Why $ker(i_*:H_0(\mathbb{S}^0)\to H_0(\mathbb{S}^2))$ is isomorphic to $\mathbb{Z}$?
I know that $H_0(\mathbb{S}^0)\cong \mathbb{Z}^2$ and $H_0(\mathbb{S}^2)\cong \mathbb{Z}$, so the problem is reduced to finding the kernel of $\mathbb{Z}^2\to\mathbb{Z}$, what is this kernel? Thank you.
Since $S^0 \cong \ast \amalg \ast$, we have a cell decomposition for it with two $0$-cells and no higher cells. Let's pick a cell decomposition for $S^2$ which has one zero cell and one two-cell. Then the cell complexes are:
$$C_\ast(S^0) = \mathbb{Z}\oplus \mathbb{Z} \leftarrow 0\leftarrow \cdots$$ $$C_\ast(S^2) = \mathbb{Z} \leftarrow 0 \leftarrow \mathbb{Z} \leftarrow \cdots $$ and we can compute their cohomology as $$H^k(S^0) = \begin{cases} \mathbb{Z}\oplus \mathbb{Z} & k=0 \\ 0 & \text{otherwise}\end{cases}$$ $$H^k(S^2) = \begin{cases} \mathbb{Z} & k=0,2 \\ 0 & \text{otherwise}\end{cases}$$
Now let's think about continuous maps $i: S^2 \to S^0$. Since $S^2$ is connected, the entirety of $S^2$ must map to a single 0-cell in $S^0$. This induces a map on 0-chains $$C_0(S^2) = \mathbb{Z} \to \mathbb{Z}\oplus\mathbb{Z} =C_0(S^0),$$ which sends $1\mapsto (1,0)$ (without loss of generality). Applying $\text{Hom}(-,\mathbb{Z})$ to the chain complexes and then taking homology, we get an induced map on the 0-th cohomology groups $$H^0(S^0) = \mathbb{Z}\oplus\mathbb{Z} \to \mathbb{Z} = H^0(S^2),$$ which is of the form $(a,b)\mapsto a$, and hence has kernel $\mathbb{Z}$.