Example 12.3.13 (0108)—The Stacks project

Example 12.3.13. Let $k$ be a field. Consider the category of filtered vector spaces over $k$ . (See Definition 12.19.1.) Consider the filtered vector spaces $(V, F)$ and $(W, F)$ with $V = W = k$ and

$F^ iV = \left\{ \begin{matrix} V & \text{if} & i < 0 \\ 0 & \text{if} & i \geq 0 \end{matrix} \right. \text{ and } F^ iW = \left\{ \begin{matrix} W & \text{if} & i \leq 0 \\ 0 & \text{if} & i > 0 \end{matrix} \right.$

The map $f : V \to W$ corresponding to $\text{id}_ k$ on the underlying vector spaces has trivial kernel and cokernel but is not an isomorphism. Note also that $\mathop{\mathrm{Coim}}(f) = V$ and $\mathop{\mathrm{Im}}(f) = W$ . This means that the category of filtered vector spaces over $k$ is not abelian.

Comments (2)

Comment #7840 by Zhenhua Wu on October 14, 2022 at 03:45

The category of even dimension vector space over a field $K$ is an easier example that is additive but not abelian, since the kernel of the linear map $K^2\to K^2,e_1\mapsto e_1,e_2\mapsto 0$ is one dimensional. The word 'filtered vector spaces' may scare people.

Comment #8064 by Stacks Project on January 20, 2023 at 10:26

The category of filtered vector spaces has kernels and cokernels but is not abelian. So it is interesting in that it shows one needs the Im $=$ Coim axiom.

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10 comment(s) on Section 12.3: Preadditive and additive categories

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