ meets every path component of $S^2$, so $H_0(A) \to H_0(S^2)$ is the identity $\mathbb{Z} \to \mathbb{Z}$. This is an isomorphism, so it follows that $$H_1(S^2,A) = H_0(S^2,A) = 0.$$
If $A$ is empty then of course $H_n(S^2,A) = H_n(S^2)$ (in particular $H_1(S^2,A) = 0$ and $H_0(S^2,A) = \mathbb{Z}$).