What is a simple example, without getting into the mess of triangulated categories, of an additive category that is not abelian?
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7 Answers
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10
There've been lots of mildly complicated examples given, but what about the category of even-dimensional vector spaces over a field?
answered Jun 18, 2014 at 19:29
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$\begingroup$ The subject is old now. But why this does not work? $\endgroup$– SovMar 5, 2018 at 23:48
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4$\begingroup$ This is not an abelian category because a linear map with odd rank (between two even-dimensional vector spaces) does not have a kernel in this category. $\endgroup$ Mar 6, 2018 at 9:33
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$\begingroup$ Why can't you have a linear map of odd nullspace between even-dimensional vector spaces? $\endgroup$ Dec 15, 2020 at 3:20
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2$\begingroup$ @linkhyrule5 You can, that’s the point. But such a map doesn’t have a kernel in the category of even-dimensional vector spaces, which means that category can’t be abelian. $\endgroup$ Dec 15, 2020 at 7:56
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1$\begingroup$ @linkhyrule5 $\begin{pmatrix}0&0\\0&0\end{pmatrix}$ doesn't factor uniquely through $k$. And you'll have the same problem for any $k$ that is not injective. $\endgroup$ Dec 16, 2020 at 11:50
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In infinite dimensions, all hell breaks loose. For example, neither the category of Banach spaces nor the category of Hilbert spaces, although additive, are abelian.
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The category of finitely generated modules over a non-Noetherian ring.
The category of filtered modules over a ring is an example given in Gelfand-Manin.
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3$\begingroup$ The category of modules over any ring is abelian. You mean the category of finitely generated modules over a non-Noetherian ring. Also, the category of filtered modules (over any ring) will do; you don't need to pass to chain complexes. $\endgroup$– Matt EAug 10, 2010 at 2:06
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$\begingroup$ @Matt E. What's the problem with filtered modules? There aren't kernels? Images? $\endgroup$ Aug 10, 2010 at 6:21
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$\begingroup$ @Matt E: Thanks. I will make the changes. $\endgroup$– user977Aug 10, 2010 at 10:54
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7$\begingroup$ The problem is that images and coimages are not isomorphic. Let's just talk about filtered vector spaces. Let V and W be one dimensional vector spaces. Filter them so that V_i=(0) for $i \leq 0$ and V_i = k for i>0, while W_i=(0) for i<0 and W_i=k for i \geq 0. The isomorphism V \to W is a map of filtered vector spaces. I leave it to you to compute that the cokernel of the kernel is 0 in degree 0, while the kernel of the cokernel is one dimensional. $\endgroup$ Aug 10, 2010 at 16:40
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4$\begingroup$ Dear Agusti, David has answered your question, so let me just add: filtered modules is a basic example of an additive category admitting kernels, cokernels, images, and coimages, but which is not abelian, because images and coimages don't coincide in general. Another such example is the category of topological vector spaces (or Banach spaces, if you like), over $\mathbb R$ or $\mathbb C$. (In general, filtrations behave a lot like topologies as a structure, and give the same kind of categorical difficulties.) $\endgroup$– Matt EAug 10, 2010 at 20:18
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There are at least two kinds of (interesting) examples.
I. When we can get an abelian category, but have to add more (co)kernels: 1) the category of free modules over a ring; 2) the category of projective modules over a ring; 3) the category of vector bundles on a topological space (if fact, 3 is a particular case of 2). (From 1 or 2 one gets abelian category of all modules over the ring, from 3 — abelian category of sheafs of vector spaces on X.)
II. When we already have (co)kernels ("category is pre-abelian") but not all mono-/epimorphisms are normal. As explained in another answer, an example is the category of filtered modules.
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$\begingroup$ The problem with free modules over a ring is that submodules of free modules need not be free? If that's the case, the category of all free modules over a PID is abelian... please correct me if I'm wrong. $\endgroup$ Jun 12, 2011 at 14:17
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1$\begingroup$ @BrunoStonek The problem with free modules is the co-kernels, not the kernels. The category of free modules over a PID only has all co-kernels if the PID is actually a field. $\endgroup$ Mar 28, 2012 at 15:12
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$\begingroup$ As an example of @ThomasAndrews's comment (though after over 10 years), one considers, in the category of free $\mathbb{Z}$-modules, the map $\mathbb{Z} \xrightarrow{\times 2} \mathbb{Z}$, then its cokernel is $\mathbb{Z}/2\mathbb{Z}$, which is not a free $\mathbb{Z}$-module, yet $\mathbb{Z}$ is indeed a PID. It seems that this example works if one starts with an arbitary nonzero element $r \in R^{\times}$ in an arbitary PID $R$. Such $r$ exists only if $R$, as a PID, is not a field.... $\endgroup$ Apr 20, 2022 at 14:10
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$\begingroup$ Conversely, when $R = k$ is a field, then we are actually in the category of $k$-modules, since all vector spaces over $k$ has a basis, then the category of $k$-modules is indeed abelian. Hope that I got this right. :) $\endgroup$ Apr 20, 2022 at 14:15
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$\begingroup$ As one may know, that every $R$-module can be written as a quotient of a free $R$-module, one may not expect that every quotient of a free $R$-module over a PID $R$ is free, and an immediate example is $\mathbb{Z}/2\mathbb{Z}$, this is how I came up with the example. So actually @BrunoStonek's question is quite inspiring. $\endgroup$ Apr 20, 2022 at 14:25
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Another nice example is the category of finite-dimensional vector bundles over a fixed base space (with bundle maps over the identity as morphisms).
