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

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.

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 $$\Bbb C$$.
• Then, maybe I am just tired today but I don't follow your second point. Let's say $\mathscr{F}$ is supported in a proper analytic subvariety $A$ of the base manifold $X$ and $s_x\in\mathscr{F}$ where $x\in A$. Then, take some holomorphic function $f$ that vanishes on $A$, then why is $(fs)_x=0$? It is true that $(fs)_y=0$ for every $y\notin A$. It is also true that $(fs)|_A=0$. But I must've missed something obvious. – YYF Oct 4 '18 at 22:23
• One more thing in my last comment: $(fs)_|A=0$ does not make much sense, since this is a section in an abstract sheaf and is not a function. – YYF Oct 4 '18 at 22:40
• But the problem is $U\cap A$ is not open. Therefore, even if I am tempted to say $(fs)|_{U\cap A}=f|_{U\cap A}s|_{U\cap A}=0$, I couldn't. – YYF Oct 5 '18 at 1:57