# How to define exact sequences in a semi-abelian category

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.
• 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 "$a/\ker(f)$"? – Ender Wiggins Aug 1 at 9:36
• 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. – Arnaud D. Aug 1 at 9:56
• 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. – Arnaud D. Aug 1 at 9:58