A standard proof of $\pi_1(\mathbb{S}^1) = \mathbb{Z}$ using universal covering spaces

Question

I am looking for "a standard proof of $\pi_1(\mathbb{S}^1) = \mathbb{Z}$ using universal covering spaces", as suggested by the book Homotopy Type Theory (p. 255). What is this proof? Where can I find it?

Answer 1

The result here is that if $\tilde X$ is the universal cover of a path-connected topological space $X$, then the group of homeomorphisms of $\tilde X$ fixing the projection to $X$, called the group of deck transformations, is isomorphic to $\pi_1 (X)$. Next, observe that $\mathbb R$ is the universal cover of $\mathbb S^1$, since $\mathbb R$ is connected and simply connected, and covers $\mathbb S^1$ by the exponential map $x \mapsto e^{2\pi i x}$. In this case, the group of homeomorphisms of $\mathbb R$ fixing the exponential map is generated by $x \mapsto x+1$ and is infinite cyclic, and therefore isomorphic to $\mathbb Z$.

Edit: This proof or a variant of it should be in Hatcher's "Algebraic Topology" in Chapter 1.

Answer 2

Let $S^1$ be the unit circle space thinking of it as unit complex numbers and pick $1+0i$ to be a base point in $S^1$. We use the fact that $\mathbb{R}$ is a covering space of $S^1$ with the map $\pi:\mathbb{R}\to S^1$ given by $\pi(x) = e^{ix}$.

Now given a closed-curved $f:I\to S^1$ based at $1$ i.e. $f(0)=f(1) = 1$ there is a lift $F:I\to \mathbb{R}$ so that $\pi \circ F = f$. Note that $\pi(0) = 1$ and so we can choose the lift $F$ in such a way so that $F(0) = 0$. In fact, by results from covering space theory, there is exactly one such $F$ and so there is a unique lift to the covering space so that $F(0)=0$.

Given two closed-curves $f,g:I\to S^1$ based at $1$ if they are homotopic then $F,G$ - their corresponding unique lifts, respectively - have the property that $F(1) = G(1)$ (this is a theorem from covering space theory). So the lifted curves both end at the same point.

This allows us to define a map $\varphi: \pi_1(S_1,1)\to 2\pi\mathbb{Z}$ by $\varphi(f) = F(1)$. This is well-defined for all of the reasons given above. First, of all $\pi(S_1,1)$ consists of classes of homotopic curves so if $f_1,f_2$ are homotopic then based on the above paragraph $F_1(1) = F_2(1)$ so that it does not matter what curve we used for the representative in the fundamental group. Second, there is only one lift with the property $F(0) = 0$, again by the above paragraph so the choice of the lifted curve is unique. Third, $F(1)$ must be an integer-multiple of $2\pi$ (note, the $\pi$ for the covering map and $\pi$ as in the real-number is badly chosen notation!) since $\pi(F(1))=f(1) = 1$ and so $e^{iF(1)}=1$ which means that $F(1) = 2\pi n$ for some $n\in \mathbb{Z}$.

This map is surjective because $f:I\to S^1$ given by $f(t) = e^{i2\pi nt}$ has the property that $\varphi(f) = n$. It remains to show that $\varphi$ is injective and that $\varphi(f\cdot g) = \varphi(f) + \varphi(g)$ - where $f\cdot g$ is the multiplication taken in $\pi_1(S_1,1)$. Try checking these results yourself. The main idea is understanding the construction of this map $\varphi$ that gives you the isomorphism.

Edit: Let me show why $\varphi$ is injective because it is easy and leave the verification that $\varphi$ is a group isomorphism for you to check. If $\varphi(f) = \varphi(g)$ then their corresponding unique lifts with $F(0)=0$ and $G(0)=0$ have the same endpoint i.e. $F(1)=G(1)$. However, $\mathbb{R}$ has the simple property that if two curves start and end at the same point then they are homotopic. Thus, there is a homotopy $h:I\times I \to R$ now simply apply $\pi$ map to get a homotopy $\pi\circ h:I\times I\to S_1$ and so $\pi(F)=f,\pi(G)=g$ are homotopic. Which shows that $\varphi$ is injective.