J.Lee:s "Topological Manifolds", prop. 7.16 (b) $\implies$ c)).

Question

There is another answer to the same proof, given here: Proof explanation, prop 7.16 "Introduction to Topological Manifolds" by Lee but my question is mostly focused on another point.

In J. Lee.s "Introduction to Topological Manifolds", we are given the following:

Let $X$ be a topological space. Suppose $f:I \to X$ is a loop based at $p \in X$, and $\tilde{f}:S^1 \to X$ is its circle representative. Then the following are equivalent:

(a). $f$ is a null-homotopic loop.

(b). $\tilde{f}$ is freely homotopic to a constant loop.

(c). $\tilde{f}$ extends to a continuous map from the closed disk to $X$.

The beginning of the proof for $b) \implies c)$ looks like this:

$\ldots$ Next, assume that $\tilde{f}$ is freely homotopic to the constant map $k:S^1 \to X$, and let $H:S^1 \times I \to X$ be the homotopy $H_0 = k$ and $H_1 = f$.

Here is my confusion; how do we know that $f$ and $k$ are homotopic, if we assume that $\tilde{f}$ and $k$ are homotopic?

Remark I: By $\tilde{f}$ being the circle representative of $f$ we mean that $\tilde{f}$ is the unique map such that $\tilde{f} \circ \omega = f$ where $\omega:I \to S^1$ is explicitly defined by $s \mapsto e^{2\pi i s}$.

Remark II: Of course, the proof does not directly state that there actually exists such a homotopy, but I think we do need the existence of one for the proof to go through (we want to show that there is an extension of $\tilde{f}$ as in c)).

Answer 1

It is a typo. The maps $f$ and $k$ have different domains $I$ and $S^1$, therefore they cannot be homotopic. You have to take the circle representative instead of $f$.

Also see Characterizing simply connected spaces for the general context.