Abelian Categories — Examples

In $\mathbf{A}\mathbf{b}$, kernels and cokernels are computed by the familiar subgroup and quotient constructions. So the categorical kernel and cokernel of $f$ are exactly $\ker (f)$ and $H\mathrm{/}\mathrm{im}(f)$.

Therefore the categorical coimage is $G\mathrm{/}\ker (f)$ and the categorical image is $\mathrm{im}(f)$. The canonical coimage-to-image comparison is the map $g+\ker (f)\mapsto f(g)$, which is the usual first-isomorphism map.

The quotient map $q:M\to M\mathrm{/}\ker (f)$ is the coimage projection of $f$, and the inclusion $\mathrm{im}(f)\to N$ is its image inclusion.

The isomorphism from [L2] identifies those two middle objects, and its composite with $q$ and the inclusion is exactly $f$. So the usual module factorization is the categorical coimage-image factorization.

The kernel of $(f,-g):M\oplus N\to P$ consists exactly of those pairs $(m,n)$ with $f(m)=g(n)$.

By [L1], that kernel is the pullback of $f$ and $g$. So the fiber product of two module maps is computed by the familiar subgroup of compatible pairs.

An $F$-vector space is exactly a left module over the ring $F$.

Therefore ${\mathbf{V}\mathbf{e}\mathbf{c}\mathbf{t}}_F$ is the special case $F\text{-}\mathbf{M}\mathbf{o}\mathbf{d}$ of [L1], so it is abelian.

The free preadditive category on the quiver $1\to 2$ is small, and an additive functor out of it is exactly the data of two abelian groups and one homomorphism between them.

Therefore the category of quiver representations is a special case of [L2] with target $\mathbf{A}\mathbf{b}$ from [L1]. So it is abelian.

The category $\mathbf{T}\mathbf{A}\mathbf{b}$ is additive: hom-sets add pointwise, the one-point group is a zero object, and finite products agree with finite coproducts because for finitely many abelian groups the direct product and direct sum carry the same topology.

Let ${\mathbb{R}}_d$ be the additive group of real numbers with the discrete topology and let $\mathbb{R}$ carry its usual topology. The identity homomorphism $\iota :{\mathbb{R}}_d\to \mathbb{R}$ is continuous, bijective, has zero kernel and zero cokernel, so its canonical coimage-to-image map is again $\iota$. But $\iota$ is not an isomorphism in $\mathbf{T}\mathbf{A}\mathbf{b}$, because the inverse map $\mathbb{R}\to {\mathbb{R}}_d$ is not continuous. Hence $\mathbf{T}\mathbf{A}\mathbf{b}$ is additive but not abelian.

Since abelian groups form an abelian category, [L1] applies to the inclusions $C\le B\le A$.

The resulting isomorphism is the familiar quotient-group map described by [L2], so the categorical statement reproduces the ordinary one without change.

By [L2], localization carries every short exact sequence of $R$-modules to a short exact sequence of $S^{-1}R$-modules.

Since both source and target are abelian by [L1], the short-exact-sequence criterion makes localization exact, which is exactly the notion in [L3].

The cited example already computes both endpoint objects explicitly: the coimage is $V$ and the image is $W$, even though both kernel and cokernel vanish.

The canonical map from coimage to image is the morphism $\iota :V\to W$ itself, and [L1] says that $\iota$ is not an isomorphism. So zero kernel and zero cokernel do not force AB2.