How do I compute this relative holomogy group? Let $A:=\{z\in S^1: z^n=1\}$, considering $S^1\subset \mathbb{C}$.
Then, how do I compute $H_k(S^1\times S^1, A\times \{1\}\cup \{1\} \times S^1)$ when $k=1,2$?
Let's consider the long exact sequence of a pair $(S^1\times S^1, A\times \{1\}\cup \{1\} \times S^1)$:
$...\rightarrow H_k(A\times \{1\}\cup \{1\} \times S^1)\rightarrow H_k(S^1\times S^1)\rightarrow H_k(S^1\times S^1, A\times \{1\}\cup \{1\} \times S^1)\rightarrow...$
The problem is when $k=1,2$. 
Consider the case $k=2$. 
Then, we have an exact sequence $0\rightarrow H_2(S^1\times S^1)\rightarrow H_2(S^1\times S^1,A\times \{1\}\cup \{1\} \times S^1)\rightarrow H_1(A\times \{1\}\cup \{1\} \times S^1)$.
Since the outer parts of the sequence are not zero, we have to find explicit forms of the inclusion & boundary maps. Using cellular homology, I know that, a generator of $H_2(S^1\times S^1)$ is $S^1\times S^1$ itself. The problem is, I don't get how the generator of $H_2(S^1\times S^1, A\times \{1\}\cup \{1\} \times S^1)$ would look like. How do I find it?
Consider the case $k=1$.
We have an exact sequence $H_1(S^1\times S^1)\rightarrow H_1(S^1\times S^1,A\times \{1\}\cup \{1\} \times S^1)\rightarrow H_0(A\times \{1\}\cup \{1\} \times S^1)$.
Since $(S^1\times S^1,A\times \{1\}\cup \{1\} \times S^1)$ is a good pair, its relative homology is the quotient of these spaces. Let $a,b$ be the most natural generators for $H_1(S^1\times S^1)$. Since $b$ is killed by taking quotient and we are quotienting $A$ too, we get $n+1$ generators, namely $a,b_1,...,b_n$. Since boundary maps take $a,b_1,....b_n$ to zero, boundary map is identically zero.  Since $b$ is killed by taking inclusion and $a$ is preserved by inclusion, we can conclude that $H_1(S^1\times S^1,A\times \{1\}\cup \{1\} \times S^1)\cong \mathbb{Z}^n$. Is my argument correct?
How do I compute this relative homology group? Thank you in advance!
 A: 
$H_2(Y)\cong 0$
$H_2(X)\cong \mathbb{Z}$, generated by $[f_1+f_2+f_3+f_4+f_5]$
$H_1(Y)\cong \mathbb{Z}$, generated by $[e_1]$
$H_1(X)\cong \mathbb{Z}^2$, generated by $[e_1]$  (which equals $[e_2]$, $[e_3]$, $[e_4]$, and $[e_5]$ in this group) and $[e_6+e_7+e_8+e_9+e_{10}]$
$H_0(Y)\cong \mathbb{Z}^5$, generated by $[v_1]$, $[v_2]$, $[v_3]$, $[v_4]$, and $[v_5]$
$H_0(X)\cong \mathbb{Z}$, generated by $[v_1]$ (which equals $[v_2]$, $[v_3]$, $[v_4]$, and $[v_5]$ in this group)
The map $H_1(Y)\longrightarrow H_1(X)$ has image $\mathbb{Z}[e_1]$, which is isomorphic to $\mathbb{Z}$.
The map $H_0(Y)\longrightarrow H_0(X)$ has kernel $\mathbb{Z}[v_1-v_2]+\mathbb{Z}[v_2-v_3]+\mathbb{Z}[v_3-v_4]+\mathbb{Z}[v_4-v_5]$, which is isomorphic to $\mathbb{Z}^4$.
Thus $H_1(X,Y)$ has a subgroup isomorphic to $\mathbb{Z}$, such that when you quotient out by it, you get something isomorphic to $\mathbb{Z}^4$. Therefore $H_1(X,Y)\cong\mathbb{Z}^5$.
The map $H_1(Y)\longrightarrow H_1(X)$ is injective, so the image of the map $H_2(X,Y)\longrightarrow H_1(Y)$ must be the trivial subgroup, so the kernel of the map $H_2(X,Y)\longrightarrow H_1(Y)$ must be all of $H_2(X,Y)$, so the image of the map $H_2(X)\longrightarrow H_2(X,Y)$ must be all of $H_2(X,Y)$, i.e. it is surjective. 
Since $H_2(Y)$ is trivial, the map $H_2(X)\longrightarrow H_2(X,Y)$ must be injective.
Therefore $H_2(X,Y)\cong \mathbb{Z}$.
$$H_2(Y)\longrightarrow H_2(X)\longrightarrow H_2(X,Y)$$
$$\swarrow$$
$$H_1(Y)\longrightarrow H_1(X)\longrightarrow H_1(X,Y)$$
$$\swarrow$$
$$H_0(Y)\longrightarrow H_0(X)\longrightarrow H_0(X,Y)$$
A: Hint: What's 
$$
H_1(A\times \{1\}\cup \{1\} \times S^1)?
$$
Hint: it's $\Bbb Z$, generated by the $S^1$ factor. 
What's 
$$
H_1(S^1\times S^1),
$$
and what are its (most natural) generators? 
Hint 2: What's $H_2$ of a 1-complex? 
