Homotopy between the trivial element of $\pi_1(S_1)$ and a single loop around the circle

Question

I am new to this topic of homotopy and the fundamental group. I have just read the proof that the fundamental group of $S_1$ is isomorphic to $\mathbb{Z}$.

There's something that has been bugging since the beginning. Can't I define an homotopy between the constant loop staying at $(1,0)$ and the loop going around the circle counter clockwise as follows? $$H(s,t) = (\cos(2\pi st), \sin(2\pi st))$$

for $t=0$ we get $H(s,0)=(\cos(0),\sin(0))$ for all $s \in [0,1]$ and $H(s,1)=(\cos(2\pi s),\sin(2\pi s))$ for all $s \in [0,1]$.

Also, $H(s,t)$ is continuous as a composition of continuous functions $p(x)=(\cos(2\pi x), \sin(2\pi x))$ with $p:[0,1]\to S_1$ and $g(s,t)=st$ where $g:[0,1]^2 \to [0,1]$.

What am I missing here?

Answer 1

You're missing that the homotopy isn't supposed to be a homotopy between paths, it's supposed to be a homotopy between loops. Which is to say, we need to have $$ H(0,t)=H(1,t) $$ for all $t\in[0,1]$ (and in some interpretations this value is required to be constant as well).

Answer 2

Welcome to MSE!

You have a great question. It's related to a common source of confusion when first learning about the fundamental group and homotopy. You proposed the homotopy $H(s,t) = (\cos(2\pi st), \sin(2\pi st))$, which is indeed continuous, but there is an issue that arises when trying to use it as a homotopy between the constant loop and the loop going around the circle counterclockwise.

The problem lies in the fact that your homotopy is not constant at the endpoint $t=1$ for all $s \in [0,1]$. Specifically, at $t=1$, the homotopy takes the loop to the point $(\cos(2\pi s), \sin(2\pi s))$ instead of the starting point $(1,0)$. This means that the endpoint of the loop changes as $t$ varies from 0 to 1, which breaks the requirement for a homotopy.

In order for a homotopy between loops to be valid, the endpoints of the loops need to remain fixed throughout the homotopy. In this case, the constant loop is based at $(1,0)$, and the loop going around the circle counterclockwise is also based at $(1,0)$. Therefore, any valid homotopy must keep the endpoints fixed, and your proposed homotopy does not satisfy this condition.

The standard homotopy for showing that the fundamental group of $S^1$ is isomorphic to $\mathbb{Z}$ involves using polar coordinates and is defined as follows: $$H:[0,1]\times[0,1]\rightarrow S^1$$ $$H(s,t)=(\cos(2πt),\sin(2πt))$$

This homotopy continuously deforms the loop going around the circle counterclockwise to the constant loop based at $(1,0)$ while keeping the endpoints fixed. This shows that the two loops are homotopy equivalent.