MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A $\delta$-functor is usually (e.g. in Grothendieck's Tohoku paper or in Cartan-Eilenberg's Homological Algebra) defined to be a pair $$((T^i)_{i\in\mathbb{Z}},((\delta^i_{\mathbb{S}})_{i\in\mathbb{Z}})_{\mathbb{S}\text{ short exact sequence}})$$ consisting of a sequence of functors $(T^i)_{i\in\mathbb{Z}}$ (with common Abelian source $\mathscr{C}$ and common additive target $\mathscr{D}$) and a family $((\delta^i_{\mathbb{S}})_{i\in\mathbb{Z}})_{\mathbb{S}\text{ short exact sequence}}$, indexed by short exact sequences in $\mathscr{D}$, of sequences of morphisms in $\mathscr{D}$ - the so-called connecting morphisms - subject to some conditions.

However, sometimes (e.g. in Rotman's Introduction to Homological Algebra (First Edition)) a $\delta$-functor is defined to be a sequence $(T^i)_{i\in\mathbb{Z}}$ of functors as above such that there exists a family $((\delta^i_{\mathbb{S}})_{i\in\mathbb{Z}})_{\mathbb{S}\text{ short exact sequence}}$ as above, subject to the same conditions as above.

One might wonder if the omission of the additional data of the connecting morphisms is done on purpose. Or, more precisely:

Are the families of connecting morphisms of a $\delta$-functor uniquely determined by its sequence of functors?

(I guess that the answer is no, and I guess it is still no if we consider only exact or even universal $\delta$-functors.)

EDIT: As Martin's example shows, the answer to the original question is no for trivial reasons. But is there an example with essentially different connecting morphisms (whatever that means, but it should rule out Martin's example, compositions with automorphisms, etc.)?

More general: Can we say that the connecting morphisms are unique up to something, and even say what something is?

Finally, concerning Martin's remark about the definition of universality: We can of course define this by using all connecting morphisms instead of a given one. Then we may wonder whether the usual notion of universality implies this stronger one...

share|cite|improve this question
You could just take $-\delta^i$ instead of $\delta^i$. The first definition (the differentials belong to the data of $\delta$-functor) is the correct one, the second is not correct. You cannot even formulate the notion of a universal $\delta$-functor with the second one. – Martin Brandenburg Nov 7 '11 at 8:33
Except when $\mathcal D$ is a $\mathbb Z/2$-category... (I guess it depends on what your counter-example quantifies over.) – Torsten Ekedahl Nov 7 '11 at 8:54

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.