Compute Homology groups $H_n(X,A)$ and $H_n(X,B)$ where $X$ is the 2-genus Torus, A is a separating Circle (as in the picture) and $B$ a non-separating Circle (as in the picture)
My attempt :
Considering $(X,A)$ to be a good pair, used $H_n(X,A) = \widetilde{H_n}(X/A)$ and (seemed to me intuitively) $X/A \simeq T \lor T$ where $T=S^1 \times S^1$, Then I computed, $H_n(X,A)= \Bbb Z^4 \text{ if n = 1 , } \Bbb Z^2 \text{ if n = 2 , } 0 $ otherwise .
Similarly, obtained, $H_n(X,B) = \tilde{H_n}(X/B)$ and (seemed to me intuitively) $X/B \simeq T \lor S^1$ where $T=S^1 \times S^1$ . Then I computed, $H_n(X,B)= \Bbb Z^3 \text{ if n = 1 , } \Bbb Z \text{ if n = 2 , } 0 $ otherwise .
Are my arguments valid? Please point out mistakes!
You do have the right groups, and have not made any false statements. In order to make things less hand-wavy, you can compute the relative homology groups (almost) formally using the long exact sequence of the pair. The algebraic machinery lets us circumvent proving that $X/B\simeq T^2 \vee S^1$, which although it's intuitively clear it's hard to write down something rigorous without doing explicit manipulation of concrete objects, or computing homotopy groups and invoking Whitehead's Theorem.
Consider the sequence for $(X,A)$. Since $H_2(A) \cong 0$ and $H_0(A)\cong H_0(X)$ we get
$$0 \to H_2(X)\to H_2(X,A) \to H_1(A) \to H_1(X) \to H_1(X,A) \stackrel{0}{\to}\dots $$
We know that $H_2(X)\cong H_1(A)\cong\mathbb{Z}$ and $H_1(X)\cong \mathbb{Z}^4$. Denote the usual generators of $H_1(X)$ by $\alpha_1, \beta_1, \alpha_2, \beta_2$, where $\alpha_i$ is the circle going around the donut hole and $\beta_i$ is the circle going around the handle. Determining the map $H_1(A)$ to $H_1(X)$ amounts to determining the homology class of $A \subset X$; but we can see this is $0$ either by noticing it is a boundary, or by seeing that it's the commutator $\alpha_1 + \beta_1 - \alpha_1 - \beta_1 = 0$. Therefore $H_1(X,A)\cong H_1(X)$, and $H_2(X,A)$ fits in a short exact sequence
$$ 0 \to \mathbb{Z} \to H_2(X,A) \to \mathbb{Z} \to 0$$
so must be $\mathbb{Z}\oplus \mathbb{Z}$.
The sequence for $(X, B)$ is even easier, since the homology class of $B$ is just $\beta_2$. In other words the map $H_1(B)\to H_1(X)$ is an isomorphism onto one of the summands, therefore $H_1(X,B)\cong \mathbb{Z}^3$ and $H_2(X,A)\cong \mathbb{Z}$.