# Oriented atlas on a circle

I'm trying to find an oriented atlas on the circle $S^1$, i.e.,

I want to find an atlas for $S^1$ such that for any two overlapping charts $(U,s)$ and $(V,t)$ of the atlas, the derivative $d s/d t$ is everywhere positive on $U\cap V$.

I can define one atlas $\mathfrak U = \{(U_i,\phi_i)\}_{i=1}^4$ for $S^1$ by letting $U_1,U_2$ be the upper and lower semicircles, respectively, and defining $\phi_i(x,y) = x$ for $i = 1,2$, and letting $U_3,U_4$ be the right and left semicircles, respectively, and defining $\phi_i(x,y) = y$ for $i = 3,4$.

However, I think this atlas is not oriented, since for example taking the overlapping charts $(U_1,x)$ and $(U_3, y)$ of $\mathfrak U$, we have $y = \sqrt{1 - x^2}$ and $$\frac{dy}{dx} = -\frac{x}{\sqrt{1-x^2}} \text,$$ which is not positive on $U_1\cap U_3$.

How can I modify $\mathfrak U$ to fix this issue? For example, should I redefine $\phi_3(x,y) = -y$, and then keep on checking the other seven derivatives? (I have to check the pairs $dy/dx$ and $dx/dy$ for each quarter-circle.)

Note that you can cover a circle (or indeed, $\Bbb S^n$ for any $n$) using just two charts (though not two graph charts, as in the question): Indeed, the stereographic projection $$(x, y) \mapsto \frac{y}{x + 1}$$ is a homeomorphism $\Bbb S^1 - \{(-1, 0)\} \to \Bbb R$ and the domain omits just one point of the circle. (This is the map that maps a point $z$ to the slope of the line through $z$ and $(-1, 0)$.)