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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.

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    $\begingroup$ 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). $\endgroup$
    – Clayton
    Oct 16, 2021 at 22:48
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    $\begingroup$ math.stackexchange.com/q/3415419 is a very similar question. $\endgroup$
    – Paul Frost
    Oct 17, 2021 at 10:06

1 Answer 1

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

$g(x,y) = \frac{1}{\sqrt{x^+y^2}}(x,y)$...

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