So I want to compute the pull back of $\omega = \frac{R^2}{\sqrt{R^2-u^2-v^2}}du \wedge dv$ under the antipodal map $f: S^2 \to S^2$ where $S^2$ is the a sphere of radisu $R$. Now I am using the hemisphere charts so for the pullback I am working in the chart $U = \{ (x,y,\sqrt{R^2 - x^2 -y^2)}\}$ with map $ \phi (x,y,\sqrt{R^2 - x^2 -y^2})) \to (x,y) = (u,v) $.
So the $z$ coordinate is positive. Then the antipodal map which sends $(x,y,\sqrt{R^2 - x^2-y^2}) \to (-x,-y,- \sqrt{R^2 - x^2-y^2}) $ can be written in coordinates as $(u, v) \to (-u, -v)$
I computed $f^*(\omega) $ as
$f^*(\omega) = (\frac{R^2}{\sqrt{R^2-u^2-v^2}} \circ f(u,v)) \cdot f^*(du) \wedge f^*(dv) = \frac{R^2}{\sqrt{R^2-u^2-v^2}} du \wedge dv $
But apparently I am suppose to get $f^*(\omega) = - \omega$. I am not sure what I am doing wrong, if you compose $\frac{R^2}{\sqrt{R^2-u^2-v^2}}$ with $f$ you get the same thing, and $f^*(du) = du$ and $f^*(dv) = dv$
Originally $\omega = i^*(xdy \wedge dz + y dz \wedge dx + z dx \wedge dy)$ under the inclusion map $i: S^2 \to \mathbb{R}^3$, which I computed in the same chart, I am pretty sure my computation for $\omega$ is correct but I dont know what I am doing wrong... to not get a negative.