# Homology or cohomology?

How do people call an additive functor from a triangulated category $C$ to an abelian one that converts distinguished triangles into long exact sequences. Does one usually call a covariant functor of this sort 'homology' and denote it by $H_i$, whereas a contravariant functor is called cohomology and is denoted by $H^i$? For example, if we consider the $i$-th homology (?) of $C$ with respect to a $t$-structure $t$ for it, is it fine to denote it by $H_i^t$?

Upd. Besides, consider a cohomological complex $\dots\to C^0\to C^1\to C^2\to \dots$. Does it have homology ($H_i(C)$) or cohomology ($H^i(C)$)?

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I've always considered that homology of a chain complex is homology, and homology of a cochain complex is cohomology. –  Bill Kronholm May 1 '11 at 16:27
Possibly you are right; but suppose that you only have the corresponding homotopy (or derived) category (so the direction of arrows is not specified); how would you call the corresponding functor? –  Mikhail Bondarko May 1 '11 at 16:48
Just for the possible amusement value, there was once an argument on this exact topic on an episode of an American television show called NCIS. Maybe this link will work:  ncisfanwiki.com/page/2.20+-+Red+Cell –  Will Jagy May 2 '11 at 4:25
From this kind of annoyance comes the convention $H_i = H^{-i}$. –  Tyler Lawson May 2 '11 at 4:25

I would say that a cohomological chain complex has cohomology, and that a homological chain complex has homology...

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Possibly you are right; but suppose that you only have the corresponding homotopy (or derived) category (so the direction of arrows is not specified); how would you call the corresponding functor? –  Mikhail Bondarko May 1 '11 at 16:47

I would say that it's cohomology. The "co" in the name seems to come from the fact that it's contravariant on topological spaces, but covariant on the category of sheaves on these spaces. Topological homology doesn't even come from a functor on sheaves, so in fact, they aren't dual concepts on the algebraic level. "Homological" chain complexes differ from cohomological only in notation.

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(CW because it's more an over-long comment than a real answer.)

I think there are too many competing normalizations to make a good choice. In lieu of sensible default, call one of them homology, and call the other cohomology and I'm sure it'll be fine. This is also probably why someone worked out the language "left-derived" and "right-derived".

Personally, I think the distinction between "homological" and "cohomological" chain complexes is artificial; what might make a chain complex "co"-chain is that it was initially zero, whereas "really-chain" complexes might be eventually zero. Of course, we're often interested in things that do both, and things that do neither --- e.g. either theory on manifolds, or geometric K theory and elliptic cohomology. Anyways, if you're playing with chain complexes, I'd always call the kernel-mod-image construction the "homology" of a chain complex. We'd reserve the right to write "cohomology of X" for the homology of a contravariant construction from X, but it's still a homology of something.

Historically, "Homology" comes from a relation that Poincaré described on submanifolds of a manifold; there are both covariant and contravariant ways to make this functorial and algebraic on the category of smooth manifolds; for eitehr one you probably have to deal with singular submanifolds eventually. The contravariant one is called cobordism[1] these days and is represented by a spectrum indexed by codimension. There is that theorem that all reasonable "cohomology" theories on spaces are corepresentable; there are reasonable representable homology theories (like homotopy), and there are homology theories that aren't representable (like singular homology of spaces, iirc). Perhaps if your things are always corepresenable it'd be reasonable to call them cohomologies.

[1] That's "co" meaning "together", not "dual". That is, two things may be "co-bordant"; not that "co-bordisms" pair with bordisms. Although, I suppose they might, anyways...

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They are not corepresentable, since the target category is far from being abelian groups.:) –  Mikhail Bondarko May 2 '11 at 7:27