In $\mathbb{R}^2$, let $V^2=\overline{B(0,1)}$. Prove $V^2$ is homeomorphic to $I\times I$ where $I=[0,1]$

Question

In $\mathbb{R}^2$, let $V^2=\overline{B(0,1)}$. Prove $V^2$ is homeomorphic to $I\times I$ where $I=[0,1]$

My work:

I know two spaces $A,B$ are homeomorphic if exists a function $f:A\rightarrow B$ homeomorphic.

Definition: Let $f:A\rightarrow B$ a function. We say f is homeomorphic is $f$ is bijective, continuous and $f^{-1}$ is continuous.

Then,

We need find a function $f$ such that $f:\overline{B(0,1)}\rightarrow[0,1]\times[0,1]$ bijective, continuous such that $f^{-1}$ is continuous.

I'm stuck here, can someone help me?

Answer 1

You can try $f(x)=x \|x\| / \|x\|$ where the two norms are different. Can you continue from there?