I saw the movie "Interstellar" a few years back, and was amazed that Cooper was able to fall from 1 AU into a black hole before his daughter turned 110. Intuitively, I would think that there are no paths in our spacetime that cross over the event horizon.
Does a geodesic exist that can take one from normal space to the other side of an event horizon before the end of time?
Does a geodesic exist that can take one from normal space to the other side of an event horizon before the end of time?
Yes.
There are geodesics that do cross the event horizon. Both the geodesics themselves and the horizon are coordinate independent features of the spacetime, so this statement is fully coordinate independent. These geodesics include timelike, null, and spacelike geodesics. In principle, Cooper could follow a timelike geodesic across the horizon, so all further discussion of geodesics will specifically assume timelike geodesics.
In the Schwarzschild spacetime all timelike geodesics pass through the event horizon and then go to the singularity. So that clearly answers the question "Does a geodesic exist that can take one from normal space to the other side of an event horizon" in the affirmative, consistent with the TLDR answer.
The only tricky or complicated part is the final part of the question "before the end of time". That phrase is ambiguous, so there are multiple ways to interpret it. Here is a list of ways to interpret that phrase and the consequence to the TLDR answer.
"before the end of time" could mean:
Before the end of proper time along the geodesic. This is my preferred interpretation because proper time is coordinate independent, so this is the meaning that makes the whole question one about the physics rather than about some arbitrary choice of coordinates. In this case, the answer is Yes. Proper time ends as the geodesic approaches the singularity, and the crossing of the event horizon is an earlier event.
Before the end of coordinate time in $X$ coordinates where $X \in \{\text{Schwarzschild, isotropic, ...} \}$. Since this is a coordinate time this will only be true when using $X$ coordinates. The $X$ coordinates do not cover the region of spacetime containing the event horizon, so the $t$ coordinate ends before the horizon. Therefore, in this case the answer is No. The event where the geodesic crosses the horizon is after the end of $X$ coordinate time.
Before the end of coordinate time in $X$ coordinates where $X \in \{\text{Kruskal–Szekeres, Lemaître, Gullstrand–Painlevé, ...} \}$. Since this is a coordinate time this will only be true when using $X$ coordinates. The $t$ coordinate in $X$ coordinates is finite at the horizon, so the horizon is in the region covered by these coordinates. Therefore, in this case the answer is Yes. The event where the geodesic crosses the horizon is before the end of $X$ coordinate time.
Before the end of coordinate time in $X$ coordinates where $X \in \{\text{Eddington-Finkelstein, ...} \}$. This will only be true when using $X$ coordinates. There is no coordinate time in $X$ coordinates. None of the coordinates are timelike, there are only spacelike and null coordinates. The coordinate chart covers the event horizon, so the event where the geodesic crosses the horizon has perfectly valid and finite coordinates in $X$ coordinates. But none of those coordinates are "time". Therefore, in this case the answer is undefined. There is no time so there is no end of time so there is no way to compare the perfectly valid event of crossing horizon with the end of time to determine if it was before or after.
