# Is there a covering space proof of $\pi_1(S^1) \cong \mathbb{Z}$ .

I'm looking at 29 pages. When he does the does the calculation of the fundmental group of circle.

Was wondering is there a easier way using covering spaces to prove this. As I'm trying to find a quick and easy way to do this. But, I can't seem to find one. I found an easy way to show that sphere and higher spheres are zero. But, that was using van kampen theorem and it won't work for the circle.

So was wondering what is a neat way to prove this.

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$t \mapsto e^{2\pi i t}$ is a covering map $\mathbb{R} \to S^1$. – t.b. Mar 12 '12 at 0:38
@t.b. I know, but don't see the point of covering spaces. Like I can't see how you use it to do any calculations. It's just heres a cover. – simplicity Mar 12 '12 at 0:41
You can use covering spaces for all sorts of things. For example, Reidemeister torsions. – Neal Mar 12 '12 at 12:00
"I'm looking at 29 pages. When he does the does the calculation of the fundmental group of circle." is not the clearest way to start a question. You mean you're looking at page 29 of Hatcher? – Matt Mar 12 '12 at 15:42
@simplicity: as an example of their utility, when is the last time you have thought of angle measure as being anything other than a real number, possibly together with the notion that it should be modulo $2 \pi$? – Hurkyl Apr 29 '12 at 13:43
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The standard proofs have the advantage of not needing a lot of extra theory. Although the covering space method is quite elegant, there is a lot of extra theory you must learn to really understand it, probably more than 29 pages. I won't try to explain all of the theory, but I will go over the argument and point out where the big gaps are with a (!). To truly understand, it is imperative that you fill in those gaps, or else you will likely make a mistake when trying to apply the same technique is a different situation. Also, the background theory is totally awesome on its own.

First of all, homotopy theory is a theory of pointed spaces, so chose base points $0\in\mathbb{R}$ and $1\in S^1(\subset\mathbb{C})$

We know that $\mathbb{R}$ is the universal covering space of $S^1$ (!), with covering map $p$ given by $p(t)= e^{2\pi it}$ (as t.b. pointed out in the comments). Intuitively, this map wraps $\mathbb{R}$ around the circle, with all of the integers lining up exactly on top of $1\in S^1$. $\mathbb{R}$ being the universal cover means that $\pi_1(S^1,1)$ is naturally isomorphic to the group of "deck transformations" of the covering $p:\mathbb{R}\rightarrow S^1$ (!). (By Deck transformation I mean an automorphisms $\alpha:\mathbb{R}\rightarrow\mathbb{R}$ which "preserves projection" i.e. $p=p\circ\alpha$.) One can verify from definition that the deck transformations in this case are just the linear translations of $\mathbb{R}$ by integer values, i.e. $p=p\circ\alpha$ iff $\alpha(x)=x+n$ for $n\in\mathbb{Z}$. This group of deck transformations is easily isomorphic to $\mathbb{Z}$. Tying it all together with the theory, we see $\pi_1(S^1,1)\cong\mathbb{Z}$

Although the proof itself is relatively straightforward there is still the issue of developing the covering space theory. This involves some very non-trivial steps: homotopy-lifting property, existence/uniqueness of universal covering spaces (for nice enough base spaces), $\pi_1(B)$ acting on a covering $p:E\rightarrow B$ by Deck transformations, ETC. I think the proof is a bit cuter, but technically it isn't any easier, you just encapsulate background theory.

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Thanks. That seems like a much better proof. I will look at it more detail tomorrow. But, yeah didn't like the proof given by lecturer. – simplicity Mar 12 '12 at 1:08
I personally like the covering space method a lot because it has very broad applications, but if you're a pragmatist and just want to know that $\pi_1(S^1)\cong\mathbb{Z}$ then I think it's way overkill. – you Mar 12 '12 at 1:14

I mention that when I was a writing a topology text in the 1960s whose 1st edition was published in 1968 and whose 3rd edition is now "Topology and groupoids" (2006), I got irritated by the fact that one had to make a detour through a bit, or a lot, of covering space theory to obtain the fundamental group of the circle, THE basic example in algebraic topology. So it was great to find that using the fundamental groupoid $\pi_1(X,A)$ on a set $A$ of base points ($2$ base points for the circle, not much more than $1$ base point) one could do this and much more; so that book moved over to giving a full account of $\pi_1(X,A)$. And the question of the use of groupoids in topology, particular higher homotopy, has for me been rewarding ever since.

My understanding is that in the intervening 44 years no other text mentions $\pi_1(X,A)$! So who is right?

Grothendieck has written: ""Choosing paths for connecting the base points natural to the situation to one among them, and reducing the groupoid to a single  group, will then hopelessly destroy the structure and inner symmetries of the situation, and result in a mess of generators and relations no one dares to write down, because everyone feels they won't be of any use whatever, and just confuse the picture rather than clarify it. I have known such perplexity myself a long time ago, namely in Van Kampen type situations, whose only understandable formulation is in terms of (amalgamated sums of) groupoids.""

May 28: I ought to explain in some detail how groupoids work for $\pi_1(S^1)$. Consider the following two pushout diagrams

where $\mathbf I\cong \pi_1([0,1], \{0,1\})$ is the groupoid with $2$ objects $0,1$ and non-identity arrows $\iota: 0 \to 1, \iota^{-1}: 1 \to 0$; the proof of this is easy since the unit interval is convex in $\mathbb R$. The first diagram shows that the circle $S^1$ is obtained from the unit interval $[0,1]$ by identifying, in the category of spaces, the two end points $0,1$. The second diagram shows that the additive group of integers is obtained from the groupoid $\mathbf I$ by identifying, in the category of groupoids, the two end points $0,1$. Note also that the groupoid $\mathbf I$ has only $4$ arrows, so it is easy to work out everything about it! This groupoid plays a role in the category of groupoids analogous to that of the integers in the category of groups.

All this seems a convincing reason for the result on $\pi_1(S^1)$, to be placed alongside the covering space proof, which is valuable for other reasons.

The above proof is also valuable for suggesting higher dimensional versions, giving, for example, $\pi_n(S^n), \pi_n(S^n \vee S^1)$.

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 I had forgotten the book Michel Zisman, Topologie algébrique élémentaire, A. Colin, 1972. – Ronnie Brown Apr 29 '12 at 16:14

Using the fact that the fundamental group is isomorphic to the group of Deck transformations of the universal covering $\mathbb R \to S^1$ certainly yields a quite elegant proof but, as "you" pointed out, it involves a lot of background theory. However you can use the covering $p: \mathbb R \to S^1, p(t) = e^{2\pi it}$ together with only the path-lifting property to define an isomorphism $\pi_1(S^1) \to \mathbb Z$.

Namely, choose $0 \in \mathbb R$ and $1 \in S^1 \subset \mathbb C$ as base points and let $[f] \in \pi_1(S^1)$ be the homotopy class of a loop $f$. By the path-lifting property, $f$ can be lifted to a path $\tilde f$ in $\mathbb R$ with $f = p \circ \tilde f$. Furthermore, $\tilde f$ can be chosen in such a way that it starts at $0$. Since $f$ is a loop the end point of $f$ is again the base point of $S^1$ so the lifted path $\tilde f$ ends at some integer $n \in \mathbb Z$ (because $\mathbb Z$ is the preimage of $1 \in S^1$ under the covering $p$). The mapping $[f] \mapsto n$ is our isomorphism $\pi_1(S^1) \to \mathbb Z$.

Of course, one still has to show that this is independent of the choice of $f$ from its homotopy class, and that the map is an isomorphism, but this is not that difficult; at least it shouldn't take 29 pages.

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