How to prove that sections of family of curves do not exist (exercise 4.7 in Harris' Algebraic Geometry)?

Question

In exercise 4.7 of his book Algebraic Geometry, Prof. Harris asked to show that there is no local section of the universal hyperplane section of a smooth plane conic or the twisted cubic. The question is about local sections of the projection on the space of hyperplanes, that is to say the dual of respectively $\Bbb{P}^2$ and $\Bbb{P}^3$

Edit More precisely, it is asked to show that, for a conic $X$ for instance, there is no open subset $U$ of $(\Bbb{P}^2)^*$, the space of hyperplanes of $\Bbb{P}^2$, such that there is a regular map from $U$ to $$\Omega_X=\{(x,H)\in X\times (\Bbb{P}^2)^* : x\in H\cap X\}$$ that is section of the second projection of $\Omega_X$ on $(\Bbb{P}^2)^*$

I could not find a proof without using equations and coordinates for the plane conic or the twisted cubic. Is there a less pedestrian & more geometric proof in the spirit of this book ? For instance, Prof. Harris mentioned that this result is an immediate consequence of his Theorem 11.14, which says that the domain of a regular map between projective varieties is irreducible if it has irreducible fibers with constant dimension and an irreducible image. I failed to see why !?

Edit Following the comment of Georges, the reference to Theorem 11.14 is just to show that $\Omega_X$ is irreducible

Answer 1

Contrary to the assertion in the exercise, there does exist a regular section of the fibration $\pi_X:\Omega_X\to X$ ( and not only a rational one!).
The variety $\Omega_X$ consists of pairs $(H,x)$ with $x\in X$ and $H$ a line $H\subset \mathbb P^2$ passing through $x$ i.e. $x\in H$ .
The projection $\pi_X$ sends such a pair to $x\in X\subset \mathbb P^2$.

A section $s:X\to \Omega_X$ is given by the following construction:
Fix a point $p\in \mathbb P^2\setminus X$ and define $s:X\to \Omega_X$ by $s(x)=(\overline {xp},x)$, where $\overline {xp}$ is the line joining $x$ to $p$.

Edit
Since brunoh tells me in his comment that the problem is about the other projection $\pi _1:\Omega_X\to (\mathbb P^2)^{\star }$, here is why that other projection has no rational section :

The morphism $\pi _1$ is ramified $2$- covering of $(\mathbb P^2)^{\star }$ and a section $s$ of that covering over an open subset $U\subset (\mathbb P^2)^{\star }$ would choose for each line $H\subset \mathbb P^2$ one of the two points of intersection $H\cap X$ of the line $H$ and the conic $X$.
Choosing the other point of intersection would yield another section $s'$ of $\pi_2$ and the closures $I,I'$ of the images of those sections $s,s'$ would give a reduction $\Omega_X=I_1\cup I_2$ of the irreducible variety $\Omega_X$, a contradiction.

New edit
The situation is crystal-clear geometrically:
If you fix a conic $X\subset \mathbb P^2$, every line $H\subset \mathbb P^2$ will cut $X$ in two points, except in the rare cases where the line is tangent to the conic, in which case there is only one (double) point.
[Rare means that the set of such tangent lines is an algebraic curve in $(\mathbb P^2)^{\star }$, called the dual conic]
The result we are investigating says that it is impossible for each non-tangent line to choose in an algebraic way one of those two points .

Answer 2

In addition to the nice answer of Georges, I would like to complete it with the other part of the exercise about the twisted cubic, following the same principle than his approach. It seems to me that the key of his answer is that having a local section $s$ of $\Omega_X$ defined on an open set $U$ with image $s(U)$ an open subset of $\Omega_X$, which is a regular map of varieties, we have to be able to build another section defined on the same $U$ with an open image $s'(U)$ disjoint of $s(U)$, and which is also a regular map of varieties. In the case of a generic smooth conic and the particular case of the twisted cubic, this can be done easily because the equations of the variety obey a natural simple symmetry (here changing the sign of one or two of the projective "coordinates"). So from $s$ we can build $s'$ immediately locally, and get a regular map. This also answer my question in one of my comment: this approach can clearly not be easily generalized to other $d$-dimensional curves.