# Relative homology groups of the solid torus relative to the torus exterior

I am trying to compute the relative homology groups $H_n(X, T)$ where $X$ is the solid torus $S^1 × D^2$ and $T$ is the subspace $S^1 × S^1$.

I have computed the homology groups of the solid torus ( $\mathbb{Z}$ for $n = 0,1$ and trivial otherwise) and of the torus ( $\mathbb{Z}\oplus\mathbb{Z}$ for $n = 1$ , $\mathbb{Z}$ for $n = 0, 2$ and trivial otherwise) but I am struggling to using the long exact sequence of a pair $(X, T)$.

I am not sure what the generators of the homology groups of $X$ will be. Any help / example of similar solution would be appreciated.

Using LES: $$H_2(X)\to H_2(X,A)\to H_1(A)\to H_1(X)\to H_1(X,A)\to H_0(A)\to H_0(X)\ldots$$ As $H_2(X)=0$, the second arrow is an injection. The third arrow, after the obvious identifications, is $(a,b)\mapsto b$ which has kernel $\{(a,0)\}$ and hence $H_2(X,A)\simeq \Bbb Z$. It is generated by the meridian $2$-cell that has boundary in $A$. The last map is an isomorphism, so the second-last is a zero map; but the third map is onto, so the fourth map is zero as well: it follows that $H_1(X,A)=0$. (Intuitively, the only candidate for $H_1(X,A)$ is a circle going around the solid torus, but this can be homotoped to the boundary).
• Thank you very much. I am able to do it this way (and much prefer it!), but the question I am looking at wants the result via the LES with explicit maps, whilst finding also explicit generators for the homology groups of $X$ which I am struggling with. Feb 5, 2017 at 14:24
• I am not sure if this is right: the image of the third arrow is $< b > \cong \mathbb{Z}$, so $H_1(X) = <b>$ and since b maps to 0 under the 4th map, which is surjective, we have $H_1(X, T) = 0$? Feb 5, 2017 at 17:07
• Oh sorry, I thought the arrow was both surjective and the 0 map hence the conclusion. So $H_1(X)$ is not $<b>$ ? I am not sure then what an explicit generator for $H_1(X)$ is. Feb 5, 2017 at 17:18
• Sorry, I was wrong and you were right: $H_1(X,A)$ is indeed zero. And $X/A$ it is not $S^1\times S^2$. I'm somehow confused of what it is. Feb 5, 2017 at 17:34