$X$ is simply connected iff every continuous $f:S^1 \rightarrow X$ has a continuous extension to the unit disc?

Question

Given that a topological space $X$ is path connected, I need to show that $X$ is simply connected iff every continuous $f:S^1 \rightarrow X$ has a continuous extension to the unit disc. My professor explained to me the proof of the $\leftarrow$ direction but I didn't quite understand it:

$\Leftarrow$: Suppose that every continuous map from $S^1$ to $X$ has a continuous extension to the unit disc $D$. Let $x_0 \in X$ and let $\sigma$ be a loop at $x_0$. We need to show that $\sigma \simeq x_0$.

Let $\phi:I \rightarrow S^1$ be given by $\phi(s) = e^{2\pi{}is}$, and let $\bar{\sigma}: S^1 \rightarrow X$ be given by $\bar{\sigma}(e^{2\pi{}is}) = \sigma(s)$. We have that $\bar{\sigma} \circ \phi = \sigma$. Also, $\bar{\sigma}$ is a continuous map from $S^1$ to $X$, so there is a continuous function $\bar{F}: D \rightarrow X$ with $\bar{F}|_{S^1} = \bar{\sigma}$. Finally, let $\Phi:I \times I \rightarrow D$ be the map $\Phi(s, t) = te^{2\pi{}is}$. Our homotopy from $x_0$ to $\sigma$ is given by $F = \bar{F} \circ \Phi$.

It is not clear to me why this is a homotopy. Obviously $F(s, 1)$ = $\bar{F}(e^{2\pi{}is}) = \bar{\sigma}(e^{2\pi{}is}) = \sigma(s)$ as desired, but we also need $F(s, 1) = F(0, t) = F(1, t) = x_0$, and it's not clear to me why this should be the case. Any help is appreciated, thank you.

Answer 1

So indeed $F(s,1)=\bar{F}(\Phi(s,1))=\bar{F}(e^{2\pi is})=\bar{\sigma}(e^{2\pi is})=\sigma(s)$.

But now you need to show that $F(s,0)=F(0,t)=F(1,t)=x_0$. So we calculate: $F(s,0)=\bar{F}(0)$ (meaning $F$ is a free homotopy between $\sigma$ and constant $\bar{F}(0)$) and $F(0,t)=F(1,t)=\bar{F}(t)$ (which is not quite enough for it to be a path homotopy). Note that $t\in[0,1]$ and we treat the interval $[0,1]$ as a subset of the complex plane $\mathbb{C}$, formally $[0,1]\times\{0\}$.

Warning: be aware that in general the existance of a free homotopy need not imply the existance of a based homotopy. We need to do some work in this particular case.

So the question is: can we choose $\bar{F}$ in such a way that $\bar{F}(t)=x_0$ for all $t\in[0,1]\subseteq\mathbb{C}$? We know that $\bar{F}(1)=x_0$ because $\bar{\sigma}(1)=\sigma(0)=x_0$.

So lets ask another question: does there exist a continuous function $\Psi:D\to D$ such that $\Psi(z)=z$ for $z\in S^1$ and $\Psi(t)=1$ for $t\in[0,1]$? If such function exists then you simply modify $\bar{F}$ by taking $\bar{F}\circ\Psi$.

And indeed since $S^1\cup[0,1]$ is closed and $\Psi$ restricted to that subset is continuous then such function exists by the (generalized) Tietze extension theorem (together with the fact that $D$ is a retract of $\mathbb{R}^2$). There might be an explicit formula but I just couldn't figure it out.

Answer 2

I would like to redefine this question as follows: given that there is a continuous $F:I\times I \rightarrow X$ with $F(0,t) = f(t)$ and $F(1,t) = x_{0}$, and given that $F(s,t)$ is a loop for all fixed $s$, prove $f$ is null-homotopic. Here, as in your case, we need to find a homotopy which fixes a basepoint, which I will choose as $f(0)$. Define $g(t) = F(1-t,0)$ (there is slight abuse of notation here, dont be alarmed) and define $g_{s}$ as the restriction of g to $[0,s]$, taken with a reparamterization so it is still a path from $[0,1]$. geometrically, our loop $f$ is deformed freely to a point, so we can choose to follow the path a point on $f$ makes as it is being deformed to a point. Now define the following loop homotopy between $f$ and $f*\bar{f}$: $ G(s,t) = f(t)*g_{s}(t) * \bar{F}(s,t) * \bar{g_{s}}(t) $. If I have mistakes in my formulation of the homotopy, here is a geometric interpratation: we know $f * \bar{f} $ is null-homotopic. we can choose to deform $\bar{f}$ continuously to a point, while still connecting it with a path to $f$. Doing this will give us a loop homotopy between $f * \bar{f}$ and $f * g * \bar{g}$, which is loop homotopic while fixing the same basepoint as before to $f$. This shows $f*\bar{f} \sim f$, but $[f * \bar{f} ] = e$ and so $f$ is null-homotopic.