It should be quite an easy question and probably there's also a certain degree of intrinsic silliness in it, but still...

Let $\mathcal{C}$ be an abelian category and let $C(\mathcal{C})$ be the category of (cochain) complexes on $\mathcal{C}$, i.e. the full subcategory of the functorial category $Fct((\mathbb{Z},\leq),\ \mathcal{C})$ given by those (covariant) functor $F$ such that, for every $n\in\mathbb{Z}$, $F(n\leq n+2)=0$. I depict an object $X\in$ Ob($\mathcal{C}$) as a sequence $$\cdots\rightarrow X^{n-1}\overset{d_{X}^{n-1}}\longrightarrow X^{n}\overset{d_{X}^{n}}\longrightarrow X^{n+1}\rightarrow\cdots$$ For such an $X$, I define the n-th cohomology object of $X$ as $$H^{n}(X):= Coker (Im(d_{X}^{n-1})\hookrightarrow Ker(d_{X}^{n})).$$ If $\bar{d_{X}^{n}}\colon Coker(d_{X}^{n})\to X^{n+1}$ is the factorization of $d_{X}^{n}$ through $Coker (d_{X}^{n-1})$, I need to prove that $H^{n}(X)\simeq Ker(\bar{d_{X}^{n}})$, without using Freyd-Mitchell's Theorem.

My attempt: let $l\colon Im(d_{X}^{n-1})\hookrightarrow X^{n}$, $j\colon Im(d_{X}^{n-1})\hookrightarrow Ker(d_{X}^{n})$, $i\colon Ker(d_{X}^{n})\hookrightarrow X^{n}$, $p\colon X^{n}\twoheadrightarrow Coker (l)$ and $q\colon Ker(d_{X}^{n})\twoheadrightarrow H^{n}(X)$ be the canonical morphisms (so that, in particular, $i\circ j=l$). I first noticed that $Coker(d_{X}^{n-1})\simeq Coker (l)$ and so $p$ is indeed also the epimorphism of $X^{n}$ onto $Coker(d_{X}^{n-1})$, so that $\bar{d_{X}^{n}}\circ p= d_{X}^{n}$. Since $p\circ i\circ j=p\circ l=0$, $p\circ i$ factors through $q$ via a unique $\gamma\colon H^{n}(X)\to Coker(d_{X}^{n-1})$ (thence $\gamma\circ q =p\circ i$). Looking at what happens in the case of modules, I am tempted to believe that such a $\gamma$ must be proven to be a kernel for $\bar{d_{X}^{n}}$. I've shown so far that $\bar{d_{X}^{n}}\circ \gamma=0$, but I'm stuck at this point. In particular, I do not manage to prove the universal property for $\gamma$ nor to show that the factorization of $\gamma$ through the kernel of $\bar{d_{X}^{n}}$ is an isomorphism (that is, as we are in the abelian context, it is both mono and epi).

Any (detailed) help (or suggestion about an easier proof) would be greatly appreciated.

  • $\begingroup$ I suppose you are asking about whether the two possible definitions of (co)homology object coincide. They do, even in semi-abelian categories: see Proposition 2.3 in [Everaert and van der Linden, Baer invariants in semi-abelian categories II]. $\endgroup$ – Zhen Lin Nov 6 '13 at 23:39
  • $\begingroup$ @ZhenLin Thanks for the link. I'll give it a look, but I think it's a little bit out of range for me at the moment. If someone (or you yourself) has explicit, direct solutions or references for what I'm asking for, I'd still appreciate it. $\endgroup$ – Marco Vergura Nov 7 '13 at 19:00

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