We say that a topological set $S$ is $contractable$ if there exists $s_{0}$ in $S$ and a continous map $K: S\times[0,1]\rightarrow S$ such that for all $s $ in $S$ we have $K(s,0) = s$ and $K(s,1)= s_{0}$
1.Show that, if $S $ is contractable, then S is simply connected.
Now I found the definition we use in our class for simply connected, which is as follows:
An open subset U ⊆ C is called simply connected if it is path-connected and if any closed curve is homotopic to a constant curve.
I was able to prove path connectedness, however, seem to be stuck in proving the second part. Some help on that would be greatly appreciated.