In Some inequalities in certain nonorientable Riemannian manifolds, seemly, the author state that : Assuming $S^2$ is 2-dimensional sphere. $(S^2,g), (S^2,\tilde g)$ are two Riemannian manifold, where $g, \tilde g$ are two Riemannian metric. Then, there is $\lambda: S^2\rightarrow \mathbb R^+$ such that $$ g(p)=\lambda(p) \tilde g(p) ~~~~~~~~ \forall p\in S^2 \tag{1} $$ (The red line of first picture below.)
I find some reference, for example, in Wiki (the second picture below), seemly, it is right. But I am not sure, since my English is poor, and I am unfamiliar with Riemann sphere. So, I want to know: whether $(1)$ is right for all metric on 2-dimensional sphere?
PS(2023-10-11): According to comments, I have the following understanding.
1st, There is a projective plane $(P^2, g_{P^2})$.
2nd, By the projection process $\pi:S^2\rightarrow P^2$, we get a metric on $S^2$. Denote it as $(S^2, g_{S^2})$.
3rd, By Uniformization theorem, there is diffeomorphism $\phi:S^2\rightarrow U^2$ and function $\lambda: U^2\rightarrow \mathbb R^+$ such that $$ (\phi^{-1})^*g_{S^2}(x)=\lambda(x)g_E(x) ~~~~~~\forall x\in U^2 $$ where $U^2$ is unit 2-sphere and $g_E$ is the induced metric of $U^2$ from Euclidean space.
4th, However, the proof also need $\forall x\in U^2, \pi\circ \phi^{-1}(x)=\pi\circ\phi^{-1}(-x)$. This is an additional requirement. Although I feel such $\phi$ is existing, but I don't know how to show it.
