the mapping class group of the disk is trivial proof Proof : 

Identify $D^2$ with the closed unit disk in $\mathbb{R}^2$. Let $\phi : D^2 \rightarrow D^2$ be a homeomorphism with $\phi_{\partial D^2}$ equal to the identity. We define,

$F(x,t) = \begin{cases} (1-t)\phi(\displaystyle\frac{x}{1-t}) &amp; \text{$0 \leq |x| &lt; 1 - t$} \\ x &amp; \text{$1 - t \leq |x| \leq 1$} 
\end{cases}$

for $0 \leq t &lt; 1$ and we define $F(x,1)$ to be the identity map of $D^2$. The result is an isotopy $F$ from $\phi$ to the identity. 

What exactly is going on here in this proof visually? What is $F(x,t)$ telling us? I am not sure what the proof is doing here.
Just imagine you extend $\phi$ to a homeomorphism $\Phi$ of the whole plane by setting $\Phi(z)=\phi(z)$ whenever $|z|\leq 1$ and $\Phi(z)=z$ whenever $|z|\geq 1$. Then $F(\cdot,t)$ is obtained by considering $\Phi$ on a large disk and scaling that disk to be of unit size. You basically compress the area where the original $\phi$ is active to a disk of radius $1-t$, outside of which $\Phi$ acts as the identity.
Every map defined on a contractible space is nullhomotopic. So$ MCG(D^n)$={$id^*$.}

*:Of course, you may work in different categories and you want your maps to be either 
isotopic , diffeomorphic, etc. So,in the broadest sense,i.e,  up to homotopy alone, $$MCG(D^n)$$ is trivial.(If  we only mention a homotopy,  this is what we want ). Now, considering isotopies,  since $D^n$ is homeo. to $\mathbb R^n$ , and $MCG (\mathbb R^n)=$ {$\pm Id$} , it all comes down to deciding if $Id$ ~ -$Id$ . But, e.g., (for n=2 )$-Id \circ -Id$ (maybe unfortunate shorthand for the isotopy $z \rightarrow ze^{i\pi t} ; t \in [0,1]$) gives you an isotopy from $-Id$ to $Id$ , so you are then done.