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?

  • 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


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.


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