It's similar to if you and the car in front of you are accelerating at the same rate, they have a small head start, and they've got someone throwing fastballs at you at 100mph. When you get to the point where you are going 100mph relative to the ground, you'll hit the ball that was thrown from exactly that spot relative to the ground. So, sure, crossing the 100mph barrier implies something interesting mathematically, but it's not something that the observer would particularly notice. The math for GR isn't exactly the same (baseballs won't redshift), and in particular there isn't even a piece of dirt to compare the photon's motion to, as the horizon is just a mathematically-defined "place", but to a first-order approximation, it's the same thing going on with photons and spacetime distortion. It's a continuous function.
There's a bit more to it than can be explained in a comment, but the main thing to know is that (as far as we know) nothing special happens at the horizon if you're falling in.
why would bob see anything? I understood that the event horizon is a threshold, not a shell that you cross and suddenly can see inside.
to see something photons have to bounce on something and reach our eyes, we stop seeing stuff inside the horizon because those photons don't bounce back and they are pulled into the singularity.
my logic said that if light can't escape the horizon, then it can't escape alice to reach bob, even if he's inside the horizon, photons can't suddenly go backwards from alice until bob, they are being pulled further inside.
The problem is that (classically) when you cross the event horizon, the photons that you emit at just that moment will _stay_ _in_ _place_ forever.
Uhhh... one of those words should go.
Let's keep it fully classical and drop "photon": we're interested in gravitational effects rather than quantum ones (uncertainty, fluctuations, tunnelling, details about scattering and more). Really what we want is something to illuminate (pardon the pun) interesting null geodesics, so a thin collimated beam -- a pencil of light -- will do.
The relevant surface here is the apparent horizon, which can be measured by infalling apparatuses, and not the event horizon, the location of which is determined by the configuration of the entire spacetime. (See Visser PRD 2014 <https://journals.aps.org/prd/abstract/10.1103/PhysRevD.90.12...> or the corresponding arxiv version <https://arxiv.org/abs/1407.7295>).
> stay in place forever
Note the region inside the shell and the downward-pointing wedge in Fig 1. of Visser 2014 is flat Minkowski spacetime. Everything in that region will work like Special Relativity, as one would expect from the shell theorem.
In particular, a pencil of light directed outwards through the apparent horizon in Fig. 1. will ride the AH down to the singularity, but a receiver intercepting the pencil of light just inside the shell would notice nothing unusual: spacetime is flat there.
Eventually the collapsing shell collides with observers floating weightlessly inside it, and they have a bad time. But they can direct a pencil of light inwards just before the shell hits them.
I don't quite understand their diagram and their point. Are they looking at a region inside the collapsing shell of material? Then there's no contradiction here, the observer won't see anything until the reach the singularity (where the space-time stops being locally flat).
Even if we consider an observer in the center of the shell, they'll only encounter the singularity after they get hit by the infalling matter.
Not quite. He will see the light emitted by _all_ of the matter that has fallen in before him, but only in an infinitely small area.