# The Fundamental Group of the Circle, Part 3

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__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

## Claim: Φ is a Homomorphism, i.e. Φ(m+n)=Φ(m)∙Φ(n)

*Notice that in our "helix model" of $\mathbb{R}$, this is a shift up or down by $|m|$.*This observation lets us conclude that $$\widetilde{\omega}_m\cdot \tau_m \widetilde{\omega}_n\;\underset{F}{\simeq}\;\widetilde{\omega}_{m+n}$$ by some homotopy $F:I\times I\to\mathbb{R}$ in $\mathbb{R}$. This becomes evident once we pause for a bit to think about what these paths really are:

- On the right-hand side, $\widetilde{\omega}_{m+n}$ is simply the line (path) from $0$ to $m+n$ in $\mathbb{R}$
- On the left-hand side, $\widetilde{\omega}_m\cdot \tau_m \widetilde{\omega}_n$ is the product of the path from $0$ to $m$ (in $\mathbb{R}$) with the path from $m$ to $n+m$ (in $\mathbb{R}$). (Note: $\tau_m \widetilde{\omega}_n$ takes the line which starts at $0$ and ends at $n$ and shifts it so that the starting point is now $m$ and the ending point is now $m+n$.) In other words, it is the union of the green and blue paths below.

So when we project each of these paths onto $S^1$, we see that the resulting paths are *also* homotopic:
$$p\circ(\widetilde{\omega}_m\cdot \tau_m \widetilde{\omega}_n)\underset{G}{\simeq}p\circ \widetilde{\omega}_{m+n}$$
where $G:I\times I\to S^1$ is $G=p\circ F$. And now we're basically done! We simply need to write down the following equalities: