I have to show that $\pi_1(D^2) = 0$. I can't use the fact that $D^2$ is convex, instead I have to show that $D^2$ is contractible and thus simply connected.
I know that:
Def: A space X is called contractible if X is homotopy equivalent with a point.
Unfortunately I have no clue how to work with this!
Q: How do I show that $D^2$ is contractible, by using the definition given above?
If $x \in D^2$, so is $tx$ for all $t \in [0,1]$. ($||x|| \le 1$ implies $||tx|| = |t|\cdot||x||\le |t| \le 1$). So $H(x,t) = (1-t)(x)$ from $D^2 \times [0,1]$ to $D^2$ is an homotopy between $1_{D^2}$ and the constant map with value $0$.