# The Fundamental Group of the Circle, Part 1

/*Algebraic Topology*section 1.1, and so I will assume the reader knows the definition of a homotopy, a loop, fundamental groups, etc.

*lost*. Don't get me wrong though - the

*ideas*aren't difficult at all, but filtering through the notation/language was a little daunting. So I decided to rewrite everything (with lots of pictures!) to fill in some of the "white space between the lines" of Hatcher's proof. Also, the ideas found here appear over and over again in a study on covering spaces (Hatcher's section 1.3), so taking our time through this material will be worthwhile. But brevity and accessibility seem to be inversely proportional in mathematics, so this series is somewhat lengthy.

*Fair warning!*

Let me begin by stating the theorem, then I'll give the outline of the proof.

**Theorem:**The map $\Phi:\mathbb{Z}\to \pi_1(S^1)$ given by $n\mapsto [\omega_n]$, where $\omega_n(s)=(\cos{2\pi ns},\sin{2\pi ns})$ for $s\in[0,1]$ is a loop in $S^1$ based at $(1,0)$, is an isomorphism. In other words $\pi_1(S^1)\cong \mathbb{Z}$.

*Outline of the proof:*

__Part 1:__ Set-up/observations

__Part 2:__ Show $\Phi$ is well defined

__Part 3:__ Show $\Phi$ is a group homomorphism

__Part 4:__ Show $\Phi$ is surjective

__Part 5:__ Show $\Phi$ is injective (Note: parts 4 and 5 require two lemmas whose proofs we will defer until part 6)

__Part 6:__ Prove the two lemmas used in parts 4 and 5

Let's get started!

## Set-up/Observations

**lift**of $\omega_n$. Finally, as a closing remark, notice that as the path $\widetilde{\omega}_n(I)$ climbs up/down the helix, its shadow, i.e. the projection $\omega_n$, winds around $S^1$. This correspondence beween the path in $\mathbb{R}$ and its shadow in $S^1$ is crucial to the proof, so I've included a couple of pictures to help the image 'stick.'

The following video illustrates the above rather nicely, especially the animation from 0:45 to 1:05. (After the 1:05 mark, the animation relates to covering spaces which we are *not* discussing in today's post.)