Monomorphisms of torsion-free sheaves induce monomorphims of their determinant line bundles.

Question

I've been trying to understand the following proof in Kobayashi's book. Let me state some relevent definitions.

Let $X$ be a complex manifold and $\mathscr{O}_X$ its structure sheaf. A coherent sheaf $\mathscr{E}$ is torsion-free if the map $\mathscr{E}\to\mathscr{E}^{**}$ is injective. It is a torsion sheaf if the above map is a zero map. A torsion-free sheaf $\mathscr{E}$ of rank $r$ has a determinant line bundle $\det\mathscr{E}=(\wedge^r\mathscr{E})^{**}$.

Every monomorphism $\mathscr{F}\to\mathscr{F'}$ between torsion-free coherent sheaves of the same rank induces a sheaf monomorphism $\det\mathscr{F}\to\det\mathscr{F'}$.

The proof goes like this:

Outside their singular sets (where coherent sheaves are not locally free), the map $\mathscr{F}\to\mathscr{F'}$ and the induced map $\det\mathscr{F}\to\det\mathscr{F'}$ are isomorphisms. Hence, the $\ker(\det\mathscr{F}\to\det{F'})$ is a torsion sheaf. Since it is a subsheaf of a torsion-free sheaf, it must be zero.

My questions are in the following:

1. Outside their singular sets, $\mathscr{F}$ and $\mathscr{F'}$ are locally free. Since they have the same rank, the map $\mathscr{F}\to\mathscr{F'}$ has a cokernel which has rank 0 and hence is a torsion sheaf. But why is it necessarily zero so that $\mathscr{F}\to\mathscr{F'}$ is an isomorphism?

2. If $\mathscr{F}\to\mathscr{F'}$ is an isomorphism outside the singular sets, then $\ker(\det\mathscr{F}\to\det{F'})=0$ outside the singular set. Then, how do I conclude that it must be a torsion sheaf? I don't know its stalks inside the singular set. To show it is a torsion sheaf, I need to make sure every stalk is a torsion module.

Thanks.

Answer 1

1. I don't see that this is true without other assumptions. Let $X=\Bbb C$ and $\mathscr{O}$ be the structure sheaf. Then $\mathscr{O}\stackrel{\cdot z}{\to}\mathscr{O}$ is a perfectly good morphism of free sheaves with cokernel the skyscraper sheaf supported at $0$ with sections $\Bbb C$.

Perhaps there should be a modification of the proof here - since the cokernel is torsion, it's supported on a closed submanifold of strictly smaller dimension, and we can throw that in with the singular set as a locus we can ignore - we will still get that the map on determinant bundles is an isomorphism on a dense open subset, which is really the thing that makes the proof work.

2. Any sheaf supported on a submanifold of strictly smaller dimension is torsion: consider the action given by multiplying by a nonzero function which vanishes on this submanifold.