$\newcommand\F{\mathcal{F}}\newcommand\G{\mathcal{G}}$Suppose $\F$ is a subsheaf with generic rank $0$ of a coherent sheaf $\G$ with generic rank $p$ on a smooth variety $X$. Is there a nonzero map from $\det(\F) \to \bigwedge^n \G$ for some $n$?
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Yes, assuming that $det(\mathcal F)$ refers to some nonzero $\wedge^k\mathcal F$, presumably the last nonzero one, and an exterior power of $\mathcal G$ is $\wedge^k \mathcal G$.
The map $\mathcal F \to \mathcal G$ naturally induces a map $\wedge^k \mathcal F \to \wedge^k \mathcal G$. This map is always injective, since we can just check this on the fiber, which is a vector space, and that's true for vector spaces. Since $det(\mathcal F)=\wedge^k(\mathcal F)$ is nonzero, and the map to $\wedge^k \mathcal G$ is injective, it must be nonzero.
answered Jul 25, 2012 at 14:25
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$\begingroup$ "This map is always injective, since we can just check this on the fiber ..." What precisely do you mean by this? For an injective homomorphism of $\mathcal{O}_X$-modules, the base change to the residue field of a point need not be injective. If the homomorphism is locally split, then I agree with your argument. In fact, I feel that the question of the OP is ill-posed, since he does not tell us what definition of "det" he is using (my guess is the one used by Mumford). $\endgroup$ Jul 26, 2012 at 14:12
The projectivity of the moduli space of stable curves. I. Preliminaries on "det'' and "Div''.of Knudsen and Mumford. $\endgroup$