TITLE: Relative homology groups of the torus
QUESTION [10 upvotes]: I have the following question to problem 2.1.17 in Allen Hatcher's "Algebraic Topology". 

Compute the groups $H_n(X,A)$ and $H_n(X,B)$ where $X$ is a closed orientable surface of genus two and $A$ and $B$ are the circles shown in the picture on page 132  of Hatcher (page 141 of the pdf). 

So far I came up with the following exact sequences (for A and B):
$$
\begin{aligned}
0&\rightarrow H_{2}(A) \rightarrow H_{2}(X) \rightarrow H_{2}(X,A)\rightarrow\\
&\rightarrow H_{1}(A) \rightarrow H_{1}(X) \rightarrow H_{1}(X,A)\rightarrow\\
&\rightarrow H_{0}(A) \rightarrow H_{0}(X) \rightarrow H_{0}(X,A) \rightarrow 0
\end{aligned}
$$
and 
$$
\begin{aligned}
0&\rightarrow H_{2}(B) \rightarrow H_{2}(X) \rightarrow H_{2}(X,B)\rightarrow\\
&\rightarrow H_{1}(B) \rightarrow H_{1}(X) \rightarrow H_{1}(X,B)\rightarrow\\
&\rightarrow H_{0}(B) \rightarrow H_{0}(X) \rightarrow H_{0}(X,B) \rightarrow 0,
\end{aligned}
$$
where $H_{2}(A) = H_{2}(B) = 0$, $H_{1}(A) = H_{1}(B) = \mathbb{Z} = H_{0}(A) = H_{0}(B)$ and for $X$ there is $H_{2}(X) = H_{0}(X) = \mathbb{Z}$ and $H_{1}(X) = \mathbb{Z}^{4}$. Furthermore I know that the mappings $H_{1}(A) \rightarrow H_{1}(X)$ is zero and that $H_{1}(B) \rightarrow H_{1}(X)$ is injective. By these I could deduce that $H_{0}(X,A) = 0$ and $H_{1}(X,A) = \mathbb{Z}^{4}$ and $H_{0}(X,B) = 0$. But I can't go on further. What about the other relative homology groups? What do I need more? Hope this question is not too trivial and apologize. Hope someone to help.
mika

REPLY [2 votes]: (I don't have enough reputation so can't reply to the comment of @man_in_green_shirt above.)
In @Lor's answer above, there should be a zero on the right hand side of,
$$ 0 \rightarrow H_1(X) \rightarrow H_1(X,A) \rightarrow 0 $$
but not because $H_0(A) = 0$.
All elements $[\alpha] \in H_1(X,A)$ get sent to the homology class in $H_0(A)$ corresponding to $\partial \alpha \in C_0(A)$. But then observe that any such $\partial \alpha \in C_0(A)$ is actually bounding a contiguous region of $A$. Then any such $\partial \alpha$ is a boundary in $C_0(A)$, and thus $\partial \alpha$ is null homologous in $H_0(A)$.
Notice that this would not be true if, for example, $A$ was not the circle shown but was instead two disjoint arcs of the circle.