# The Fundamental Group of the Circle, Part 2

**EDIT 11/22/16: ***I encourage the reader to skip this post and proceed directly to Part 3. Part 2 merely contains the justification of a shortcut that we never actually use in the remainder of this series. In particular, we are certainly *not

*proving 'well-definedness,' even though that's what I claim. (Apologies, dear readers!) For more, see my explanation in the comments below.*

This week we continue our proof from Hatcher's *Algebraic Topology *that the fundamental group of the circle is isomorphic to $\mathbb{Z}$. Recall our outline:

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 5)

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

Last week we defined the map $\Phi:\mathbb{Z}\to\pi_1(S^1)$ by $n\mapsto[\omega_n]$ where $\omega_n:[0,1]\to S^1$ is the loop given by $s\mapsto (\cos{2\pi ns}, \sin{2\pi ns})$. Our goal for today is to prove that $\Phi$ is well-defined, and the proof is quite simple.

## Claim: Φ is Well-Defined

Recall that $\omega_n=p\circ\widetilde{\omega}_n$ is a loop from $I$ to $S^1$ where $\widetilde{\omega}_n:I\to\mathbb{R}$ is a path (a straight line, actually) from $0$ to $n$. We wish to show that equivalence class $[\omega_n]$ doesn't depend on its representative. In other words, we wish to show that $\omega_n$ doesn't depend on the path $\widetilde{\omega}_n$ from $0$ to $n$.

To this end, let $\widetilde{\gamma}_n$ be any *other* path from $0$ to $n$ in $\mathbb{R}$. Then $\widetilde{\omega}_n\simeq \widetilde{\gamma}_n$ are homotopic by the straight line homotopy $F:I\times I\to\mathbb{R}$ where $F(s,t)=(1-t)\widetilde{\omega}_n+t\widetilde{\gamma}_n$.

Set $\omega_n'=p\circ \widetilde{\gamma}_n$. If we can show the existence of a homotopy $G:I\times I\to S^1$ such that $$\omega_n=p\circ \widetilde{\omega}_n\;\underset{G}{\simeq}\; p\circ\widetilde{\gamma}_n=\omega_n',$$ then we'll be done since we can conclude $[\omega_n]=[\omega_n']$! But this is easy enough since we can simply let $G=p\circ F$ be the projection of the homotopy $F$ in $\mathbb{R}$ down to $S^1$:

And indeed $G$ is a homotopy since it's continuous (as both $p$ and $F$ are) and

- $G(s,0)=p(F(s,0))=p(\widetilde{\omega}_n(s))=\omega_n(s)$
- $G(s,1)=p(F(s,1))=p(\widetilde{\gamma}_n(s))=\omega_n'(s)$.

Therefore $[\omega_n]=[\omega_n']$ and so $\Phi$ is indeed well-defined.

Next time, we'll show that $\Phi$ is in fact a group homomorphism.