Parallelizing lines Let $n \geq 1$ be an integer, and $L_1,\ldots,L_n$ be $n$ lines in $\mathbb{R}^3$ which are pairwise disjoint. Is it possible to move all $n$ lines continuously so that they never cross, and so as to send the line $L_k$ to $\{k\} \times \mathbb{R} \times \{0\}$?
This seems instinctively true, but I don't see any slick proof right now. The problem is that the space of affine lines is $4$-dimensional, while "crossing a given line" is $3$-dimensional, so codimension $1$. Hence, if I try to move each line once at a time, it seems entirely possible that a set of badly placed lines cages off a portion of space, and that the remaining lines can't be moved around freely.
Bonus question : what is the least number of pairwise disjoint lines $(L_k)$ I need so that the space of lines which cross no $L_k$ is disconnected?
 A: I really liked this question! The wikipedia page for configurations of skew lines has a reference to this beautiful paper by Viro & Viro that -among other things- answers the first question positively!
Here is how the idea worsk: Consider two parallel planes in general position (that is, not paralled to any of the lines). In the first plane fix the points of intersection of the lines and the plane. In the second plane make a mark (relative to that plane) at each point the lines intersect it. Then move the second plane vertically far away from the first and allow the lines to move as their point of intersection with the second plane should always be the corresponding marked point.
You can also think about it as drilling a small hole in the second plane, so that when it moves it forces the whole configuration of lines to change with it. In a sense the moving plane is like a comb that moves all lines in parallel position.
Now, we need to check that during this process the lines would never intersect. It is enough to check for a single pair of lines. This isn't very difficult to do:
First assume the planes are parallel to the $x,y$-plane. Now call the fixed points in the first plane $A=(a_1,a_2,0)$ and $B=(b_1,b_2,0)$ and the moving marked points $C=(c_1,c_2,t)$ and $D=(d_1,d_2,t)$. Now, the lines $AC$ and $BD$ intersect if the following system has a solution in $x$ and $y$ $$a_1+x(c_1-a_1)=b_1+y(d_1-b_1)$$ $$ a_2+x(c_2-a_2)=b_2+y(d_2-b_2)$$ $$0+x(t-0)=0+y(t-0)$$ Now this forces $x=y$ and then $$\frac{b_1-a_1}{c_1-a_1-d_1+b_1}=\frac{b_2-a_2}{c_2-a_2-d_2+b_2}$$ But notice that this does not depend on $t$! That is, if the lines didn't initially cross, they will never cross during the combing motion imposed by the moving second plane.
