Take coefficients in a field, so as to not have the correction from Tor.

I am thinking about the functor sending a topological space $X$ to its cohomology ring $H^*(X)$. So specifically, I am thinking of a functor from the category of Topological spaces to the opposite category of graded commutative rings.

Then if ethical I would like to think of Kunneth in this way: it sends a product of spaces into the product in the $($Graded-commutative rings$)^{op}$ (which gives the tensor product of the cohomology rings).

Does this actually work? Or is there some other reason why one should expect it to commute with products?

I wouldn't be overly surprised if there was a combinatorial way to take a graded-commutative ring $R$ and build a simplicial complex out of it, but I don't know. (Take $dim_{\mathbb{K}} R_n$ cells of dimension n, glue them in some way respecting the product in this ring?)

(Over more general coefficients maybe there is some fancy derived version?)

Motivation: I was asking because I saw somebody explain on stackexchange yesterday how one can think of the formula $\wedge^k(V \oplus W) = \oplus_{i + j = k} \wedge^i (V)\otimes \wedge^k (W)$ as a consequence of the functor $\wedge$ being left adjoint to the functor taking a graded commutative ring to its degree 1 piece. I liked that.

  • $\begingroup$ Functors from where to where? Be more specific. Help people help you. $\endgroup$ – Thomas Andrews Jan 26 '16 at 17:14

No, at least not in any obvious sense; it's not hard to give counterexamples showing that it does not preserve most limits, e.g. equalizers or pullbacks, or for that matter infinite products. (Depending on where you take these limits you run into further problems, e.g. the homotopy category of topological spaces itself has very few limits.)

Abstractly, taking cohomology is a homotopical version of taking a set $X$ to the set of functions $X \to k$ where $k$ is a commutative ring (which it reduces to in the special case of discrete spaces). This means we can hope that it has good behavior with respect to colimits (and, homotopically speaking, this is true), but no reason to expect that it has good behavior with respect to limits.

It's already an interesting question why the functor $X \mapsto k^X$ sends (finite) products to tensor products (Edit: for finite sets; for infinite sets the better-behaved functor is "homology," or taking the free $k$-module). One way to think about this is to think about a finite set $X$ as the coproduct of $|X|$ copies of $1$, and then use the fact that finite products of $k$-modules are the same as finite coproducts, together with the fact that both finite products of sets and tensor products of $k$-modules distribute over coproducts.

  • 1
    $\begingroup$ Minor corrrection: $X\mapsto k^X$ only preserves finite products for finite sets, and in fact preserves all finite limits on finite sets. This fact is indeed interesting and seems very special to commutative rings. In particular, it is closely related to the fact that commutative rings have a robust notion of "idempotents" (or dually, that affine schemes have a robust notion of "connectedness"); for some related discussion see this question on MO and my answer to it. $\endgroup$ – Eric Wofsey Jan 27 '16 at 6:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.