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).

  • $\begingroup$ 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. $\endgroup$ – ech-93 Feb 5 '17 at 14:24
  • $\begingroup$ 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 $H_1(X, T) = 0$? $\endgroup$ – ech-93 Feb 5 '17 at 17:07
  • $\begingroup$ 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. $\endgroup$ – ech-93 Feb 5 '17 at 17:18
  • $\begingroup$ 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. $\endgroup$ – Peter Franek Feb 5 '17 at 17:34
  • $\begingroup$ Ah great, no problem. Thank you very much for your help. Yes I'm having trouble thinking about what it is also... $\endgroup$ – ech-93 Feb 5 '17 at 18:35

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.