Computing Relative Homology I'm sort of uncertain about how to compute relative homologies, so I would like some feedback on my work. As I understand it, you entirely use the idea that they form long exact sequences.
Suppose $X = S^1 \times [0, 1]$, the cylinder. And let $A = S^1 \times \{0, 1\}$ the upper and lower boundary circles. We want to compute $H_n(X,A)$ for every $n$.
So we use the idea that the homology groups form a long exact sequence $$...H_n(A) \rightarrow H_n(X) \rightarrow H_n(X, A) \rightarrow H_{n-1}(A) \rightarrow H_{n-1}(X) .....$$
Next since $X$ deformation retracts to a circle, $H_n(X) = \mathbb{Z}$ for $n = 1, 2$ and trivial otherwise. Similarly, as $A$ is a disjoint union of two circles, $H_n(A) = \mathbb{Z} \oplus\mathbb{Z}$ for $n = 1,2$ and trivial otherwise. This means that $H_{n}(X,A) = 0$ if $n \geq 4$. For $n = 3$,
$$.. 0 \rightarrow H_3 (X, A) \rightarrow \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$$
is what we obtain. Which means that $H_3(X, A)$ is a subgroup of $\mathbb{Z} \oplus \mathbb{Z}$. Moreover, the map $i_{*} : \mathbb{Z} \oplus \mathbb{Z} \rightarrow \mathbb{Z}$ is induced by the inclusion $i : C_2(A) \rightarrow C_2(X)$.
However, I'm not sure which properties of $i_{*}$ are to be used to get what $H_3(X, A)$ is exactly. I'm facing similar problems for the other $H_n(X,A)$ ($n = 0,1,2$).
I think I could get it done after looking at a concrete example but such examples are surprisingly hard to find. Most examples, like the ones in Hatcher, are trivial and quick. I'm sure this one is easy as well and I'm just being an idiot but I would still appreciate the help.
 A: To start: $H_n(X) = \mathbb{Z}$ if $n=0, 1$, zero otherwise; and $H_n(A) = \mathbb{Z} \oplus \mathbb{Z}$ if $n=0, 1$, zero otherwise. Furthermore, you also need to know that the map $H_1(A) \to H_1(X)$ sends both $\mathbb{Z}$ summands to $\mathbb{Z}$ by the identity: $(a,b) \mapsto a+b$. Well, maybe you don't need this precise computation, but at least you need to know that the map $H_1(A) \to H_1(X)$ is onto. After all, as you point out, $X$ deformation retracts onto a circle, and so the inclusion of either end of the cylinder induces the identity map on $\mathbb{Z}$.
From there, you can use exactness to see that the map $H_1(X) \to H_1(X,A)$ is zero, and then you can compute $H_2(X,A)$.
A: You have already received hints to do the computation using the homology LES. Here is another way to compute the relative homology group: if A is deformation retract of a neighborhood in X, then $H_n(X, A) \approx \widetilde{H}_n(X/A)$. In our case, this is indeed true as $S^1 \times [0, \frac12) \sqcup S^1 \times (\frac12, 1]$ is a neighborhood in $X$ that deformation retracts to $A$.
Observe that $X/A$ can be obtained by collapsing the two boundary circles. This is like a 2-sphere with its north and south poles identified, which is homotopy equivalent to $S^2 \vee S^1$. It should be easy to compute the reduced homology of this.
$H_0(X, A) = \widetilde{H}_0(S^2 \vee S^1) = 0$
$H_1(X, A) = \widetilde{H}_1(S^2 \vee S^1) = \mathbb{Z}$
$H_2(X, A) = \widetilde{H}_2(S^2 \vee S^1) = \mathbb{Z}$
$H_k(X, A) = \widetilde{H}_k(S^2 \vee S^1) = 0$ for $k \ge 3$.
A: For $n \ge 3$ we have an exact sequence
$$0 = H_n(X) \to H_n(X,A) \to H_{n-1}(A) = 0$$
which shows $H_n(X,A) = 0$.
Let $S^1_k = S^1 \times \{ k \}$ for $k= 0,1$. Let $i : A \to X$ and $i_k : S^1_k \to A$ denote inclusions. The maps $u_k : S^1 \to S^1_k, u_k(x) = (x,k)$, are homeomorphisms. The projection $p : X \to S^1$ and the maps $j_k : S^1 \to X, j_1(x) = (x,k)$, are homotopy equivalences.
Since the $S^1_k$ are the path components of $A$, the map
$$h  : H_n(S^1_0) \oplus H_n(S^1_1) \to H_n(A), h(a,b) =  (i_1)_*(a) + (i_2)_*(b)$$ is an isomorphism. Moreover
$$\psi : H_n(S^1) \oplus H_n(S^1) \to H_n(S^1_0) \oplus H_n(S^1_1), \psi(c,d) = ((u_1)_*(c),(u_2)_*(d))$$
is an isomorphism.
For $n = 0,1$ let us compute $\phi = p_* i_*h \psi: H_n(S^1) \oplus H_n(S^1) \to H_n(S^1)$. Let $g$ one of the two generators of $H_n(S^1)$. Obviously $\phi(g,0) = p_*i_*h\psi(g,0) = p_*i_*(i_1)_*(u_1)_*(g) = g$. Similarly $\phi(0,g) = g$. Thus $\phi(ag,bg) = (a+b)g$. We conclude that $\phi$ is onto with $\ker \phi \approx \mathbb Z$. Thus also $i_*$ is onto with $\ker i_* \approx \mathbb Z$. Hence $j_* : H_n(X)\to H_n(X,A)$ is the zero map because $\ker j_* =\operatorname{im} i_* = H_n(X)$.
The exact sequence
$$0 = H_2(X) \to H_2(X,A)  \stackrel{\partial}{\to} H_1(A) \stackrel{i_*}{\to} H_1(X)$$
shows that $H_2(X,A) \approx \operatorname{im} \partial = \ker i_*  \approx \mathbb Z$.
The exact sequence
$$H_1(X)  \stackrel{0}{\to} H_1(X,A) \stackrel{\partial}{\to} H_0(A) \stackrel{i_*}{\to} H_0(X)$$
shows that $\ker \partial = \operatorname{im} 0 = 0$, i.e. $\partial$ is injective. But $\operatorname{im} \partial = \ker i_*  \approx \mathbb Z$, thus $H_1(X,A) \approx \mathbb Z$.
The exact sequence
$$H_0(X) \stackrel{j_*}{\to}  H_0(X,A) \to 0$$
shows that $j_*$ is surjective. But $j_*$ is the zero map, thus $H_0(X,A) = 0$.
