1
$\begingroup$

Let $\mathbb{K}$ be a suitable field and $(\mathcal{Ch}_{\mathbb{K}},\otimes)$ the "standard" symmetric monoidal category of chain complexes over that field, with the symmetric tensor product, induced by the commutativity constraint $a\otimes b = (-1)^{|a||b|}b\otimes a$.

Considering graded vector spaces as chain complexes with trivial differential, the homology is an endofunctor $H: \mathcal{Ch}_{\mathbb{K}} \to \mathcal{Ch}_{\mathbb{K}}$ that maps any chain complex to its homology.

A key feature of $H$ is, that it is a strict symmetric monoidal functor. Now my question is: In what sense (if any) is $H$ a symmetric comonoidal functor?

$\endgroup$
2
  • 1
    $\begingroup$ Why do you think that it has a comonoidal structure at all? $\endgroup$
    – HeinrichD
    Mar 30, 2017 at 16:33
  • $\begingroup$ "In what sense (IF ANY) is $H$ symmetric comonoidal" implies, that I don't think that. $\endgroup$
    – Bobby
    Mar 30, 2017 at 16:58

1 Answer 1

0
$\begingroup$

The monoidal functor structure is given by the Künneth isomorphism $H(C\otimes D)\cong H(C)\otimes H(D)$. This is an isomorphism, so you can feel free to read it as either a monoidal or a comonoidal functor.

$\endgroup$

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.