I have been following a course on GR that at one point discusses the metric derived for the outside of a physical, non rotating uncharged massive object with spherical symmetry. For this situation I have seen the Schwarzschild metric derived, in Scharzschild coordinates.
From the form of the metric, it is observed that bad things may happen at $r = 2GM$ and $r = 0$, but this comes with the warning that since the metric is coordinate dependent, one should check the scalar contractions of the Riemann curvature tensor to see if any physical bad things happen. The course proceeds to show that this is the case for $r=0$ but not for $r=2GM$.
Then, light cones are studied by finding out what happens with the geodesic equation for massless particles, in Schwarzschild coordinates, and they appear to close up as $r$ nears $2GM$. This suggests that in these coordinates, light can reach the horizon, but not cross it.
A coordinate transformation to Eddington-Finkelstein coordinates is then used to show that the metric in these coordinates is benign at $r=2GM$ and an analysis of the light cones in these coordinates show that one edge of the light cone, the one that is oriented radially inward to the center of the geometry, appears to be unchanged from flat spacetime while light oriented radially outward has its edge of the light cone tipping over until at the horizon, this is aligned with the horizon.
This analysis shows that light can reach into the horizon, but cannot escape the region.
With this introduction and context, my question is how interpret the different analysis of the light cones in these two different coordinate systems. On the one hand, when done in Schwarzschild coordinates, light cones close up and align with the horizon, while in E-F coordinates it is clear that there are geodesics into the horizon, only none that come out.
To be clear, I’m not asking whether light can or cannot enter an horizon formed by a black hole. I’m looking for some clarity on why the analysis in Schwarzschild coordinates cannot be trusted (fully), but apparently the one done in the Eddington-Finkelstein coordinates can.