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can you tell me examples of additive categories which have morphism that has no kernel and morphism has no cokernels.

if you tell me reference which provide this kind of examples it will be great.thanks.

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2 Answers 2

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A typical example is the category of vector bundles on a (non-pathological) topological space.

The corresponding algebraic example is the category of finitely generated projective modules over some (non-pathological) commutative ring.

Notice: If $K \to P$ is a kernel of $P \to Q$ in the category of f.g. proj. $R$-modules, then this is also the kernel in the category of all abelian groups. The reason is that $\hom(R,-)$ is left exact and that it identifies with thr forgetful functor. Therefore, it suffices to find an example of a map between f.g. projective modules whose "usual" kernel (i.e. the one in the category of abelian groups) is not projective. Unfortunately, there is no such argument for cokernels, which will be more complicated.

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  • $\begingroup$ thank you very much,that was great. $\endgroup$
    – kpax
    Sep 28, 2014 at 18:57
  • $\begingroup$ Well, since the opposite category of an additive category is also additive. You may get an example for cokernels easily. $\endgroup$
    – Jon Snow
    Oct 10, 2023 at 11:16
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The opposite category of an additive category is also additive, whence I would only show the kernel case. Let's consider the category $\mathfrak C$ of injective modules over a commutative ring $R$. Let $\varphi:A\to B$ be a morphism in $\mathfrak C$, if $\mu:K\rightarrowtail A$ is the kernel in $\mathfrak C$. Let $\mu':K'\to A$ be the kernel in the category of $R$-modules. As $K$ is an injective module, there is a morphism $\epsilon:A\to K$ such that $1_K=\epsilon\mu$, by splitting lemma, $K$ is a direct summand of $A$. Assume $A=A'\oplus K$. As $K'$ is the kernel of $\varphi$ in the $\mathfrak M_R$, there is a unique morphism $\varphi':K\to K'$ such that $\mu=\mu'\varphi'$. As $A=A'\oplus K$, $\mu$ is nothing but $\{0,1_K\}$. If we assume that $\mu':K'\to A'\oplus K$ is $\{f,g\}$, we have $\{0,1_K\}=\{f,g\}\varphi'$, whence $f=0\varphi'=0$. But in many cases, we can find three injective modules $A',K$ and $B$ and a morphism $A'\oplus K\to B$ whose kernel does not look like $\{0,g\}$ (the direct products of injective modules are injective).

For instance, consider $R=\mathbb Z$, $A'=K=B=\mathbb Q/\mathbb Z$ and $A'\oplus K\to B$ is defined by $\langle1,-1\rangle$, i.e. $(p/q+\mathbb Z,p'/q'+\mathbb Z)\mapsto(p/q-p'/q')+\mathbb Z$.

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