# 7.05 Fundamental group of the circle

## Video

Below the video you will find accompanying notes and some pre-class questions.

## Notes

### Fundamental group of the circle

(0.00) In this section, we will see a proof that $$\pi_1(S^1,1)=\mathbf{Z}$$.

Given an integer $$n$$, define the loop $$\delta_n(t):=e^{2\pi int}$$.

Define the homomorphism $$F\colon\mathbf{Z}\to\pi_1(S^1,1)$$ by $$F(n)=[\delta_n]$$. We need to check:

• this is a homomorphism,
• $$F$$ is injective,
• $$F$$ is surjective.

#### $$F$$ is a homomorphism

(1.36) $$F(0)=[\delta_0]$$ and $$\delta_0(t)=e^0=1$$ is the constant loop, so $$F$$ takes the identity to the identity.

(2.23) To see that $$F(m+n)=F(m)\cdot F(n)$$, we need to check that $$\delta_n\cdot\delta_m\simeq\delta_{m+n}$$ [NOTE: I got my concatenation the wrong way around in the video]. Let $$p\colon\mathbf{R}\to S^1$$ be the covering map $$p(x)=e^{ix}$$ and let $$\tilde{\delta}_n$$ be the unique lift of $$\delta_n$$ with the initial condition $$\tilde{\delta}_n(0)=0$$, in other words $$\delta_n(t)=e^{i\tilde{\delta}_n(t)}$$.

(4.03) Let $$\tilde{\delta}_m$$ be the unique lift of $$\delta_m$$ to the same covering space with $$\tilde{\delta}_m(0)=\tilde{\delta}_n(1)$$. The concatenation $$\tilde{\delta}_n\cdot\tilde{\delta}_m$$ makes sense and is a lift of $$\delta_n\cdot\delta_m$$.

(5.23) Finally, let $$\tilde{\delta}_{m+n}$$ be the unique lift of $$\delta_{m+n}$$ to the same covering space with $$\tilde{\delta}_{m+n}(0)=0$$.

(6.28) We have

\begin{align*} \tilde{\delta}_{m+n}(t)&=2\pi(m+n)t)\\ \tilde{\delta}_m(t)&=2\pi mt\\ \tilde{\delta}_n(t)&=2\pi(nt+m) \end{align*}

so $$\tilde{\delta}_n\cdot\tilde{\delta}_m$$ and $$\tilde{\delta}_{m+n}$$ are paths which start at $$0$$ and end at $$2\pi(m+n)$$.

(9.16) Since $$\mathbf{R}$$ is simply-connected, any two paths with the same endpoints are homotopic rel endpoints, so $$\tilde{\delta}_m\cdot\tilde{\delta}_n\simeq\tilde{\delta}_{m+n}$$ via a homotopy $$H$$. The homotopy $$p\circ H$$ is then a homotopy $$\delta_n\cdot\delta_m\simeq\delta_{m+n}$$.

#### $$F$$ is injective

(10.30) As $$F$$ is a homomorphism, it suffices to show that $$n\in\ker(F)$$ implies $$n=0$$. We therefore want to show that if $$n\neq 0$$ then $$\delta_n$$ is not homotopic to the constant loop.

(11.18) For this, we will show that, for a suitable covering space, the monodromy $$\sigma_{\delta_n}$$ is nontrivial. Let $$p\colon\mathbf{R}\to S^1$$ be the covering space $$p(x)=e^{ix}$$. For each $$2\pi k\in 2\pi \mathbf{Z}=p^{-1}(1)$$, the path $$\tilde{\delta}_n(t)=2\pi(nt+k)$$ is a lift of the loop $$\delta_n$$ with initial condition $$2\pi k$$, so $\sigma_{\delta_n}(2\pi k)=\tilde{\delta}_n(1)=2\pi(n+k).$ We see that the monodromy $$\sigma_{\delta_n}\colon 2\pi\mathbf{Z}\to 2\pi\mathbf{Z}$$ is $$2\pi k\mapsto 2\pi(k+n)$$, which is nontrivial if $$n\neq 0$$.

#### $$F$$ is surjective

(15.07) Given a loop $$\gamma\colon[0,1]\to S^1$$ based at $$1\in S^1$$, we want to find an $$n$$ such that $$\delta_n\simeq\gamma$$. Take the same covering space $$p\colon\mathbf{R}\to S^1$$ and lift $$\gamma$$ to get a path $$\tilde{\gamma}$$ in $$\mathbf{R}$$ with $$\tilde{\gamma}(0)=0$$ (so $$\gamma(t)=\exp(i\tilde{\gamma}(t))$$). If $$n=\tilde{\gamma}(1)/2\pi$$ then $$n\in\mathbf{Z}$$. Let $$\tilde{\delta}_n$$ be the unique lift of $$\delta_n$$ with $$\tilde{\delta}_n(0)=0$$. Then $$\tilde{\delta}_n$$ and $$\tilde{\gamma}$$ are paths in $$\mathbf{R}$$ connecting $$0$$ to $$2\pi n$$. Because $$\mathbf{R}$$ is simply-connected, these paths are homotopic via a homotopy $$H$$, which implies that $$\delta_n\simeq\gamma$$ via the homotopy $$p\circ H$$.

## Pre-class questions

1. What was the key property of the covering space $$p\colon\mathbf{R}\to S^1$$ which made this proof work?