Homology of $S^1 \times (S^1 \vee S^1)$

This solution relies on the long exact reduced homology sequence of a NDR pair (Hatcher's Theorem $2.13$), and provides different approach to the problem from @tsho's solution.

Let us call $$\underbrace{\Huge{\mathsf x} \normalsize\times S^1}_{A}~~\subset~~ \underbrace{\Huge{\propto}\normalsize\times S^1}_{B}~~\subset~~ \underbrace{\Huge{\infty}\normalsize \times S^1}_{X}$$ All three pairs $(B,A),~(X,A),~(X,B)$ are Neighborhood Deformation Retracts, as Hatcher puts it, "good pairs". Also, it is obvious that $A$ is homotopy equivalent to the circle, and $B$ to the torus. Let us write the long exact reduced homology sequence for the the good pairs $(X,A)$ and $(X,B)$ : the morphism of pair given by the inclusion $(X,A)\hookrightarrow (X,B)$ gives following commutative diagram $$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/A)&\to&\tilde{H}_1(A)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/A)&\to 0\\ &\downarrow&&\Vert&&\downarrow&&\downarrow&&\Vert&&\downarrow\\ 0\to& \tilde{H}_2(B)&\to&\tilde{H}_2(X)&\to&\tilde{H}_2(X/B)&\to&\tilde{H}_1(B)&\to&\tilde{H}_1(X)&\to &\tilde{H}_1(X/B)&\to 0 \end{array}$$ Now $X/A$ is the wedge sum of two pinched spheres $P$ (the space studied in the previous question), and $X/B$ is simply a pinched sphere, and it follows that $\tilde{H}_*(X/B)\simeq\tilde H_*(P)$ and $\tilde{H}_*(X/A)\simeq\tilde H_*(P)\bigoplus \tilde H_*(P)$ where the isomorphism is given by the map $(i_*^+,i_*^-)$ where $i^+$ (resp. $i^-$) are the inclusions of $P$ as the upper (resp. lower) pinched sphere in $X/A$.

Since a pinched sphere is homotopy equivalent to a sphere with a diameter attached to it, which in turn is homotopy equivalent to the wedge sum of a sphere and a circle, we have $\tilde H_*(P)\simeq \Bbb Z\oplus\Bbb Z$ concentrated in degree $1$ and $2$. We can now replace the above diagram by the following simpler one

$$\begin{array}{c}0\to &0&\to&\tilde{H}_2(X)&\to&\Bbb Z\oplus \Bbb Z &\stackrel{\gamma}{\to}&\Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z\oplus \Bbb Z &\to 0\\ &\downarrow&&\Vert&&~~\downarrow\beta&&\downarrow&&\Vert&&\downarrow\\ 0\to& \Bbb Z &\to&\tilde{H}_2(X)&\stackrel{\alpha}{\to}& \Bbb Z &\to& \Bbb Z\oplus \Bbb Z &\to&\tilde{H}_1(X)&\to & \Bbb Z &\to 0 \end{array}$$

From the left side of this diagram, it follows that $\tilde H_2(X)$ is a subgroup of $\Bbb Z\oplus \Bbb Z$ containing a copy of $\Bbb Z$, so $\tilde H_2(X)\simeq\Bbb Z$ or $\Bbb Z\oplus\Bbb Z$. Let us assume $\tilde H_2(X)\simeq \Bbb Z$ and try to derive a contradiction.

Since $\Bbb Z$ is torsion free, we must have $\alpha=0$. The vertical map $\beta$ is onto as it corresponds to collapsing the lower copy of $P$ inside $P\vee P\simeq X/A$ to a point, and thus $\beta$ is the projection onto the first factor. The commutativity of the diagram then forces the image of $\tilde H_2(X)$ to lie inside $\Bbb Z\oplus 0\subset \Bbb Z\oplus \Bbb Z$. However, there is an obvious self-homeomorphism of $X$ interchanging the upper and lower toruses of $X$ which passes to the quotient, and permutes the two factors $\Bbb Z \oplus \Bbb Z=\tilde H_2(X/A)$ (and possibly adds a sign). Thus, the image of $\tilde H_2(X)$ inside $\Bbb Z \oplus \Bbb Z$ is contained in $\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$, but this contradicts the injectivity of the top left arrow. The same argument works when we replace $B$ with $B'=T(B)$ where $T$ is the self-homeomorphism of $X$ that interchanges the two circles in the wedge sum $S^1\vee S^1$. The new map $\beta'$ is the projection onto the second factor, so the map $\tilde H_2(X)\to \Bbb Z\oplus\Bbb Z$ sends $\tilde H_2(X)$ into $\ker(\beta)\cap\ker(\beta')=\Bbb Z\oplus 0\cap 0\oplus \Bbb Z=0$ contradicting injectivity.

