Compute the homology groups $H_n(X, A)$ when $X$ is $\mathbb{S}^2$ or $\mathbb{S}^1\times \mathbb{S}^1$ and $A$ is a finite set of points in $X$.
I am trying to understand the following solution given in http://staff.ustc.edu.cn/~msheng/AT2018/exercises/solution7.pdf.
I already know that $H_2(\mathbb{S}^2,A)\cong \mathbb{Z}$ and if $n>2$ then $ H_n(\mathbb{S}^2,A)=0$ but I still do not know why $H_0(\mathbb{S}^2,A)=0$ and $H_1(\mathbb{S}^2,A)\cong\mathbb{Z}^{\oplus m-1}$. I know $\Delta: H_1(\mathbb{S}^2,A)\to H_0(A)$ is injective and so $H_1(\mathbb{S}^2,A)\cong Im (\Delta)=ker(i_*)$ where $ i_*: H_0(A)\to H_0(\mathbb{S}^2)$ but how can I find $ker(i_*)$ if $H_0(A)\cong\mathbb{Z}^{\oplus m}$ and $H_0(\mathbb{S}^2)\cong \mathbb{Z}$?
On the other hand, to calculate $H_0(\mathbb{S}^2,A)$, we know that $\pi_*: H_0(\mathbb{S}^2)\to H_0(\mathbb{S}^2,A)$ is surjective and so $H_0(\mathbb{S}^2)/ker(\pi_*)\cong H_0(\mathbb{S}^2,A)$ then, for exactitude we have $H_0(\mathbb{S}^2,A)\cong H_0(\mathbb{S}^2)/Im(i_*)$. What is $Im(i_*)$?
I know that $H_2(\mathbb{S}^1\times\mathbb{S}^1,A)\cong \mathbb{Z}$ and if $n>2$ then $H_2(\mathbb{S}^1\times\mathbb{S}^1,A)=0$, but how can I get that $H_1(\mathbb{S}^1\times\mathbb{S}^1,A)\cong\mathbb{Z}^{\oplus m+1}$ and $H_0(\mathbb{S}^1\times\mathbb{S}^1,A)=0$.
Thank you!
It is well-known that $H_0(Z)$ is a free abelian group with one canonical generator $g_P$ for each path component $P$ of $Z$. If we have a map $f : Z \to Z'$, then $f(P)$ is contained in a unique path component $P'$ of $Z'$ and we have $f_*(g_P) = g_{P'}$.
This shows that $i_* : H_0(A) \to H_0(X)$ is surjective if $X$ is path-connected and $A \ne \emptyset$. We conclude $H_0(X,A) = 0$.
For each pair $(X,A)$ the kernel $\text{ker} (i_*)$ is a subgroup of a free abelian group and therefore itself a free abelian group. In case $X$ is path-connected and $A$ has $m$ path components we can say more. $H_0(A)$ is a free abelian group with $m$ generators $g_j$ which are mapped by $i_*$ to the single generator $g$ of $H_0(X)$. That is, $i_*(\sum_{j=1}^m a_j g_j) = (\sum_{j=1}^m a_j) g$. Thus $$\text{ker} (i_*) = \{ \sum_{j=1}^{m-1} a_j g_j - (\sum_{j=1}^{m-1} a_j)g_m \mid (a_1,\dots,a_{m-1}) \in \mathbb Z^{m-1} \} \approx \mathbb Z^{m-1} .$$
Obviously the long exact sequence of any pair $(X,A)$ produces a short exact sequence $$0 \to H_1(X)/\text{im}(i_*) = H_1(X)/\text{ker} (j_*) \stackrel{j_*}{\to} H_1(X,A) \stackrel{\Delta}{\to} \text{im}( \Delta) = \text{ker} (i_*) \to 0 $$ where $j : X = (X,\emptyset) \to (X,A)$ denotes inclusion of pairs. Since $\text{ker} (i_*)$ is a free abelian group, this sequence splits. To see this, choose generators $x_\iota$ of $C = \text{ker} (i_*)$ and pick $y_\iota \in B = H_1(X,A)$ such that $\Delta(y_\iota) = x_\iota$. Then there exists a unique group homomorphism $\phi : C \to B$ such that $\phi(x_i) = y_i$. We have $(\Delta \phi)(x_i) = x_i$, hence the homomorphism $\Delta \phi$ is the identity. This means the above short exact sequence splits. We conclude $$H_1(X,A) \approx (H_1(X)/\text{im}(i_*)) \oplus \text{ker} (i_*) .$$
We can now finish answering your question. In both case we have $H_1(A) = 0$, hence $\text{im}(i_*) = 0$ so that $$H_1(X,A) \approx H_1(X) \oplus \mathbb Z^{m-1} .$$
For $X = S^2$ we have $H_1(X) = 0$, hence $$H_1(X,A) \approx \mathbb Z^{m-1} .$$
For $X = S^1 \times S^1$ we have $H_1(X) = \mathbb Z^2$, hence
$$H_1(X,A) \approx \mathbb Z^2 \oplus \mathbb Z^{m-1} \approx \mathbb Z^{m+1} .$$