The Incluson Map $S_1\to S_1\times S_1$ Induces an Injection in First Homology. Let $T=S^1\times S^1$ be the torus and $A=S^1\times \{x_0\}$ be the "vertical" circle in the usual depiction of the torus as a tyre tube sitting "horizontally".

Let $i:A\to T$ be the inclusion map. I want to show that the induced map $i_*:H_1(A)\to H_1(T)$ is an injection.

I have been successful at doing so but my solution is very ad hoc and it is just that I got lucky. I am looking for some insights or alternate proofs.
Here is my solution.
Since $H_2(A)=0$, the long exact sequence of the pair $(X, A)$ reads
$$0\rightarrow H_2(T)\xrightarrow{j_*} H_2(T, A)\xrightarrow{\partial} H_1(A)\xrightarrow{i_*} H_1(T)\rightarrow \cdots$$
Using cellular homology I found $H_2(T)=\mathbf Z$. To find $H_2(X, A)$ doesn't look easy to me. But here is a way to do it. We note that $(X, A)$ is a good pair. Thus $H_2(X, A)\cong H_2(T/A)$. The following diagram shows that $T/A$ is homotopy equivalent to the wedge sum of a sphere with a circle.

The fact we are using here is that is $A$ is a contractible subcomplex of a CW-complex $X$, the the quotient map $X\to X/A$ is a homotopy equivalence.
So we deduce that $H_2(T, A)=\mathbf Z$. Also, $H_1(A)$ is clearly $\mathbf Z$. The long exact sequence becomes
$$0\rightarrow \mathbf Z\xrightarrow{j_*} \mathbf Z\xrightarrow{\partial} \mathbf Z\xrightarrow{i_*} H_1(T)\rightarrow \cdots$$
To show that $i_*$ is injective, it is enough to show that $j_*$ is an isomorphism. IF such is not the case, then the image of $j_*$ is $n\mathbf Z$ for some $n>0$, meaning $\mathbf Z/n\mathbf Z$ embeds in $\mathbf Z$. This is not possible. Thus $j_*$ is an isomorphism and we are done.
 A: The projection $p:(x,y)\in T\to (x,x_0)\in A$ onto the first factor has the property that $p\circ i$ is the identity of $A$. It follows that the composition $$H_1(A)\xrightarrow{i_*} H_1(T)\xrightarrow{p_*}H_1(A)$$ is the identity of $H_1(A)$ and, in particular, an injective map. Of course, it follows that $i_*$ itself is injective.
A: Here are a couple more systematic ways to see this.  First, you can notice that $A$ is a retract of $T$, via the map $r:S^1\times S^1\to S^1\times\{x_0\}$ defined by $r(x,y)=(x,x_0)$.  It follows by functoriality that the homology of $A$ is a direct summand of the homology of $T$, and in particular the induced map is injective.
Second, you can compute everything quite explicitly and straightforwardly in terms of cellular homology.  There is a standard CW-complex structure on $T$ which realizes $T$ as a quotient space of a square, with a single $0$-cell, two $1$-cells, and one $2$-cell.  The cellular boundary maps are all $0$ for this cell structure.  Moreover, your circle $A\subset T$ is just the union of the $0$-cell and one of the two $1$-cells.  So the map $i:A\to T$ induces a map on the cellular chain complexes which is just the inclusion of the subcomplex consisting of the multiples of these two cells.  Since the cellular boundary maps are $0$, the homology is just isomorphic to the chain groups, and so the induced map on homology is injective because the induced map on the chain complexes was injective
