Reverse Toponogov triangle comparison

Question

See the wiki page https://en.wikipedia.org/wiki/Toponogov%27s_theorem

One consequence of the Toponogov comparison Theorem is that if the sectional curvature of a manifold $M$ is pinched below by a number $\delta$. Let $pqr$ be a geodesic triangle, i.e. a triangle whose sides are geodesics, in $M$, such that the geodesic $pq$ is minimal. Let $p′q′r′$ be a geodesic triangle in the model space $M_\delta$, i.e. the simply connected space of constant curvature $\delta$, such that the length of sides $p′q′$ and $p′r′$is equal to that of $pq$ and $pr$ respectively and the angle at $p′$ is equal to that at $p$.

Then $d(q,r) \le d(q',r').$

The wiki then claim that "When the sectional curvature is bounded from above, a corollary to the Rauch comparison theorem yields an analogous statement, but with the reverse inequality."

Is this true? If so, can someone please provide a reference to this? It seems a bit weird to me since the statement of Rauch Theorem is symmetric, if this reverse statement is a corollary of Rauch Theorem, why isn't Toponogov Theorem also a corollary?

Any comments and references are welcome.

Answer 1

Rauch's comparison local result, while Toponogov's comparison is global.

For the upper curvature bound an analog of Toponogov's comparison holds only locally and indeed it follows from Rauch's comparison. There is a gloabal version for upper comparison --- the so-called (generalized) Hadamard--Cartan theorem [9.65+9.67 in our book]: For the curvature bound $\kappa\le 0$ it has an addition assumption that space is simply connected. If $\kappa=1$ then one has to assume that any closed curve shorter than $2\cdot\pi$ can be contracted in the class of closed curves shorter than $2\cdot\pi$. (The case $\kappa>0$ can be reduced to $\kappa=1$ by rescaling.)