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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?

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Welcome to MathOverflow!

Question 1: this can be checked locally, on affine opens or local rings, and then becomes an exercise: if $0 \to M' \to M \overset{\pi}{\to} M'' \to 0$ is an exact sequence of $R$-modules, then the submodule of $M$ generated by $M'$ and $\pi$-preimage of torsion in $M''$ has torsion free quotient. (E.g. for finitely-generated $\mathbb{Z}$-modules a submodule which has torsion-free quotient is called primitive, and every submodule can be enlarged to become primitive one by this construction.)

Question 2: the rank claim can be checked locally again; it follows from the fact that torsion modules have rank zero, by definition. For the degree claim, we compute using additivity of degree on short exact sequences $$ \deg(\overline{\mathcal{E}}) = \deg(\mathcal{F}) - \deg(\mathcal{(\mathcal{E}/\mathcal{F}})/\mathcal{T}) = \deg(\mathcal{F}) - \deg(\mathcal{\mathcal{E}/\mathcal{F}}) + \deg(\mathcal{T}) \ge \deg(\mathcal{E}) $$ where we used that the degree of a torsion-sheaf is non-negative (in fact, positive if it's nonzero).

Hopefully this argument also hints what happens to preimages of sheaves in general...

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    $\begingroup$ Wonderful answer! Thanks a lot :) $\endgroup$
    – John117
    Commented Feb 3, 2021 at 9:38

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