If we examine the situation before the black hole has formed, and after it has evaporated, we need not be concerned about the peculiarities of general relativity. Baryonic matter goes in; mostly photons (and gravitons also) come out. This is the "Joe point of view" wherein Joe has been destroyed in the black hole. This must be correct, so what is wrong with the "Mary point of view" in which Joe lingers forever just above the event horizon? In Joe's rest frame, he emits a finite number of photons before crossing the horizon, and for him, crossing the horizon is not particularly eventful. Mary detects these photons within a finite amount of her proper time, and after the last one, there are no more. Photon number is conserved under propagation and Doppler shift in general relativity. How could Mary test the hypothesis that Joe never actually fell into the black hole? She might send in a beam of lightand look for the scattering off of Joe. But on a space-time diagram, when the probe light reaches the event horizon, Joe is already inside. If the light does catch up with and scatter off Joe, the scattered light is trapped inside the horizon, and never gets back to Mary. Where does the notion of "lingering just outside the horizon" come from. In the elementary textbook description for the Schwartzchild black hole, we have coordinates r and t that are the usual radius and time far from the black hole. In these coordinates, Joe's orbit r(t) approaches the Schwartzchild radius as t approaches infinity, and so he seems to linger there forever. But just because we name something "t" doesn't make it into a time. Inside the horizon, r becomes timelike, and t becomes spacelike, while on the horizon itself, the metric in these coordinates becomes singular. The black hole singularity is not so much "at the center" as it is in the future. Joe is compelled to fall into the singularity for the same reason that we are all compelled to move into the future, whatever that reason may be. -- Gene