If exist a extension of $f$ to a map $D^2\to X$, then $f: S^1\to X$ is homotopic to a constant map

Question

Every map $f: S^1\to X$ is homotopic to a constant map, if and only if, exist a extension of $f$ to a map $D^2\to X$.

Here, $X$ is an arbitrary topological space. I managed to prove the direction $\Rightarrow$, for reference I will also leave some links for similar answers. (1), (2), (3), (4), (5).

I know that there is a list of equivalences, in this case:

The following three conditions are equivalent:

1. Every map $S^1\to X$ is homotopic to a constant map, with image a point.
2. Every map $S^1\to X$ extends to a map $D^2\to X$.
3. $\pi_1(X, x_0) = 0$ for all $x_0\in X$.

As I said, the direction I prove is $1\Rightarrow 2$, and the direction I'm having trouble proving is $2\Rightarrow 1$. From the list of equivalences, it would be enough to prove $2\Rightarrow 3$ and then $3\Rightarrow 1$, but I would like to know the direct demonstration of $2\Rightarrow 1$.

Answer 1

Consider a continuous function $f : S^1 \to X$, and let $\widetilde{f\,} : D^2 \to X$ one of its extensions. Now note that if $z \in S^1$ and $t \in [0,1]$, $$|(1-t)z+t| \leq (1-t)|z|+t = 1$$ and so we can define $H : S^1 \times [0,1] \to X$ by $H(z,t) := \widetilde{f\,}((1-t)z+t)$. Finally, observe that $H(-,0) = f$ and that $H(-,1)$ is the constant map $z \mapsto f(1)$.

Answer 2

Thinking of $D^2$ as consisting of concentric circles with shrinking radii, we can use a homotopy $H(re^{i \theta}, t) := \tilde{f}(tre^{i \theta})$, where $\tilde{f}$ is the extension of $f$.