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Let $\mathcal{A}$ be an abelian category and let $Ch(\mathcal{A})$ be the category of homologicaly, non-negatively graded chain complexes in $\mathcal{A}$. The sequence of homology functors $H_n:Ch(\mathcal{A})\to \mathcal{A}$ is a (in fact, the prototypical) $\delta$-functor. My question is:

Is it true that $(H_n)_{n\in \mathbb{N}}$ is a universal $\delta$-functor?

Intuitively, this of course should be the case, but I couldn't find a direct argument. What I was able to show is that, if $\mathcal{A}$ has enough projectives, then $Ch(\mathcal{A})$ has enough projectives and we can show that the homology functors are the derived functors of $H_0$ and thus by the genral theory a universal $\delta$-functor. This method requires the identification of projectives in $Ch(\mathcal{A})$ and a (simple) spectral sequence argument for the double chain complex obtained from the projective resolution of a complex in $Ch(\mathcal{A})$. It also gives a bit more as one can extend the argument to show that the total derived functor of $H_0$ is quasi isomorphic to the identity.

I don't worry much about the "enough projectives" hypothesis, but I would like to see a direct argument for the seemingly tautological fact that homology is a universal $\delta$-functor.

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By the effecability criterion, it suffices to show that any complex $X$ can be written as a quotient of a complex with vanishing higher homology. In the unbounded setting, you can take the cone over $\Sigma^{-1} X$, which is even contractible, and for non-negative chain complexes, you need to truncate this cone at $0$.

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  • $\begingroup$ I see. A projective cover is just a special case of an acyclic cover which is enough (and exists without any assumptions on the category) and then you can take an acyclic resolution and the argument goes without change. I guess that the "effecability criterion" (which I am not familiar with) is an elementary way to bypass the spectral sequence argument, right? Thanks! $\endgroup$ – KotelKanim Feb 20 '16 at 12:14
  • $\begingroup$ It's a criterion for universality of $\delta$-functors, you find it for example on Wikipedia or in Hilton Stammbachs book on homological algebra. $\endgroup$ – Hanno Feb 21 '16 at 7:22
  • $\begingroup$ @KotelKanim Do you have any further questions? $\endgroup$ – Hanno Feb 25 '16 at 6:19
  • $\begingroup$ No, everything is perfect. I just forgot to accept and then didn't have access for a while... thanks for the answer! $\endgroup$ – KotelKanim Feb 25 '16 at 16:24

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