There are a lot of questions about crossing the EH (event horizon) of a black hole on this site.
Some of them suggest, that when you cross the horizon, nothing special happens, you don't even notice crossing the horizon, and some suggest that it is even impossible to detect the horizon locally.
Nothing special happens to the observer as they cross the event horizon.
In your co-ordinate system you will notice nothing unusual.
What do you feel when crossing the event horizon?
There will be no discontinuity in behaviour at the event horizon.
Taking selfies while falling, would you be able to notice a horizon before hitting a singularity?
Now there are others, who suggest that inside the horizon, everything, including light must move towards the singularity, the singularity becomes a point in time (future).
So inside the horizon even a light ray directed outwards actually moves inwards not outwards.
How does light behave within a black hole's event horizon?
https://arxiv.org/abs/2002.01135
Is the event horizon locally detectable?
Based on the first one, when two astronauts cross the EH together, their walkie talkie (or radio) could keep working.
Based on the second one, this is not so clear. Obviously, outside the horizon, the radio still works, because EM waves from the sender still spread spherically, and would still reach the receiver. But once you cross the horizon, the curvature becomes so extreme, that the escape velocity exceeds the speed of light. Thus, EM waves would not spread spherically anymore, but only towards the singularity. Based on this, the EM waves from the sender might not be able to reach the receiver anymore, this the radio stops working when crossing the EH.
Just to make it clear, I am asking about two astronauts, co-moving, falling in together, and will the radio stop working between the two of them?
Question:
Does the radio (between two co-moving astronauts) stop working when crossing the event horizon?
Best Answer
The infalling observer who is free falling with negative escape velocity v=-c√(rs/r) will receive redshifted signals from the far away observer all the way down to the singularity (if he falls in with less than the escape velocity the signal he receives might as well be blueshifted).
The far away observer will receive redshifted signals from the infalling observer until the end of time, although the last signal he receives at the end of eternity will be the infinitely redshifted signal the infalling observer sent when he crossed the horizon.
All the signals the infalling observer sends after he crossed the horizon will not make it out since their dr/dt<0 inside the horizon (and dr/dt=0 for an outgoing signal right at the horizon).
In this simulation of a freefalling observer (red) who emits a signal (36 photons with 10° separation, the photons are depicted green) at r=rs/2 (t=0.8619286) in Raindrop coordinates you see that the radially inward directed photons move faster towards the singularity than the free falling observer, and the outward directed ones slower.
Edit: to adress the question in the comment I updated the animation to show a second observer who crosses the horizon with a delay of Δt=0.1GM/c³ and also emits a signal when he is between the horizon and the singularity to make it obvious that both observers catch each others signal.
So two free fallers are able to exchange light signals if their separation is not too large; if you send a signal right after you crossed the horizon it might not reach an observer that is right before the singularity (and vice versa), but an observer close below rs/2 can communicate with an observer close above rs/2.
However, the observer above rs/2 will receive the signal only when he himself has already fallen below the radius where the lower observer was when he emitted the signal (the signal directed at him still travels inwards, but slower than himself), while the lower observer will be overtaken by the radially inward directed photon emitted by the higher observer: