1
$\begingroup$

I have a problem with the definition of exact sequences in non-necessarily abelian categories. In this nLab page it is written that exact sequences can be defined in semi-abelian categories. My problem is: how can one claim that if $g\circ f=0$ then $\mathrm{im}(f)\subseteq \ker(g)$ (or even just that there exists a canonical morphism $\mathrm{im}(f)\to\ker(g)$)?

Let me explain in more detail: if $$ a\stackrel{f}{\longrightarrow} b\stackrel{g}{\longrightarrow} c $$ and $g\circ f=0$ in a semi-abelian category $\mathcal{A}$, then there exists a unique $\tilde{f}:a\to\ker(g)$ such that $k_g\circ \tilde{f}=f$, where $k_g:\ker(g)\to b$. Write $k_f:\ker(f)\to a$. Since $k_g$ is monic, $$ 0=f\circ k_f = k_g\circ \tilde{f}\circ k_f $$ implies that $\tilde{f}\circ k_f = 0$ and hence there exists a unique morphism $\hat{f}:\mathrm{coim}(f)\to \ker(g)$ such that $\hat{f}\circ c_{k_{f}} = \tilde{f}$, where $c_{k_f}:a\to \mathrm{coker}(k_f)=\mathrm{coim}(f)$. Now, without knowing that $\mathrm{coim}(f)\cong \mathrm{im}(f)$, how do I relate $\mathrm{im}(f)$ and $\ker(g)$?

I tried a different approach as well. In a semi-abelian category we have the canonical decomposition $$ a\stackrel{c_{k_f}}{\longrightarrow} \mathrm{coim}(f) \stackrel{\bar{f}}{\longrightarrow} \mathrm{im}(f) \stackrel{k_{c_f}}{\longrightarrow} b, $$ where $c_f:b\to \mathrm{coker}(f)$ and $k_{c_f}:\ker(c_f)=\mathrm{im}(f) \to b$. Since $g\circ f=0$ and $c_{k_f}$ is epi, we deduce that $g\circ k_{c_f} \circ \bar{f} = 0$, but again: without knowing that $\bar{f}$ is at least epi, how do I relate $\mathrm{im}(f)$ and $\ker(g)$?

$\endgroup$
2
$\begingroup$

By definition, a semi-abelian category (or homological) is regular, so every arrow $f:A\to B$ factorizes as a regular epimorphism $p_f:A\to Im(f)$ followed by a monomorphism $m_f:Im(f)\to B$. This $I$, or more precisely the subobject $m_f:Im(f)\to B$, is by definition the image of $f$. Then if you have a sequence $$A\stackrel{f}{\longrightarrow} B \stackrel{g}{\longrightarrow} C$$ such that $g\circ f=0$, your factorization $f=k_g\circ \widetilde{f}$ shows that $Im(f)\subset Ker(g)$, in the sense that you must have a morphism $j:Im(f)\to Ker(g)$ such that $m_f=k_g\circ j$ (you can just take $j=m_{\widetilde{f}}$). Then the sequence is exact at $B$ if this $j$ is an isomorphism, which is equivalent to the condition that $\widetilde{f}$ is a regular epimorphism (because the factorization is unique up to a unique appropriate isomorphism) and that $m_f$ is the kernel of $g$.

In a homological category, one can prove that every regular epimorphism is the cokernel of its kernel, which implies that your $\overline{f}$ is always a monomorphism, and thus also that a morphism has zero kernel if and only if it is a monomorphism. So the image is really what you call the coimage; what you call the image, i.e. the kernel of the cokernel of $f$, is generally less useful, because not every monomorphism in a semi-abelian category is a kernel. In fact your image is the smallest kernel containing $m_f$, so if $m_f$ is a kernel then it coincides with your definition of image.

$\endgroup$
  • $\begingroup$ Let me see if I understood your answer: are you saying that the problem lies in the fact that the definition of $\mathrm{im}(f)$ in a semi-abelian category is different from the definition of $\mathrm{im}(f)$ in an abelian one? And that your $m_f$ coincides with the composition $k_{c_f}\circ\bar{f}$? So that what I am really doing when I do homological algebra in a semi-abelian category is to consider the "quotient" of $\ker(g)$ by the "image" (in $b$) of "$a/\ker(f)$"? $\endgroup$ – Ender Wiggins Aug 1 at 9:36
  • $\begingroup$ Well, in an abelian category the cokernel of the kernel and the kernel of the cokernel are always isomorphic, possibly by definition. In a semi-abelian category this need not be true, and it turns out that the cokernel of the kernel is more interesting than the kernel of the cokernel; so we keep that as the image. $\endgroup$ – Arnaud D. Aug 1 at 9:45
  • $\begingroup$ Ok, thank you very much. $\endgroup$ – Ender Wiggins Aug 1 at 9:54
  • $\begingroup$ Defining homology objects in semi-abelian categories is a bit weird, since the equivalent definitions mentioned in this answer are no longer equivalent. It also means that if you define the homology object as the cokernel of $\widetilde{f}$, then it could be zero without your sequence being exact. $\endgroup$ – Arnaud D. Aug 1 at 9:56
  • $\begingroup$ But again, the problem disappears if the image of $f$ is really a kernel; and this is the case, for example, for the chain complex obtained by normalizing a simplicial object. See also this paper for more information. $\endgroup$ – Arnaud D. Aug 1 at 9:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.