Comment by machina_ex_deus

First of all, kruskal coordinates show beyond doubt that the event horizon is just a regular null hypersurface that the observer wouldn't notice crossing locally. (Of course if you look around, at the moment of crossing into the event horizon you see everything else that was falling into it unfreeze and continue crossing).

If you want to take into account the evaporation of the black hole, then you should look at something like the vaidya metric. The mass function is a function of the ingoing Eddington coordinate v, which takes on a specific value when you cross the event horizon, and so you observe the black hole at a specific mass as you cross the event horizon. Contradicting your layman understanding of time dilation for the observer relative to the black hole.

Once you cross the horizon, the r coordinate becomes timelike, and so you are forced to move to decreasing r value just like a regular observer is forced to move to increasing t value. Your entire future, all your future light cone is within the black hole and it all terminates at the singularity. Minewhile, the t coordinate is space like which is what gives you space like separation from the mess that had happened in the original gravitational collapse. You wouldn't be blasted by a frozen supernova like you have said.

You can kind of say the universe splits at the event horizon, the time like coordinate changes from t to r and the future of the black hole branch of the universe is permanently cut off from the rest of the universe.

In rotating and charged black holes it is different, and you observe the evaporation of the black hole once you cross the Cauchy horizon. If the black hole is eternal (because someone kept feeding radiation to the black hole, maybe by reflecting the hawking radiation inwards), then you would in fact see timelike infinity as you reach the Cauchy horizon, so this time like infinity is quite physical. You would need to avoid being vaporized by blue shifted incoming radiation.