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 $\ker(g)$?


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.

  • $\begingroup$ 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

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