If the base space is not too simple (the interval suffices), then this category is not (pre-)Abelian because there are, in general, neither kernels nor cokernels. Intuitively speaking, the obvious candidate for the kernel of a bundle map need not form a vector bundle because its dimension need not be locally constant.
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$\begingroup$ what's the problem with this one? $\endgroup$– user325Aug 10, 2010 at 16:38
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$\begingroup$ Sorry if I overlooked something, but why isn't this abelian? $\endgroup$– user325Aug 10, 2010 at 17:26
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$\begingroup$ @Soarer: I have added clarification to my answer. I deleted my previous comment because it contained a false statement. $\endgroup$– RasmusAug 10, 2010 at 18:03
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There's a slight modification of an example that almost works but that was deleted: take the category of Hausdorff topological abelian groups. The coimage of a morphism in this category is its image in the ordinary sense (edit: topologized via the quotient topology), but the image is the closure of the coimage (exercise) (edit: topologized via the subspace topology), so the two don't need to agree in general.
(Edit, 6/21/22: In fact the difference between the subspace and quotient topologies means the category of topological abelian groups already works, no Hausdorff assumption necessary.)
As Matt E mentions in a comment, adding topologies, like adding filtrations, is an easy way to cause this sort of thing to happen. Similarly we could take Hausdorff topological vector spaces, etc.
answered Jun 18, 2014 at 17:48
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9$\begingroup$ You can argue that the category is not abelian by providing a morphism which is monic and epic, but not an isomorphism: the identity from $\mathbb{R}$ with the discrete topology and $\mathbb{R}$ with the usual metric topology is such a mono-epi not iso. Any nondiscrete topological abelian group works, too. $\endgroup$– egregJun 18, 2014 at 17:54
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$\begingroup$ I'm confused. In topological group, isomorphism theorem holds, so coim is canonically isom to I'm, why the image is closure of coim ? $\endgroup$ Jun 21, 2022 at 18:15
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$\begingroup$ @afortiori: that's not the first isomorphism theorem; the statement there requires the additional assumption that the map is open. Check that if $\mathbb{R}_d$ denotes $\mathbb{R}$ with the discrete topology then the abstract isomorphism $\mathbb{R}_d \to \mathbb{R}$ has the property that its coimage is $\mathbb{R}_d$ and its image is $\mathbb{R}$ so the two are not homeomorphic (and this map isn't open). $\endgroup$ Jun 21, 2022 at 19:04
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1$\begingroup$ @afortiori: again, that's not the first isomorphism theorem. The map there is required to be surjective; the full statement of the first isomorphism theorem is about the image of a not-necessarily-surjective map, and in that generality the isomorphism theorem can fail because the image of a Banach space map is not necessarily closed. $\endgroup$ Jun 22, 2022 at 7:46
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A quite explicit example coming from quantum algebra.
Consider the ring $K_h:=\mathbb K[[h]]$ of formal power series with coefficients in the field $\mathbb K$ (take for example $K=\mathbb R)$. Let $\mathcal C_f$ be the category of topologically free $K_h$ modules, i.e. all those $K_h$-modules that are isomorphic to modules of the form $M[[h]]$, denoting by $M$ any $K$-vector space. Morphisms $\varphi : M[[h]]\rightarrow N[[h]]$ are formal power series $\sum_{i\geq 0} \varphi_i h^i$ with $\varphi_i: M\rightarrow N$ morphism of $K$-vector spaces. If $\varphi : M[[h]]\rightarrow N[[h]]$ with $\varphi=\sum_{i\geq 0} \varphi_i h^i$ and $\psi : N[[h]]\rightarrow Q[[h]]$ with $\psi=\sum_{i\geq 0} \psi_i h^i$ are morphisms in $\mathcal C_f$, then their composition is the morphism $\psi\circ\varphi$ with power series expansion
$\sum_{i+j\geq 0}(\psi_j\circ\varphi_i) h^{i+j}$.
$\mathcal C_f$ is additive with biproduct $M[[h]]\oplus_h N[[h]]:=(M\oplus N)[[h]]$; it is not abelian because the cokernel of the inclusion (which is a morphism in $\mathcal C_f$)
$i: M\rightarrow M[[h]]$, $m\mapsto i(m)=(m,0,0,\dots)$
i.e. $coker(i)=hM[[h]]$, is not an object in $\mathcal C_f$. In fact, there exists no $\mathbb K$-vector space $N$ and isomorphism $\rho: hM[[h]] \rightarrow N[[h]]$ in $\mathcal C_f$; if such a morphism $\rho$ existed, then it would not be an isomorphism as any object in $N[[h]]$ of the form $(n,0,0,...)$ does not belong to the image of $\rho$ for $n\neq 0$.
