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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.

Wyatt Gregory
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    Consider the disc $D^2$ as a cone over $S^1$ with vertex $0$. A map $f:D^2 \to X$ is literally a homotopy between $f\vert \partial D^2$ and $f\vert {0}$. – anomaly Sep 12 '18 at 01:46
  • @anomaly So since $\bar{F}$ restricted to $S^1$ is $\sigma$, this shows that $\sigma$ is homotopic to $\bar{F}(0)$? – Wyatt Gregory Sep 12 '18 at 02:17
  • There is a relative homeomorphism $(D^2,S^1)\cong (I^2,\partial I)$. – Tyrone Sep 12 '18 at 09:14
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    @anomaly it is not "literally" a homotopy, a cone is not "literally" $I\times I$. You would need to show that this "cone function" (that is not guaranteed to preserve endpoints) actually induces a homotopy. As I can see OP is interested in details. Your comment is not a solution, it just moves the burden of the construction to another area. – freakish Sep 12 '18 at 11:09
  • Of course it is--- it would be pedantic to insist otherwise--- and the OP can fill in the details for him- or herself. – anomaly Sep 12 '18 at 13:10
  • @anomaly Of course it isn't. Your $f$ function doesn't preserve the base point. Note that path homotopy has to fix the base point at every level, not only at $0$ and $1$. The concept of path homotopy is stronger then simply homotopy. OP even asked about that directly. Your $f$ has to be tweaked like in my answer. And from that perspective introducing cones is completely useless and confusing only, solving nothing actually. And if the OP was able to fill in gaps then he would not ask the question. – freakish Sep 12 '18 at 14:19
  • @anomaly You may want to read this: https://math.stackexchange.com/questions/2458349/base-point-homotopy-vs-free-homotopy-example – freakish Sep 12 '18 at 14:23
  • Why do you want an explicit homotopy? The point is that the disc is a contractible space. If every map from the circle factors through a contractible space it must be nullhomotopic. That's it. – Exit path Sep 12 '18 at 14:41
  • @leibnewtz you may be right. But the question is explicitly about the given construction and why it works. – freakish Sep 12 '18 at 15:09
  • @freakish That's why I didn't post my comment as an answer. It just seems unnecessary to do so much work when it's a two line proof – Exit path Sep 12 '18 at 16:11

2 Answers2

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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.

freakish
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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.

Tom Ariel
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