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Let $K$ be a field. For a group $G$ we write $K[G]$ for the group ring of $G$.

Given group homomorphisms $F \to G, H \to G$ is the canonical map $$ K[F \times_G H] \to K[F] \times_{K[G]} K[H] $$ an isomorphism, where the fiber product $K[F] \times_{K[G]} K[H]$ is taken in the category of hopf algebras over $K.$

Is it already true that the free K-vector space functor from the category of sets to the category of cocommutative coalgebras over K preserves fiber products?

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    $\begingroup$ This seems false. If G is trivial, on the left side you have the tensor product of group rings and in the right the direct product $\endgroup$
    – Benjamin Steinberg
    Commented Jun 3, 2021 at 19:45
  • $\begingroup$ @BenjaminSteinberg is the product in Hopf algebras really just the direct product? I can't think of a natural comultiplication on that. (Analogously, in the commutative case, the coproduct in affine group schemes is not given by disjoint union, as the latter does not have a natural group structure.) $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 3, 2021 at 20:02
  • $\begingroup$ @R.vanDobbendeBruyn sorry I hadn't caught that it was taken in hopf algebras. That's out of my pay grade $\endgroup$
    – Benjamin Steinberg
    Commented Jun 3, 2021 at 20:06
  • $\begingroup$ It seems the tensor product is the product so I withdraw my objection. $\endgroup$
    – Benjamin Steinberg
    Commented Jun 3, 2021 at 20:15
  • $\begingroup$ One possible way to prove something like this is if the free functor $F \colon \mathbf{Gp} \to \mathbf{Hopf}_K$ has a left adjoint. It does have a right adjoint given by grouplike elements, but I have no idea if it has a left adjoint (this would imply much more limits commute, which is maybe too optimistic). $\endgroup$
    – R. van Dobben de Bruyn
    Commented Jun 3, 2021 at 21:10

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