proof of criterion for conformal mapping of Riemannian spaces

In this attachment, we prove that the a mapping $f$ of Riemannian (orpseudo-Riemannian) spaces $(M,g)$ and $(N,h)$ is conformal if and only if$f^{*}h=sg$ for some scalar field $s$ (on $M$).

The key observation is that the angle $A$ between curves $S$ and $T$ whichintersect at a point $P$ is determined by the tangent vectors to these two curves(which we shall term $s$ and $t$) and the metric at that point, like so:

$$\cos A=\frac{g(s,t)}{\sqrt{g(s,s)}\sqrt{g(t,t)}}$$

Moreover, given any tangent vector at a point, there will exist at least onecurve to which it is the tangent. Also, the tangent vector to the image ofa curve under a map is the pushforward of the tangent to the original curveunder the map; for instance, the tangent to $f(S)$ at $f(P)$ is $f^{*}s$. Hence,the mapping $f$ is conformal if and only if

$$\frac{g(u,v)}{\sqrt{g(u,u)}\sqrt{g(v,v)}}=\frac{h(f^{*}u,f^{*}v)}{\sqrt{h(f^{*}u,f^{*}u)}\sqrt{h(f^{*}v,f^{*}v)}}$$

for all tangent vectors $u$ and $v$ to the manifold $M$. By the way pushforwardsand pullbacks work, this is equivalent to the condition that

$$\frac{g(u,v)}{\sqrt{g(u,u)}\sqrt{g(v,v)}}=\frac{(f^{*}h)(u,v)}{\sqrt{(f^{*}h)(u,u)}\sqrt{(f^{*}h)(v,v)}}$$

for all tangent vectors $u$ and $v$ to the manifold $N$. Now, by elementaryalgebra, the above equation is equivalent to the requirement that thereexist a scalar $s$ such that, for all $u$ and $v$, it is the case that$g(u,v)=s{h}^{*}(u,v)$ or, in other words, $f^{*}h=sg$ for some scalarfield $s$.