# Is the cohomology of a topological operad a cooperad?

For cohomology with coefficients in a field $F$ the map $H^\cdot(X;F) \otimes H^\cdot(Y;F) \to H^\cdot(X \times Y;F)$ of the Kunneth theorem is an isomorphism of algebras over $F$. I am correct in thinking that this together with the fact that cohomology with a contravariant functor, implies that the cohomology of a topological operad is a cooperad, the dual of a notion of a cooperad?

If this is not true, where does my reasoning fail? Does it at least hold on the level of vectorspaces over $F$? If this is true, why aren't there many references about this construction? It seems that the additional algebra structure would give you some additional information.

Furthermore, using the Thom isomorphism for the cohomology, would this also imply that there is a notion of string cohomology dual to string topology?

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There is one little caveat, though: you need either to assume your spaces in the operad are of finite type, or work in the category of topological vector spaces and completed tensor products. The reason is that the Künneth theorem in cohomology only holds for spaces of finite type if you use the regular tensor product. Consider, for a counterexample, an infinite wedge of spheres.

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Do you a reference this way of considering cohomology groups as topological vector spaces? What is the natural topology on them? Is it the weak-* topology with respect to the homology-cohomology pairing? – skupers Feb 23 '10 at 14:57
It's the topology of a pro-finite dimensional vector space. You have to take field coefficients. Let's assume we work in the category of CW-complexes. Every CW-complex X is the directed colimit of its finite subcomplexes, and this gives an inverse system after applying cohomology. The limit of this system is the cohomology of X. (There are no phantom maps.) – Tilman Feb 24 '10 at 6:38

Yes the cohomology of an operad over a field is a cooperad . Indeed over a field F, the cohomology functor H* is a monoidal functor (in a contravariant sense.) More precisely the category of topological spaces Top equipped with the cartesian product × is a symmetric monoidal category (the unit being the one point space *). Similarly the category of graded vector spaces is symmetric monoidal when equipped with the tensor product ⊗ (the unit being the ground field concentrated in degree 0 that we denote by F). Then Kunneth theorem tells you that you have a map H(X)⊗ H(X) --> H(XxY), and moreover it is an isomorphism. Also H(∗) is isomorphic to the field F. This is exactly the meaning of your functor being monoidal (which here as to be interpreted in a contravariant sense.) In other words your functor commutes, up to canonical isomorphisms, with the two monoidal structures × and ⊗.

A consequence of this suppose given a topological operad, that is a sequence (X(n))n≥0 of spaces together with structure maps μ:X(k)xX(n1)x...xX(nk) --> X(n1+...+nk) satisfying the associativity condition for an operad (and also a unit belonging to X(1) and some compatible action of the symmetric groups). Then, applying the monoidal functor H you get a cooperad (HX(n))n≥0 in the category of graded vector spaces. Indeed composing H(μ) with the Kunneth isomorphism you get a map
H(X(n1+...+nk)) --> H(X(k)xX(n1)x...xX(nk)) ≅ H(X(k))⊗H(X(n1))⊗...⊗H(X(nk))
satisfying the coassociativity condition of a cooperad (with also a counit and Σ-equivariance.)

You can also try to do the same at the level of (co)chains. However the cochain functor is not monoidal (it is weakly "comonoidal"). So it is not true that the cochains of an operad is a genuine cooperad (although it is amost a cooperad). But the functor of singular chains is monoidal and so it is true that the chains of a topolgical operad is an operad of chain complexes.

Maybe it is because this lack of genuine structure at the cochain level that one do not consider often the cooperad structure on the cohomology. Anyway it would not give more structure that the homology and it is often easier to work with this.

The reason for which this is not in the litterature is that it is folklore that a monoidal covariant/contravariant functor turns operad into operads/cooperads.

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You consider the symmetric monoidal category of graded vector spaces, thereby forgetting about the cup product. Does the statement fail if we use the category of algebras over $F$ with tensor product, trying to remember the cup product? – skupers Feb 23 '10 at 14:52
You can also do tis in the category of graded algebra, so you get indeed a cooperad of algebras using the cup products. Notice however that if you do this in homology you get an operad of coalgebras (the coalgebra structure in homology is dual to the algebra structure of cohomology). Therefore you dot not really gain more structure passing from homology to cohomology. As Tilman pointed out homology is actually better when you deal with space of not finite type (whch is actually rare in the usual applications) – Pascal Lambrechts Feb 24 '10 at 8:30