As a consequence, $$\tilde H_2(X)\simeq \Bbb Z\oplus\Bbb Z$$

To finish the proof, we note that by the standard theory of finitely generated abelian groups, the quotient of $\Bbb Z\oplus\Bbb Z$ by a subgroup $S$ isomorphic to $\Bbb Z\oplus\Bbb Z$ is the product of two cyclic groups, and cannot be a subgroup of $\Bbb Z$ unless the subgroup $S$ is all of $\Bbb Z\oplus\Bbb Z$. This forces the top left arrow $\tilde{H}_2(X)\to\Bbb Z\oplus \Bbb Z $ to be an isomorphism, and $\gamma=0$. The top sequence then degenerates to a short exact sequence $$0\to\Bbb Z \to\tilde{H}_1(X)\to \Bbb Z\oplus \Bbb Z \to 0$$

Consequently, $$\tilde{H}_1(X)\simeq \Bbb Z\oplus\Bbb Z\oplus\Bbb Z$$

Firstly, I believe that $H_1(\mathrm{Torus})=\mathbb{Z}^2$ hence the sequence should be $$0 \xrightarrow 0 \mathbb{Z}^2 \xrightarrow f \tilde{H}_2(X) \xrightarrow g \mathbb{Z} \xrightarrow h \mathbb{Z}^2 \oplus \mathbb{Z}^2 \to \tilde{H}_1(X) \to 0.$$ Now, you need to determine at least one map to solve this. In Mayer Veatoris sequence, f is $(t,s) \mapsto i_*t+i_*s$, g is $t\mapsto \partial t$, and h is $t \mapsto (i_*t,-i_*t)$ where all of my $i_*$'s are the homomorphims related to the appropriate canonical inclusions. (At this point you should try solving the question yourself and check the answer for feedback).

We'll start by using the exactness of the sequence: $\operatorname{im}0=0=\operatorname{ker}f$ means $f$ is injective, $\operatorname{im}f=\operatorname{ker}g$ tells us nothing yet, and $\operatorname{im}g=\operatorname{ker}h$.

It will be most natural to determine $h$, since its the only one whose both domain and range are known. So, $h$ takes a generator of the circle, embeds it in each of the tori, and the embedded circle inside each of the tori we know to be a generator of one of his $\mathbb{Z}$'s. ($H_1(T)=\mathbb{Z}^2$, a $\mathbb{Z}$ for the "horizontal" circle and a $\mathbb{Z}$ for the "vertical" circle, in our case, the generator of the original $\mathbb{Z}$ is being mapped to the horizontal circle, that is if you draw the torus "horizontally" as you normally would, and draw $X$ as two tori ("tires") stacked on top of each other).

So we can write $h(1)=((1,0),(1,0))$ and so $\operatorname{im}g=\operatorname{ker}h=0$, hence $g=0$, so $\operatorname{im}f=\operatorname{ker}0=$ entire group, meaning $f$ is surjective and hence an isomorphism. So $H_2(X)=\mathbb{Z}^2$. $H_1$ could be obtained similarly or by using $H_1=\operatorname{Ab}(\pi _1)$.

If your coefficients are in a field, then you can use the Künneth formula, which says that if $X=Y \times Z$ $$ H^i(X) = \bigoplus_{j+l=i} H^j(Y) \otimes H^{l}(Z). $$

So, for example to calculate $H^1(X)$, there are two terms in sequence, namely when $(j,l)=(0,1)$ and when $(j,l)=(1,0)$. So $$ H^1(X) = H^1(Y) \otimes H^0(Z) \bigoplus H^0(Y) \otimes H^1(Z).$$

In your case, $Y=S^1$ and $Z=S^1 \lor S^1$, so that $$ H^1(X) = H^1(S^1) \otimes H^0(S^1 \lor S^1) \bigoplus H^0(S^1) \otimes H^1(S^1 \lor S^1)$$ $$ = k \otimes k \bigoplus k \otimes(k^2) = k \oplus k^2 = k^3.$$