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I have the following example in my lectures notes:
In $(\mathbb{R}^n,\varepsilon_n)$ , where $\varepsilon_n$ is the usual euclidean topology or the one induced from it. Let $ B=B(O,1)$ be the open ball with center the origin $O$ and radious $1$ and $Q=\{x \in \mathbb{R}^n| | x_i| <1, i=1,...,n \}$ the n-dimensional cube of center $O$ and side $ 2 $ (or in other words the open ball of center $O$ and radious $r$ with rispect to the metric $d_{L^{\infty}}$) We want to prove $ B$ is homeomorphic to $Q$

Let $x=(x_1,...,x_n) \in \mathbb{R}$, we have that $\|x\|_{L^{\infty}}=d_{L^{\infty}}(x,O)=max\{|x_i|| i=1,...,n\}$. Let $k(x)=\frac{\|x\|_{L^{\infty}}}{\|x\|}$, and let's consider the mapping $f:(\mathbb{R}^n,\varepsilon_n) \rightarrow (\mathbb{R}^n,\varepsilon_n)$ defined by: $$f(x)=\begin{cases} k(x)x & \text{if } x \neq 0, \\ 0 &\text{if } x=0\end{cases}$$

Now, $f$ is a homeomorphism from $(\mathbb{R}^n,\varepsilon_n)$ to itself such that $f(Q)=B$, that is $f$ is open

In fact $ 1/\sqrt(n) \leq k(x) \leq 1$ and $k(\alpha x)=k(x)$ for each $x \neq O$ and $ \alpha \in \mathbb{R}^n\setminus\{0\}$, then $ \lim_{x \to O}f(x)=O$ and $f$ is continuous over $\mathbb{R}^n$

Furthermore the mapping $$g(x)=\begin{cases} x\frac{1}{k(x)} & \text{if } x \neq 0, \\ 0 &\text{if } x=0\end{cases}$$

is defined and continous over $\mathbb{R}^n$ and $g\circ f=f \circ g = I_{\mathbb{R}^n}$, then $g=f^{-1}$ and f is a homeomorphism. Besides $\|f(x)\|=\|x\|_{L^{\infty}}$ and $\|g(x)\|_{L^{\infty}}=\|x\|$, so $f(Q)=B$

I need some help filling in the details . To prove f is an homeomorphism I have to prove it is continuous, bijective and open.

(1) continuity.

The relation $ 1/\sqrt(n) \leq k(x) \leq 1$ comes from the equivalence of the $L^p$ metrics:

I know $\|x\|_{L^{\infty}} \leq \|x\|_{L^p} \leq n^{1/p}\|x\|_{L^{\infty}}$ for $p=2$ and $\|x\|_{L^2}=\|x\|$ I get $ 1/\sqrt(n) \leq k(x) \leq 1$. Also, the expresion $k(\alpha x)=k(x)$ is straightforward, but then I can't figure out how to show $ \lim_{x \to O}f(x)=O$

(2)injectivity For the injectivity I tried the ususal approach: $f(x_1)=f(x_2)$, so $\frac{\|x_1\|_{L^{\infty}}x_1}{\|x_1\|} = \frac{\|x_2\|_{L^{\infty}}x_2}{\|x_2\|}$ but then I don't know how to get to $x_1=x_2$

(3) surjectivity For the surjectivity I need to show the equation $\frac{\|x\|_{L^{\infty}}x}{\|x\|}=y$ has a solution for any $y \in \mathbb{R}^n$, but I can't figure out how to solve for $x$

(4)openness

I don't get how $\|f(x)\|=\|x\|_{L^{\infty}}$ and $\|g(x)\|_{L^{\infty}}=\|x\|$ manage to conclude $f(Q)=B$

I was trying to do like this: By definition of image $f(Q)=\{y|f(x)=y, x \in Q\}$ I take $x \in Q \iff\|x\|_{L^{\infty}}<1 \iff$ $\|x\|_{L^{\infty}}= \|\frac{1}{k(x)}y\|_{L^{\infty}}=\frac{1}{k(x)}\|y\|_{L^{\infty}}<1 $ $\implies \|y\|_{L^{\infty}}<k(x)\leq 1$ and I am getting nowhere, I should be getting to something like $\|y\|<1$ but instead I have the infinity norm there

Any help would be much appreciated

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  • $\begingroup$ What is $\varepsilon_n$ in $(\mathbb R^n,\varepsilon_n)$ in your problem statement? $\endgroup$ – kimchi lover Jul 7 '20 at 15:18
  • $\begingroup$ @kimchi lover Its the usual euclidean topology $\endgroup$ – J.C.VegaO Jul 7 '20 at 15:20
  • $\begingroup$ Thanks. Edit your post to say so. You have $\varepsilon_n, \|\cdot\|$ and $\|\cdot\|_{L^2}$, all meaning the same thing. May I suggest you replace them all with $\|\cdot\|_2$, and so on? $\endgroup$ – kimchi lover Jul 7 '20 at 15:21
  • $\begingroup$ @kimchi lover all right, $\varepsilon_n$ is the euclidean topology. I think others denote like $\tau_e$, for other two you are right $\endgroup$ – J.C.VegaO Jul 7 '20 at 15:27
  • $\begingroup$ @kimchi lover Actually , I did it on purpose to see the conection of $p$ with the inequality in which it appears as an exponent: $\|x\|_{L^{\infty}} \leq \|x\|_{L^p} \leq n^{1/p}\|x\|_{L^{\infty}}$, then I did put $\|x\|_{L^2}=\|x\|$, $\endgroup$ – J.C.VegaO Jul 7 '20 at 15:29
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Continuity

$\lim_{x\rightarrow 0}\|k(x)x\|=\lim_{x\rightarrow 0} \frac{\|x\| \ \|x\|_{L^\infty}}{\|x\|}=0$

Injectivity and surjectivity

I would suggest the following aproach:

Prove that: $x\in Q$ implies that $f(x)\in B$ $x\in B$ implies that $g(x)\in Q$

Then prove that $g$ is the inverse of $f$, i.e. that $f(g(x))=x$ and $g(f(x))=x$.

This is enough to show that $f$ is a bijection from $Q$ to $B$.

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  • $\begingroup$ Not sure how $x\in Q$ implies that $f(x)\in B$ , $x\in B$ implies that $g(x)\in Q$ together wih having the inverse implies f is bijective $\endgroup$ – J.C.VegaO Jul 7 '20 at 17:51
  • $\begingroup$ Functions with an inverse are bijective. math.stackexchange.com/q/1215365/15624 $\endgroup$ – Angela Pretorius Jul 8 '20 at 4:04
  • $\begingroup$ ok, and what about these: " $x\in Q$ implies that $f(x)\in B$ , $x\in B$ implies that $g(x)\in Q$" what are they for? $\endgroup$ – J.C.VegaO Jul 8 '20 at 8:50

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