# Homeomorphism between $S^1$ and $K$ (square in the taxicab metric)

Let us consider $$\mathbb{R}^2$$ in the euclidean metric, and let $$S^1 = \left\{(x,\,y) \in \mathbb{R}^2 \colon x^2 + y^2 = 1\right\}$$ and $$K \doteq \left\{(x,\,y) \in \mathbb{R}^2 \colon \lvert x \rvert + \lvert y \rvert = 1\right\}$$ as subspaces. I have to prove that $$S^1$$ and $$K$$ are homeomorphic.
My obvious guess was to define $$f \colon S^1 \longrightarrow K$$ as $$f(x,\,y) \doteq \left(\frac{x^2}{x^2 + y^2},\,\frac{y^2}{x^2 + y^2}\right).$$ It is continuous, since both $$u(x,\,y) \doteq \frac{x^2}{x^2\,+\,y^2}$$ and $$v(x,\,y) \doteq \frac{y^2}{x^2\,+\,y^2}$$ are continuous in $$S^1$$, and $$f(x,\,y) \in K$$, for all $$(x,\,y) \in S^1$$, but I'm not sure how to prove it's bijective and it's inverse $$g$$ such that $$g(u,\,v) = (x,\,y)$$ is continuous as well.
My guess is that it fails to be injective, since $$f(a,\,b) = f(x,\,y)$$ would imply $$ay = bx$$, and we can't guarantee the uniqueness of the solution. Any ideas will be appreciated, thanks in advance.

• I’m not sure if this works exactly as you want, but here is my idea: I would suggest considering the ray emanating from the origin at an angle of $\theta$ for any value of $\theta\in[0,2\pi)$. Any such ray intersects $K$ and $S^1$ in exactly one point. Then define the function $f$ to map from one point in $S^1$ to its corresponding point in $K$. Intuitively, this is clearly a homeomorphism (though it may not be as easy to prove, so I’m just leaving my idea as a comment). Oct 16, 2021 at 22:48
• math.stackexchange.com/q/3415419 is a very similar question. Oct 17, 2021 at 10:06

Use $$f(x,y) = \frac{1}{|x|+|y|}(x,y)$$, from $$S^1$$ to $$K$$ with inverse
$$g(x,y) = \frac{1}{\sqrt{x^+y^2}}(x,y)$$...