Homeomorphism VS Curvature

Question

Following a comment on the answer in this question: Sphere homeomorphic to plane?
I would like to ask: if a sphere(or any surface) is locally homeomorphic to a plane, why doesn't that show us how flat the sphere(or any object) locally is?

Answer 1

It's because curvature of a surface or a curve isn't conserved under homeomorphisms/diffeomorphisms. If that was the case, then any regular curve or surface would have zero curvature, because they are locally homeomorphic to an interval or ball in $\mathbb R ^2$. Hence it would be a useless measure of curvature.

So in short - when you are considering embedded surfaces into euclidean space the embedding itself defines the curvature of the surface. Any homeomorphic transformation of the embedding may change the curvature.

In the general setting of manifolds its an object called a "connection" that dictates what the curvature is - but that is a whole other story.

Answer 2

The point is that being locally homeomorphic is qualitative : you know that it is "locally flat", but you don't know how. The curvature is a quantitative measure : you need to actually exhibit an embedding and do computation on it.

This is a bit like saying that a function $f$ is differentiable tells you nothing about how it actually grows, just that $f$ is locally "linear". You have a whole bunch of differentiable functions that behave very differently locally, from constant ones to exponentials. You need to compute $f$'s derivative to decide if $f$ is more like an horizontal line ($f'(x)=0$), a diagonal line ($f'(x)=1$) or grows even faster ($f'(x) > 1$).