Show that $\pi_1(D,b)$ is a trivial group I would like to proof that $\pi_1(D,b)$ is a trivial group, where $D:=\{\langle x,y\rangle:x^2+y^2\leq1\}$, and $b$ is the base point of $D$. For that, let $f:[0,1]\rightarrow D$ be any continuous map such that $f(0)=f(1)=b$ and $e:[0,1]\rightarrow D$ as $e(v):=b, \ \forall \ v\in[0,1]$. Then, I think I have to find $H:[0,1]^2\rightarrow D$ such that $H(v,0)=f(v)$, $H(v,1)=e(v)$ and $H(\{0,1\},t)=b$, $\forall \ v,t\in[0,1]$. Using another way, my teacher suggested us taking $F:D\times[0,1]\rightarrow D$ given by $F(r,\theta,t)=(rt,\theta)$, to reduce every loop in $D$ to a point, I guess. But I couldn't understand that and to get $H$ as above seems more straightforward to me. I'd appreciate some help, please. Thank you!

Consider the map $H : [0,1]^2 \to D$ given by $$H(s,t) = (1-t)f(s) + tb$$ for all $s,t\in [0,1]$. 
