Let $X$ be a connected smooth projective curve over an algebraically closed field $K$. Let $\mathcal{F}$ be a locally free sheaf on $X$ and $\mathcal{E}$ a subsheaf of $\mathcal{F}$, which is again locally free since $dim(X)=1$. Let $E,F$ be the corresponding vector bundles associated with $\mathcal{E},\mathcal{F}$ respectively.

**Definition:** The *vector subbundle of $F$ generically generated by $E$* is a vector subbundle $\bar{E}\subset F$, which is the vector bundle associated with the locally free sheaf $\bar{\mathcal{E}}:=\pi^{-1}\bigg(\mathcal{T}\big(\mathcal{F}/\mathcal{E}\big)\bigg)$, where $\pi:\mathcal{F}\rightarrow\mathcal{F}/\mathcal{E}$ is the projection and $\mathcal{T}\big(\mathcal{F}/\mathcal{E}\big)$ is the torsion subsheaf of the quotient.

## Question 1

Why is $\mathcal{F}/\bar{\mathcal{E}}$ torsion-free (and hence locally-free)?

By using this property we obtain that $\bar{E}$ is a vector subbundle of $F$

## Question 2

Why do the following equations: $\quad rk(\bar{\mathcal{E}})=rk(\mathcal{E})\qquad deg(\bar{\mathcal{E}})\ge deg(\mathcal{E})\quad$ hold? Do they hold in general when we are dealing with inverse image sheaf or they are a special case?