from Elementary Differential Geometry By Andrew Pressley
A parametrized curve $\overline {\gamma}\colon(\overline {\alpha},\overline {\beta})\to R^n$ is a reparametrization of a parametrized curve $ {\gamma}\colon( {\alpha}, {\beta})\to R^n$ if there is a smooth bijective map $\phi\colon(\overline {\alpha},\overline {\beta})\to( {\alpha}, {\beta})$ (the reparametrization map) such that the inverse map $\phi^{-1}\colon( {\alpha}, {\beta})\to(\overline {\alpha},\overline {\beta})$ is smooth and $\overline{\gamma}(\overline{t})=\gamma(\phi(\overline{t}))$ for all $\overline{t}\in(\overline {\alpha},\overline {\beta}).$
the circle $x^2 + y^2 = 1$ has a parametrization $\gamma(t)=(cost,sint)$. Another parametrization is ${\overline{\gamma}(t)}=(sint,cost)$ (since $sin^2t + cos^2t = 1$).
To see that $\overline{\gamma}$ is a reparametrization of $\gamma$, we have to find a reparametrization map $\phi$ such that $(cos\phi(t),sin\phi(t)) = (sin t, cost)$.
One solution is $\phi(t)=\pi/2 — t$.
My question is,for $\phi$ to be a solution $\phi$ must be bijective smooth map.I accept $\phi$ is smooth.I can also proved this is injective.But I am unable to show that $\phi$ is surjective.How can I prove this?