# Thread: Disc extension (Fundamental group question)

1. ## Disc extension (Fundamental group question)

Let $f:S^1\rightarrow X$. Show that $[f]=1 \in \pi_1(X)$ iff $f$ extends to the unit disc $D^2$

Am I right in thinking that the fundamental group of this is trivial, so the identity map is homotopic to any "loop" $f:S^1\rightarrow X$? Any hints are greatly appreciated.

2. Originally Posted by skamoni
Let $f:S^1\rightarrow X$. Show that $[f]=1 \in \pi_1(X)$ iff $f$ extends to the unit disc $D^2$

Am I right in thinking that the fundamental group of this is trivial, so the identity map is homotopic to any "loop" $f:S^1\rightarrow X$? Any hints are greatly appreciated.
Denote an element of $S^1$ by $e^{i\theta}$. If $[f]=1 \in \pi_1(X)$ then there is a homotopy $h:[0,1]\times S^1\to X$ such that $h(0,e^{i\theta}) = x_0$ (a fixed point in X), and $h(1,e^{i\theta}) = f(e^{i\theta})$. Then the map $re^{i\theta}\mapsto h(r,e^{i\theta})$ is an extension of f to the unit disc.

That construction is essentially reversible. Given an extension of f to the disc, you can use it to construct a homotopy from f to a constant map.