# 7.05 Fundamental group of the circle

## Video

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

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## 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

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

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

## Navigation

- Previous video:
**7.04 Homotopy lifting, monodromy**. - Next video:
**7.06 Group actions and covering spaces, 1**. - Index of all lectures.