Extending to a disc means fundamental group is trivial

Question

Working on this problem.

If any $f:S^1\rightarrow X$ extends to a $F:D^2\rightarrow X$, then $\pi_1(X, x_0)$ is trivial.

We can turn any loop $f^\prime:I \rightarrow X$ in $\pi_1(X, x_0)$ into an $f:S^1\rightarrow X$ by defining $f(x) =f^\prime \left(\frac{1}{2\pi}\operatorname{Arg}(x) \right) $.

I want to finish without explicitly showing that $f$ is homotopic to a constant map. (This is part of the "the following are equivalent proof" with that as part (a) so I would like to avoid assuming that or proving it midway. I should be able to do it with the extension map only.) I think I am supposed to use induced maps here and make some kind of diagram but I'm not sure how.

Answer 1

$D^2$ is contractible, i.e. $D^2 \simeq \star$, where $\star$ is the one-point space. So $f$ is homotopic to a map which factors through $\star \to \star$, i.e. $f \simeq g$ where $g : D^2 \to \star \to X$, and the map $g$ must be constant.

I've left out a lot of details, let me know if you need a push in the right direction.

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More details: The fact that $D^2 \simeq \star$ is easy to see; let $u : D^2 \to \star$ be the unique map to the one-point space and let $v : \star \to D^2$ pick out the basepoint. Then $v \circ u \simeq \mathrm{id}_{D^2}$ via a linear homotopy.

But $v \circ u$ is constant, so let $g = f \circ v \circ u$. Then $f$ is constant, and $g \simeq f$ by composing the linear homotopy above with $f$.

Answer 2

I don't imagine there's any way simpler than just writing down an explicit formula for the homotopy from $f'$ to a constant map: $$H(x,t) = F\bigl( (1-t) \, e^{2 \pi i x} + t \bigr) $$ You can think of this homotopy $H$ as being factored as a composition of $F$ with the quotient map $[0,1] \times [0,1] \to D^2$ given by the formula $(x,t) \mapsto (1-t) \, e^{2 \pi i x} + t$.