A motivating example for showing locally free sheaves failing to be an abelian category

Question

It's a paragraph in Vakil's FOAG, page 394, where the text is

14.2.6. The problem with locally free sheaves / vector bundles (as opposed to vector spaces).

Recall that $\mathscr{O}_{X}$-modules form an abelian category: we can talk about kernels, cokernels, and so forth, and we can do homological algebra. Similarly, vector spaces form an abelian category. But locally free sheaves (i.e., vector bundles), along with reasonably natural maps between them (those that arise as maps of $\mathscr{O}_{X}$-modules), don't form an abelian category. As a motivating example in the category of manifolds (cf. Warning 14.2.10), consider the map of the trivial line bundle on $\mathbb{R}$ (with coordinate $t$) to itself, corresponding to multiplying by the coordinate $t$. Then this map jumps rank, and if you try to define a kernel or cokernel you will get confused.

I can see "informally" that the map jumps rank: at $t=0$, the map has image a vector space of rank $0$ rather then $1$. But rigorously I cannot see where is the problem in this motivating example. Taking $\mathscr{O}_\Bbb{R}$ to be the sheaf of smooth functions. Then the trivial line bundle $\Bbb{R}$ is just $\mathscr{O}_\Bbb{R}$ as an $\mathscr{O}_\Bbb{R}$-module. And I want to calculate the kernel and cokernel for the map $\mathscr{O}_\Bbb{R} \to \mathscr{O}_\Bbb{R}$ defined by multiplying $t$. Then I get the kernel is just $0$, and the cokernel still is $\mathscr{O}_\Bbb{R}$. Both of them are still locally free sheaves(vector bundles). If so, then where is the "confused" point mentioned in the text? For this motivating example, it seems the kernel and cokernel still locally free sheaves? Or did I compute them wrong?

A related thread maybe Quasicoherent sheaves as smallest abelian category containing locally free sheaves. But after reading it, I still has no idea where is the problem in the quoted text.

Thank you in advance.

Answer 1

I think I would disagree with the quoted text. Just because the cokernel of the map in the category of all sheaves is not locally free, does not necessarily mean the map cannot have a cokernel in the restricted category of locally free sheaves.

In the particular example given, in fact I claim that 0 is a cokernel of the given morphism, within the restricted category of locally free sheaves. To see this, suppose we have some morphism $\varphi : \mathscr{O}_{\mathbb{R}} \to \mathscr{E}$, where $\mathscr{E}$ is locally free and $\varphi \circ (t\cdot) = 0$. Then for any section $f \in \mathscr{O}_{\mathbb{R}}(U)$, we must have $t \varphi(f) = \varphi(tf) = 0$. But since $\mathscr{E}$ is locally free, it cannot have any nontrivial $t$-torsion sections, so we must have $\varphi(f) = 0$ as well. Since $f$ was arbitrary, we see that $\varphi = 0$.

On the other hand, by a similar argument, we can see that even though $t\cdot$ is a monomorphism in this category, it cannot be the kernel of any morphism of $\mathscr{O}_{\mathbb{R}}$ to a locally free sheaf. This would be one example of a failure of the axioms of an abelian category.

(As an alternative example where a kernel in fact does not exist, suppose we instead take multiplication by $g(t) = \begin{cases} 0, & t \le 0; \\ e^{-1/t}, & t > 0. \end{cases}$ Then it should not be too hard to show that if we had a kernel of this map, its restriction to $(-\infty, 0)$ would have to have rank 1, while its restriction to $(0, \infty)$ would have to be trivial. As a hint, in this argument, it could be useful to use morphisms $\mathscr{O}_{\mathbb{R}} \to \mathscr{O}_{\mathbb{R}}$ given by multiplication by a bump function with support contained in some open subset avoiding 0. But that mismatch of rank cannot happen for any locally free sheaf on $\mathbb{R}$.)