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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?

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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 2:06
    
@Matt E. What's the problem with filtered modules? There aren't kernels? Images? –  a.r. Aug 10 '10 at 6:21
    
@Matt E: Thanks. I will make the changes. –  user977 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 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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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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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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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
    
@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
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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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what's the problem with this one? –  Soarer Aug 10 '10 at 16:38
    
Sorry if I overlooked something, but why isn't this abelian? –  Soarer Aug 10 '10 at 17:26
    
@Soarer: I have added clarification to my answer. I deleted my previous comment because it contained a false statement. –  Rasmus Aug 10 '10 at 18:03
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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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