Universal property of the tautological line bundle

Question

Let $X$ be a smooth, projective variety and $\pi:E \to X$ be a projective bundle of rank $r$ i.e., each fiber of $\pi$ is a projective space of dimension $r$. Let $F \subset E$ be a projective sub-bundle of $E$, over $X$ of codimension $1$ i.e., the fiber over each point of $\pi|_F$ is of dimension $r-1$. Then,

1) Is the pull-back of the tautological line bundle $\mathcal{O}_E(-1)$ to $F$, the tautological line bundle $\mathcal{O}_F(-1)$?

2) Is the line bundle associated to $F$, $\mathcal{O}_E(-F)$ isomorphic to the tautological line bundle $\mathcal{O}_E(-1)$?

Answer 1

Answer to (1) is yes:

It is better to use Grothendieck's convention (you have to dualize all the bundles and work with quotients and not with sub bundles) and the proof is clear (cfr. Hartshorne book):

If $Y$ is a variety. A map $g:Y\to{\Bbb P}(E)$ is a triple $(f;L;j)$ where $f:Y\to X$ is a morphism, $L$ is a line bundle on $Y$ and $j:f^\ast(E)\to L$ is a surjection of vector bundles. Moreover, $g^\ast({\cal O}(1))=L$. In particular, if $p_E:{\Bbb P}(E)\to X$, you have a tautological surjection $p^\ast(E)\to {\cal O}(1)$.

Consequently, if you have a quotient $E\to F$ of $E$, then, since on $\Bbb P(F)$ you have a double surjection $p_F^\ast(E)\to p_F^\ast(F)\to \cal O(1)$, the conclusion follows.

Answer to (2) is NO:

If you have a quotient as before, you have an exact sequence

$$0\to N\to E\to F\to 0$$ and by adjunction you can verify that ${\cal O}_E(F)={\cal O}_E(1)\otimes p_E^\ast(N^\ast)$.