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What is a simple example, without getting into the mess of triangulated categories, of an additive category that is not abelian?

Kandinskij
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7 Answers7

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There've been lots of mildly complicated examples given, but what about the category of even-dimensional vector spaces over a field?

Jeremy Rickard
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  • The subject is old now. But why this does not work? – Sov Mar 05 '18 at 23:48
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    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. – Jeremy Rickard Mar 06 '18 at 09:33
  • Why can't you have a linear map of odd nullspace between even-dimensional vector spaces? – linkhyrule5 Dec 15 '20 at 03:20
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    @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. – Jeremy Rickard Dec 15 '20 at 07:56
  • I'm using the definition of the kernel of an arrow $f$ as another arrow $k$ s.t. $f k = 0$ the zero arrow (and s.t. all other such arrows factor uniquely through it); so it seems like I don't need an actual object within the category, I just need a map whose image (in the usual linear algebra sense) is the appropriate space? – linkhyrule5 Dec 15 '20 at 14:49
  • @linkhyrule5 I don't think I understand what you mean by "I don't need an actual object within the category". If $k$ is an arrow in the category of even-dimensional spaces, then its source needs to be an even-dimensional space. – Jeremy Rickard Dec 15 '20 at 15:45
  • Sure, but it doesn't have to be surjective -- it can map into a strict subset of its target object. Then if we define $k(x) = \begin{cases}0 & x \notin \ker f \ 1 & \text{otherwise}\end{cases}$, then $f k = 0$ ... though I suppose I haven't checked the universal property yet, huh. – linkhyrule5 Dec 16 '20 at 01:28
  • @linkhyrule5 I'm not quite sure what map $k$ you have in mind (the definition you give is not linear). But if $U\xrightarrow{k}V\xrightarrow{f}W$ are linear maps between even-dimensional vector spaces with $fk=0$, and $f$ has odd rank (so the usual kernel of $f$ is odd-dimensional), then either existence or uniqueness will fail in the universal property defining a kernel in the category of even-dimensional spaces. – Jeremy Rickard Dec 16 '20 at 11:32
  • Sure it's linear, though I suppose it's not manifestly linear. But, hmm... Concretely, the simplest example would be something like, let $f = \begin{pmatrix} 1& 0\ 0 & 0\end{pmatrix}$; then $k = \begin{pmatrix}0 & 0 \ 0 & 1\end{pmatrix}$ is certainly a linear map between even-dimensional spaces s.t. $f k = 0$, and furthermore $k$ spans the nullspace of $f$. This is in some regards a nice property of two dimensions, but even if we increase the number of dimensions, as long as $k$ spans nullspace $f$, any other $k^\prime$ with $f k^\prime = 0$ should be some other matrix times $k$? – linkhyrule5 Dec 16 '20 at 11:42
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    @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. – Jeremy Rickard Dec 16 '20 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.

G. Rodrigues
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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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    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.

     – Matt E Aug 10 '10 at 02:06
  • @Matt E. What's the problem with filtered modules? There aren't kernels? Images? – Agustí Roig Aug 10 '10 at 06:21
  • @Matt E: Thanks. I will make the changes. –  Aug 10 '10 at 10:54
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    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. – David E Speyer Aug 10 '10 at 16:40
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    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.) – Matt E Aug 10 '10 at 20:18
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    Yup. Hausdorff topological abelian groups is another example like this. – David E Speyer Aug 10 '10 at 21:03
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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.

Grigory M
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  • 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. – Bruno Stonek Jun 12 '11 at 14:17
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    @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. – Thomas Andrews Mar 28 '12 at 15:12
  • 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.... – Hetong Xu Apr 20 '22 at 14:10
  • 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. :) – Hetong Xu Apr 20 '22 at 14:15
  • 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. – Hetong Xu Apr 20 '22 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.

Rasmus
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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.

Qiaochu Yuan
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    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. – egreg Jun 18 '14 at 17:54
  • I'm confused. In topological group, isomorphism theorem holds, so coim is canonically isom to I'm, why the image is closure of coim ? – Poitou-Tate Jun 21 '22 at 18:15
  • https://math.stackexchange.com/questions/415735/first-isomorphism-theorem-for-topological-groups – Poitou-Tate Jun 21 '22 at 18:15
  • @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). – Qiaochu Yuan Jun 21 '22 at 19:04
  • Where open mapping condition came from ? Morphism in topological group does not require open mapping condition.. – Poitou-Tate Jun 22 '22 at 02:01
  • @afortiori: the third paragraph in the question you linked to says "Now, I suppose that $f$ is open." – Qiaochu Yuan Jun 22 '22 at 02:03
  • @Qiaochu Yuan Thank you, you are correct. I want to prove category of etale(φ,Γ)module is abelian category. But I'm stuck with proving 'coim→imp' is isom part. Do you have any ideas ? If needed, I will open new question page. – Poitou-Tate Jun 22 '22 at 02:14
  • @afortiori: I don’t know the definition but either way you should ask a new question and put the definition etc. there. – Qiaochu Yuan Jun 22 '22 at 04:27
  • Category of Banach space is also not abelian because coimf→imp is not isom, but in this page, first isomorphism theorem holds, https://math.stackexchange.com/questions/2768437/first-isomorphism-theorem-for-banach-spaces – Poitou-Tate Jun 22 '22 at 04:34
  • Where am I missing ? In the link, additional condition is supposed ? – Poitou-Tate Jun 22 '22 at 04:34
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    @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. – Qiaochu Yuan Jun 22 '22 at 07:46
  • There's also an issue of which category of Banach spaces you work in. The category $\text{Ban}$ of Banach spaces and continuous linear maps is additive, but I'm not actually sure it even has kernels and cokernels; the issue is that because an isomorphism in this category is not required to be an isometry (this is the notion of isomorphism used in the linked question) it's not clear what norm to put on a limit or colimit. You can fix this by working in the category $\text{Ban}_1$ of Banach spaces and short maps (maps of norm $\le 1$), which is complete and cocomplete, but not additive. – Qiaochu Yuan Jun 22 '22 at 07:51
  • Anyway, please ask a new question. – Qiaochu Yuan Jun 22 '22 at 07:52
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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 $n\neq 0$.

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