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

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.

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Comment #7840 by Zhenhua Wu on

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