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I am computing the singular homology of spheres by induction.

In the process, I have come across the following short exact sequence

$$0 \to H_1(S^1) \to \mathbf Z^2 \to \mathbf Z \to 0.$$ I wonder if there is a more direct way to conclude that $H_1(S^1) = \mathbf Z$ than my argument:

it is a split short exact sequence of modules over a ring (for example, $\mathbf Z$ is a free module, so the third map has a section) ; therefore, $H_1(S^1) \oplus \mathbf Z$ is isomorphic to $\mathbf Z^2$. But $\mathbf Z^2$ is a free module over a principal domain, hence, so is $H_1(S^1)$. Because of the direct sum, I conclude that $H_1(S^1)$ has rank $1$.

I have the impression that it is overkill.

Thanks.

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  • $\begingroup$ I get the impression you are asking for a simpler way to prove $H_1(S^1)=\mathbb{Z}$. Maybe you should edit the title? The title makes a reader think the question asks a proof that what the title says is true, or at least made me think so, and the question is entirely different to my eyes. $\endgroup$
    – MickG
    Sep 25, 2014 at 17:08
  • $\begingroup$ I am not sure why the confusion is a problem, but certainly; I don't know what to replace the title with though, do you have any suggestion? Feel free to edit if you are able to. $\endgroup$
    – user25784
    Sep 25, 2014 at 17:26
  • $\begingroup$ Because I saw an answer that just replicated your proof of that, and I guess that is NOT what you want. Also, someone may come here attracted by a question he thinks he can answer and be very disappointed :). And a question with a misleading title is anyway bad form IMHO. Anyway, I'd suggest «Simple way of proving $H_1(S^1)\cong\mathbb{Z}$» or the likes. $\endgroup$
    – MickG
    Sep 25, 2014 at 17:37
  • $\begingroup$ I hope this is acceptable. $\endgroup$
    – user25784
    Sep 25, 2014 at 17:58
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    $\begingroup$ Since $\mathbf{Z}$ is free, the sequence splits, so $\mathbf{Z}^2\cong H_1(S^1)\oplus\mathbf{Z}$. A subgroup of a free group is free; counting ranks says $H_1(S^1)$ is free of rank $1$. Why should it be overkill? $\endgroup$
    – egreg
    Sep 25, 2014 at 18:13

2 Answers 2

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$H_1(S^1)$ is the kernel of the map $\mathbb{Z}^2\to\mathbb{Z}$. Every such homomorphism is of the form $(x,y)\mapsto ax+by$ for some $a,b\in\mathbb{Z}$. In this case the map is surjective, so $\gcd(a,b)=1$, and the kernel is the infinite cyclic group generated by $(b,-a)$.

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  • $\begingroup$ I like this one a lot. $\endgroup$
    – user25784
    Sep 25, 2014 at 19:03
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You could tensor it with $\mathbb{Q}$ (or any field of characteristic 0) to show that $H_1(S^1)$ is isomorphic to $\mathbb{Z} \oplus F$ for some finite abelian group $F$.

The reason for this is that you end up with the short exact sequence of vector spaces $$ 0 \to H_1(S^1) \otimes \mathbb{Q} \to \mathbb{Q}^2 \to \mathbb{Q} \to 0 $$ and so dimension counting tells you that $H_1(S^1) \otimes \mathbb{Q} = \mathbb{Q}$. Now, by the fundamental theorem of finitely generated abelian groups (ok, this is perhaps a little bit of overkill---there may be a simpler reason for this), we know that $H_1(S^1) \cong \mathbb{Z}^{\oplus n} \oplus F$ for some finite abelian group $F$. The above argument tells us that $n = 1$.

Why is the fundamental theorem of finitely generated abelian groups relevant? Well, I suppose it depends on how you define $S^n$. If you are doing so as a CW complex, or anything like that, you know that (due to compactness) there are finitely many cells in each dimension, which lets you know that this theorem is applicable.

It is not too hard to show that $F$ must be zero, since the exact sequence above gives an injection $F \to \mathbb{Z} \oplus \mathbb{Z}$.

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  • $\begingroup$ Could you expand on why $H_1(S^1)$ would be isomorphic to $\mathbf Z \oplus F$? If you tensor by a field, the resulting sequence should be exact, but I fail to see what it entails. $\endgroup$
    – user25784
    Sep 25, 2014 at 17:25
  • $\begingroup$ I hope this is more clear now. $\endgroup$
    – Simon Rose
    Sep 25, 2014 at 17:30
  • $\begingroup$ It is. Thank you! I will wait a few hours to see if someone comes along with an elementary proof; otherwise, I will accept your answer. $\endgroup$
    – user25784
    Sep 25, 2014 at 17:56
  • $\begingroup$ This answer is even more overkill than what user25784 already mentioned in his question. $\endgroup$ Sep 25, 2014 at 18:38
  • $\begingroup$ @MartinBrandenburg - Yeah, I kind of realized that while fleshing out the details. The other proof is much simpler. $\endgroup$
    – Simon Rose
    Sep 26, 2014 at 6:57

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