I have the following problem:
Let $$M:=([-1,1]\times\{-1\})\cup([-1,1]\times\{1\})\cup(\{-1\}\times[-1,1])\cup(\{1\}\times[-1,1])\subset\mathbb{R}^2.$$ Show that M is homeomorph to $S^1\subset\mathbb{R}^2$.
I found the functions $$f:S^1\rightarrow M \quad x\rightarrow \frac{x}{||x||_\infty}$$ and $$f^{-1}:M\rightarrow S^1 \quad x\rightarrow\frac{x}{||x||_2}.$$
But I have the problem to show that $f \circ f^{-1}= id_M ,f^{-1}\circ f=id_{S^1}$ Can someone help me with this problem?
Since $S^1=\{x:\|x\|_2=1\}$, the equation $f^{-1}\circ f=id$ holds, as the image of $x\in S^1$ keeps being a positive scalar multiple of $x$, i.e. stays on the ray of $x$, and it ends up with length $1\,$ (in $\|\_\|_2$ norm).
The same works for the other equation as well, once that $f$ is verified to take values indeed in $M$, because every ray starting from the origin meets $M$ in exactly one point.
If you want to see some calculation as well for $f\circ f^{-1}$, assume $x=(a,b)\in M$, say on the side $1=a\ge b\ge -1$. Then we get $$f(f^{-1}(1,b))\ =\ f\left(\frac 1{\|x\|_2},\,\frac b{\|x\|_2}\right)\ = \ (1,b) $$ as now $f$ divides its argument by $\left\|\left(\dfrac 1{\|x\|_2},\,\dfrac b{\|x\|_2}\right)\right\|_\infty=\dfrac 1{\|x\|_2}$, i.e. multiplies it by $\|x\|_2$.
The other sides of $M$ are analogous.