Loop isn't contractible - compared to a path that looks like a loop

Question

Sanity check:

If $\gamma:I\to \Bbb S^1$ is given by $\gamma(t)=e^{2\pi it}$, then this is homotopic to a constant map. I.e. we can find a continuous map $H:I\times I\to\Bbb S^1$ such that $H(1,x)=\gamma(x)$ and $H(0,x)=c_1(x)=e^0$. In fact we can take $H(t,x)=e^{2\pi i xt}$ I suppose.

When one says the `loop' isn't contractible, it is important that we are consider a map $l:\Bbb S^1\to\Bbb S^1$, in which case a homotopy would be a continuous map $H:\Bbb S^1\times I\to \Bbb S^1$, in which case my map above isn't such a map, since $H([0],x)=e^0$ and $H([1],x)=e^{2\pi it}$ but $[0]=[1]$ in $\Bbb S^1=I/\{0,1\}$.

Answer 1

Yes, that's correct.

There are two common ways to formalize loops.

1. As maps $\gamma: I \to X$ with $\gamma(0) = \gamma(1)$.

2. As maps $\gamma : \mathbb{S}^1 \to \mathbb{S}^1$.

In the first case we're usually interested in homotopy relative to $\delta I$. That is, homotopies $H : \gamma_1 \simeq \gamma_2$ where $H(0,x)= \gamma_1(0) = \gamma_2(0)$ and $H(1,x) =\gamma_1(1) = \gamma_2(1)$ for all $x$, i.e. $H$ doesn't move the endpoints of the paths at all.

It's a bit more involved than the option 2, but option 1 makes it easier to handle things like base points.

Answer 2

What you're missing is the concept of path homotopy, which requires that $H : I \times I \to \mathbb S^1$ is constant on the subset $I \times \{0\}$ and on the subset $I \times \{1\}$. The correct statement is that $\gamma$ is not path homotopic to a constant path. It's an abuse of terminology to drop the word "path" from the phrase "path homotopic" in that last sentence, although it's a somewhat common abuse (in fact, one sees an example of this very abuse in the wikipedia link provided).