Motivation for curved spacetime on General Relativity I'm trying to motivate the theory of general relativity. Concretely, using which arguments is it deduced that spacetime must have the structure of a curved Riemannian manifold?
From the strong equivalence principle I deduce that it is taken as an axiom that free-falling frames are locally intertial, but I don't know how to go from there to curvature.
I also have seen as an argument that in a free-falling frame, "free" particles do trajectories that are only localy straight, but this doesn't convince me as a rigorous argument.
 A: 
I'm trying to motivate the general relativity theory. Concretely upon which arguments is deduced that spacetime must have a structure of curved Riemannian manifold.

First, you consider spacetime without gravity. You notice that inertial objects have worldlines which are straight lines in spacetime and accelerometers measure how much a worldline bends in spacetime, with an accelerometer reading of zero corresponding to an inertial object’s straight worldline. Inertial objects at rest with respect to each other have parallel worldlines and they never intersect.
Second, you extend those ideas to a spacetime with gravity. Objects in free fall have accelerometers which read zero, so their worldlines are straight. But two objects in free fall initially at rest with respect to each other can eventually intersect. So we have straight lines which are initially parallel to each other but eventually intersect. This is impossible in a flat spacetime, but easily happens in a curved spacetime.
That is the motivation for curved spacetime. Taking seriously the idea that an accelerometer reading of zero indicates a straight worldline.
A: We don't have any better way to define straightness than by saying that the world-line of an inertially moving test particle is straight. By the equivalence principle, we have to count free-falling particles as inertial. (We can't say they're acted on by a gravitational force, because there is always a frame in which that force is zero.)
Once you accept this, the geometry is automatically non-Euclidean.
A: There is no a mathematical compelling reason: curvature is permitted by local observations about free falling bodies and by the experimental identification of gravitational and inertial mass. The point is that the consequences of the assumption are experimentally confirmed. If we assume that curvature enters the picture and geodesics (with curvature) describe free falling bodies, one finds corrections to the Newtonian description which are more in agreement with experimental data than the Newtonian description itself. An example is the perihelion precession of mercury. The same reasoning permits us to identify $g_{00}$ with the Newtonian gravitational potential when the deviation from Minkowki spacetime are small and one adopts natural coordinate frames. With this identification for instance we immediately have an explanation of the observed (and used in GPS devices) red shift effect.
