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Suppose I have $a\in]0;1[$ and $\psi:\Bbb R\to \Bbb R^2 , t\to (t-a\ sin\ t,1-a\ cos\ t)$. My job is to prove that $\psi$ is an injective immersion and whose inverse is continuous (there's a technical term for it but I don't know it, in my country it directly translates to a "dive").

I managed to prove that it is in fact an injective immersion but for the inverse function I can't seem to do it. Here's my attempt:

$$x=t-a\ sin\ t \quad , \quad y=1-a\ cos\ t$$ $$cos^2\ t+sin^2\ t=1$$ $$t=x\pm\sqrt{a^2-(1-y)^2}$$

But that's as far as I got. Should I consider the plus or minus sign? Why? And how do I formalize the inverse function $\psi ^{-1}$?

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Let $C:=\psi({\mathbb R})$ be the cycloid, considered as a subspace of the plane. It is not possible to represent $\psi^{-1}\!\!:\> C\to{\mathbb R}$ explicitly in terms of elementary functions. But we can argue as follows:

Let two points ${\bf z}_1$, ${\bf z_2}\in C$ be given, and put $\psi^{-1}({\bf z}_i)=:t_i$. We may assume $t_2>t_1$. Then $$|{\bf z}_2-{\bf z}_1|=|{\bf z}(t_2)-{\bf z}(t_1)|\geq\bigl|x(t_2)-x(t_1)|=\int_{t_1}^{t_2}x'(t)\>dt\geq (1-a)(t_2-t_1)\ .$$ It follows that $$\bigl|\psi^{-1}({\bf z}_2)-\psi^{-1}({\bf z}_1)\bigr|=|t_2-t_1]\leq{1\over1-a}|{\bf z}_2-{\bf z}_1|\ ,$$ proving that $\psi^{-1}$ is even Lipschitz continuous on $C$.

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