Show that a function in manifolds is not a globally diffeomorphism. Considerer $\mathbb{R}$ and $S^1$, then both are $1$- dimensional manifolds i try to show that $F:\mathbb{R} \rightarrow S^1$ given by $F(t)=(\cos (t), \sin (t) )$ is a local diffeomorphism but is it not global diffeomorphism.
These is my atemp:
First let $$U_1=S^1 - \{(1,0)\}$$ $$U_2=S^1-\{(-1,0)\}$$ $$\varphi_1 :U_1 \rightarrow  \mathbb{R}$$ $$\varphi_1 (x,y)=\frac{-y}{1-x}$$and $$\varphi_2 :U_2 \rightarrow  \mathbb{R}$$ $$\varphi_2 (x,y)=\frac{y}{1+x}$$
Note that $\varphi_1$ is the composition between the stereographic projection and the rotation of $\frac{\pi}{2}$ and $\varphi_2$ is the composition between the stereographic projection and the rotation of $\frac{-\pi}{2}$, is crearly see that $\varphi_1$, $\varphi_2$ are diffeomorphic in $U_1$ and $U_2$ respectively then
$$\mathcal{A}_1=\{(U_1, \varphi_1), (U_2, \varphi_2)\}$$ Is a Atlas for $S^1$, Now considerer the trivial atlas for $\mathbb{R}$,  $\mathcal{A}_2=\{(\mathbb{R},id)\}$.
Then if $F$ are global diffeomorphism then
$$\varphi_k \circ F \circ id^{-1}=\varphi_k \circ F \circ id$$
is smooth for $k=1,2$ but
$$\varphi_2 \circ F \circ id=\frac{\sin(t)}{1+\cos(t)}$$
is not smooth then $F$ is not a global diffeomorphism.
NOTE: I know that are other forms to see that, but on this occasion, as an indication I had to work specifically with $$U_1=S^1 - \{(1,0)\}$$ and  $$U_2=S^1-\{(-1,0)\}$$
Any suggestion or help i will be very grateful.

Trivially, $F(0)=F(2\pi)$, hence $F$ is not even injective and thus cannot be a global diffeomorphism.
Besides, for all $t\in\mathbf{R}$, $F'(t)=(-\sin(t),\cos(t))\neq (0,0)$, since $\cos(t)^2+\sin(t)^2=1$, therefore using the inverse function theorem implies that $F$ is a local diffeomorphism around $t$.

Commenting on your attempt, I don't really get why you are claiming that $t\mapsto\frac{\sin(t)}{1+\cos(t)}$ is not smooth. It clearly is, since it is a quotient of smooth maps.
