Homeomorphism from the solid Unit Cube to the solid Unit Ball? I have an explicit function from one to the other which is bijective. $$f(x) = \text{max}\lbrace |x_{i}|\rbrace_{0\leq i\leq n} \cdot \frac{x}{||x||}$$How do I know this is continuous? I see the intuition that the ray lines are continuous, but how about 'adjacent rays'. Also an explicit proof of continuity would be nice. 
 A: This is close. You have to tweak your function a bit, you did not define it for $0\in\mathbb{R}^n$. Let $I=[-1,1]^n$, $B=\{v\in\mathbb{R}^n\ |\ \lVert v\rVert\leq 1\}$. Then put
$$f:I\to B$$
$$f(v)=\begin{cases}
0 &\text{if }v=0 \\
\frac{\lVert v\rVert_{\infty}}{\lVert v\rVert}\cdot v &\text{otherwise}
\end{cases}$$
where $\lVert v\rVert_\infty=\max(|v_1|,\ldots,|v_n|)$ is the max norm. Note that with the max norm we can reformulate $I=\{v\in\mathbb{R}^n\ |\ \lVert v\rVert_\infty\leq 1\}$. And so a cube is nothing else than a closed ball but in a different norm.
This function is automatically continuous at any non-zero vector, because it is a composition of the following continuous functions: the identity $id(v)=v$, the scalar multiplication $m(r,v)=r\cdot v$ and both norm functions (all norms on $\mathbb{R}^n$ are continuous with respect to the Euclidean topology).
So the only question is whether this function is continuous at $0$? And this follows from the observation that any two norms on $\mathbb{R}^n$ are equivalent. Meaning there are positive constants $C_1,C_2>0$ such that
$$C_1\leq\frac{\lVert v\rVert_\infty}{\lVert v\rVert}\leq C_2$$
for any $v\in\mathbb{R}^n\backslash\{0\}$. In this particular case $C_1=1$ and $C_2=\sqrt{n}$ will do. And so if $v_m\to 0$ then $\frac{\lVert v_m\rVert_\infty}{\lVert v_m\rVert}\cdot v_m\to 0$.
With that it is easy to see that $f$ has a continuous inverse. Just take $g(0)=0$ and $g(v)=\frac{\lVert v\rVert}{\lVert v\rVert_\infty}\cdot v$.
Side note: There's nothing special about $\lVert v\rVert$ and $\lVert v\rVert_\infty$ norms here. This can be generalized to any two norms on $\mathbb{R}^n$.
