First I'll go through my working. Throughout we assume the homology groups of the torus and circle are known.

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Let $X=S^1 \times S^1$ be the torus, and $A=S^1 \times \{1\}$. The following is part of a long exact sequence:
$$
\dots \to H\_n(X) \to H\_n(X,A) \to H\_{n-1}(A) \to \dots
$$
For $n\geq 3$, we have $H\_n(X) = H\_{n-1}(A) = 0$, so $H\_n(X,A) = 0$. Using the 'reduced version' of the sequence, we have
$$
\dots \to H\_2(A) \to H\_2(X) \to H\_2(X,A) \xrightarrow{f} H\_1(A) \xrightarrow{i\_\*} H\_1(X)\qquad\qquad\qquad\qquad
$$
$$
\qquad\qquad\qquad\qquad\qquad \to H\_1(X,A)\to \tilde{H}\_0(A) \to \tilde{H}\_0(X) \to \tilde{H}\_0(X,A) \to 0.
$$

**Step 1.** Now, $A, X$ are path connected so $\tilde{H}\_0(A) =\tilde{H}\_0(X) = 0$. So $\tilde{H}\_0(X,A)=0$ by the long exact sequence, which gives us that $H\_0(X,A) \cong \mathbb{Z}$.

**Step 2.** The map $i\_\*: H\_1(A) \to H\_1(X)$ takes $[\alpha] \mapsto [i(\alpha)]$ where $i$ is the inclusion. In other words, it is injective, and so $0 = \ker i\_\* = \text{im}\:f$, and so $f$ is the zero map. This then leaves two short exact sequences, and implies that $H\_2(X,A) \cong H\_2(X) \cong \mathbb{Z}$.

**Step 3.** It is an immediate consequence of the other remaining short exact sequence that $H\_1(X,A) \cong H\_1(X)/\text{im }i\_\* \cong \mathbb{Z} \oplus \mathbb{Z}/\mathbb{Z} \cong \mathbb{Z}$.

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What I'm looking for are mainly comments on if this was done properly, but I also have two specific queries:

**1.** Have I used the reduced homology 'version' of the long exact sequence correctly? I'm *very* iffy with reduced homology...

**2.** My reason for $i\_\*$ being injective was a little handwavy. Is it always the case that a homomorphism induced by inclusion of spaces $A\hookrightarrow X$ is injective?