Is the Schwarzschild singularity stretched in space as a straight line?

Solution 1:

Given your update, the $r=const$ slices are uniform in $t$:

$$ \mathrm{d}\tau = \sqrt{\left(1- \frac{r_s}{r}\right)^{-1}} \,\mathrm{d}t = (\mathrm{const}) \cdot \mathrm{d}t$$

So you get an ordinary Euclidean line.

Solution 2:

It's not an easy task to define topology and even harder to define geometry of your singularity. You can consider your singularity to be the set of time-like geodesics ending in finite time. You would probably want to identify some of those geodesics with some others, if they become too "close" to each other. And then somehow put a topology on this set. But how would you do that?

Mathematics starts with definitions, unless you can provide a definition that you want to use, your question is too vague for a mathematician. In fact, one can give definitions to get both answers: a point and a $3$-dimensional surface.

The intuition for the latter is similar to yours and comes from the Penrose diagram (which can be drawn nicely using Krushkal coordinates). I can't speak for everyone, but my intuition for the point comes from the fact that the event-horizon is a time-like sphere of constant finite size. While the universe I don't imagine as a single point at $t=0$. But all of this is imprecise.