Calculating ${H_n(X,A)}$, where $X$ is the torus and ${A}$ is $n$ points, using long exact sequences

Question

I am really struggling to understand what's going on with this question. The question is to find the relative homology groups ${H_k(X,A)}$, where $X$ is the torus and $A$ is n-points on it's surface.

So, I know we start off by finding ${H_k(X)}$ and ${H_k(A)}$. It's clear to see that $$ H_k(A) = \left\{ \begin{array}{cc} \mathbb{Z}^n,&k=0\\ 0&\text{otherwise} \end{array} \right. $$ and we have $$ H_k(X) = \left\{ \begin{array}{cc} \mathbb{Z},&k=0,2\\ \mathbb{Z}\oplus\mathbb{Z},&k=1\\ 0&\text{otherwise} \end{array} \right. $$ so we have a long exact sequence: $$ ...\longrightarrow H_2(A){\longrightarrow}^{i_2^*} H_2(X)\longrightarrow^{j_2^*}H_2(X,A)\longrightarrow^{\partial}H_1(A)\longrightarrow^{i_1^*}H_1(X)\longrightarrow^{j_1^*}H_1(X,A)\longrightarrow...\longrightarrow H_0(X,A)\longrightarrow 0 $$ we focus on trying to find ${H_0(X,A)}$ first, so focus on the last part of the sequence $$ H_0(A)\longrightarrow^{i_0^*}H_0(X)\longrightarrow^{j_0^*}H_0(X,A)\longrightarrow^{\partial} 0 $$ i.e. $$ \mathbb{Z}^n\longrightarrow^{i_0^*}\mathbb{Z}\longrightarrow^{j_0^*}H_0(X,A)\longrightarrow^{\partial} 0 $$ now - apparently each generator in ${\mathbb{Z}^n}$ get's sent to the generator of ${\mathbb{Z}}$. But why? I don't see why this is true. Can somebody explain? Thank you.

Answer 1

What do $H_0(X)$ and $H_0(A)$ represent? Each element of $H_0(\cdot)$ is a different path component. Since the torus is path connected, it only has one path component, and each path component of $A$ (i.e. each point) gets sent to the path connected component of the torus. So, each generator in $\mathbb{Z}^n$ gets sent to the generator of $\mathbb{Z}$.

Answer 2

You may observe that by definition of singular homology, $H_0 (X)$ is the free abelian group generated by the path-connected components of $X$ (i.e. $H_0(X) = \mathbb{Z} \pi_0(x) $). Analogously, $H_0(A)$ is the free abelian group generated by the path-connected components of $A$, i.e. the $n$ points. Now $i_{0*}$, being induced by the inclusion, sends every point of $A$ onto its path-component in $X$, but since $X$ is path-connected this is exactly the generator of $H_0(X)$.