What does it mean, geometrically for two surfaces to be locally isometric? I've heard it said that if two surfaces have the same first fundamental form, or if they're locally isometric, then they "behave metrically in the same way", or variants on that theme. The textbook I'm reading says that it's "intuitively clear" that the cylinder is isometric to the plane, because "by cutting a cylinder along a generator we may unroll it onto part of a plane". Well, couldn't I do something similar to cut and flatten out any surface onto a plane?
What is the geometric meaning of the existence or non-existence of a local isometry?
 A: Of course we all know that a cylindrical surface can be  unrolled onto a plane, hence is locally isometric to the euclidean plane ${\mathbb E}^2$. If differential geometry would not confirm this fact we would have given it up long ago.
But it is something different with a $2$-sphere: You cannot map it locally isometrically to ${\mathbb E}^2$. In order to prove this "geometrically" (i.e., without setting up a heavy differential geometric apparatus) we have to produce a certain invariant which is different for $S^2$ and ${\mathbb E}^2$. Such an invariant is, e.g., the circumference of a circle of radius $0<r\ll 1$. In ${\mathbb E}^2$ this circumference is $2\pi r$, but on a sphere of radius $1$ it is $2\pi\sin r$, whereby we measure $r$ as length of a geodesic arc on $S^2$. This discrepancy shows that it is impossible to map $S^2$ isometrically to ${\mathbb E}^2$ even locally.
A: If $S$ and $S'$ are two surfaces, and if $g$ is the metric on $S$ and $g'$ the metric on $S'$, $S$ is locally isometric to $S'$ (by $\phi: S \rightarrow S'$) in a neighborhood $V$ of $x \in S$, if:
$$\forall y \in V, \forall h,h' \in TS_y,~g'_{\phi(y)}(D\phi_y(h),D\phi_y(h'))=g_y(h,h')$$
where $g_y$ is the metric at the point $y$ and $TS_y$ is the tangent space at $y$.
Maybe should I write "metric tensor":
https://en.wikipedia.org/wiki/Metric_tensor
A: 
Well, couldn't I do something similar to cut and flatten out any surface onto a plane?

Picture a sphere made of rubber (like a racquetball or tennis ball).  Cut out a hunk of it and lay it on the table.  Do all parts of the rubber touch the table?  
Since the ball has some convexity to it, either the rubber is going to touch in the center and be bowed up around the edge, or touch around the edge and bubble up in the center.  
Of course, you could squish the rubber and force it to lie flat against the table.  But now you've changed the metric (you have literally stretched certain points of the rubber further away from each other).  
So I think it's not true that any surface can be cut and flattened onto a plane. 
This is a small-scale version of the problem of drawing accurate maps of the world.  Cartographers tried to create flat representations of the surface of the Earth while preserving distance.  As it turns out, this is impossible.
