homology groups of a torus How can I find the homology group of a torus without using cellular homology and the CW complex ? in other words , how can i calculate the homology groups of a torus using only relative homology ? I appreciate your help 
 A: Consider the pair $(\Bbb T, A)$ where $A \subset \Bbb T$ is one of the longitudal circles in $\Bbb T$. We thus have the long exact sequence
$$\cdots \to H_{k+1}(\Bbb T) \to H_{k+1}(\Bbb T, A) \stackrel{\partial}{\to} H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \stackrel{\partial}{\to} H_{k-1}(A) \to \cdots$$
$(\Bbb T, A)$ has homotopy extension property, as it's a CW-pair. Thus,  we have $H_\bullet(\Bbb T, A) \cong H_\bullet(\Bbb T/A) \cong H_\bullet(S^2 \vee S^1)$ which we know is $\Bbb Z$ if $k = 1, 2$ and is trivial otherwise.
The whole point of choosing this pair is that there is a retract $r : \Bbb T \to A$ which, after applying the $H_\bullet$ functor, gives a left-inverse for the map $H_\bullet(A) \to H_\bullet(\Bbb T)$ induced by the inclusion, which proves injectivity of this map. Thus, the snake maps $\partial$ are all zero
Thus, at each $k$, the long exact sequence reduces to the split (the section being $H_\bullet(r)$) exact sequence
$$0 \to H_k(A) \to H_k(\Bbb T) \to H_k(\Bbb T, A) \to 0$$
Thus, $H_k(\Bbb T) \cong H_k(A) \oplus H_k(\Bbb T, A)$ which is $\Bbb Z \oplus \Bbb Z$ for $k = 1$, $\Bbb Z$ for $k = 2$ and trivial otherwise.

To add, $H_\bullet(S^2 \vee S^1)$ can be computed in a similar vein, without using Mayer-Vietoris sequence, by taking the CW-pair $(S^2 \vee S^1, S^2)$. There is the retract $S^2 \vee S^1 \to S^2$ by collapsing the circle to the wedged point, which gives a left-inverse for $H_\bullet(S^1) \to H_\bullet(S^2 \vee S^1)$ induced from inclusion as well as a section for the short exact sequence.
