Edit: According to the comment of Todd Trimble, I revise the question.
What are some examples of functors $F$ on the category of Abelian groups or category of rings which satisfy $$F(G\times H)\cong F(G)\otimes_{\mathbb{Z}} F(H)$$
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2$\begingroup$ I feel this question could stand more structure. First, let's use an isomorphism $\cong$ in place of an equality; otherwise it's hard to make sense of the question. Then: should the isomorphism be natural? Should we be demanding more than mere naturality: should the isomorphism get along (be compatible) with associativity or symmetry isomorphisms? Do you care only about existence of an isomorphism (that satisfies to-be-specified properties), or should the choice of isomorphism be part of the structure considered? Etc. $\endgroup$– Todd TrimbleCommented Jun 3, 2019 at 0:22
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$\begingroup$ @ToddTrimble Thank you for your very helpful comment. Yes you are right. We should consider isomorphisms rather than equality. I revise the question. But I was not thinking to some naturalitiy properties. But the naturality you pointed out make the question more meaningfull. $\endgroup$– Ali TaghaviCommented Jun 3, 2019 at 0:28
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$\begingroup$ @ToddTrimble to be honest the Leray Hirsch theorem, the Kunneth formula and the functor $X\mapsto C(X)$ on topological space were an indirect motivation for this question. $\endgroup$– Ali TaghaviCommented Jun 3, 2019 at 0:32
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$\begingroup$ @ToddTrimble could you please suggest some naturality properties which I can add to my question? $\endgroup$– Ali TaghaviCommented Jun 3, 2019 at 0:33
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1$\begingroup$ Naturality in the arguments $G$ and $H$ follows the well-known definition (compatibility between the isomorphism $\phi_{G, H}: F(G \times H) \to F(G) \otimes F(H)$ and morphisms $G \to G', H \to H'$ as expressed in the form of commutative squares). For further "coherence conditions", see ncatlab.org/nlab/show/monoidal+functor and ncatlab.org/nlab/show/symmetric+monoidal+functor, which include consideration of not just the monoidal products but also the monoidal units $1, \mathbb{Z}$. By the way, don't you mean to replace "on" with "to" in the question? $\endgroup$– Todd TrimbleCommented Jun 3, 2019 at 1:14
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1 Answer
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The first thing that comes to mind is the symmetric algebra functor
$$A \mapsto Sym_{\mathbb Z}(A) = \oplus_{n \in \mathbb N} Sym^n_{\mathbb Z}(A) = \mathbb Z \oplus A \oplus (A\otimes A/ \Sigma_2) \oplus \dots $$
This may be regarded as a functor from abelian groups to commutative rings (it is left adjoint to the forgetful functor); or by composing with the forgetful functor it may be regarded as an endofunctor of abelian groups.
One way to think about this is to observe that $\oplus$ is the binary coproduct of abelian groups and $\otimes$ is the binary coproduct of commutative rings, so any left adjoint functor $Ab \to CRing$ must take $\oplus$ to $\otimes$.
Of course, many variants immediately present themselves. One may precompose with any additive functor and postcompose with any monoidal functor. For instance, $A \mapsto Sym_{\mathbb Q}(A) = \mathbb Q \oplus (\mathbb Q \otimes A) \oplus \dots$ also has the desired property.
- For another example, take the group ring functor $A \mapsto \mathbb Z[A]$, which again can be regarded as either a functor $Ab \to CRing$ or $Ab \to Ab$. Here $\mathbb Z[A]$ is the free abelian group on the underlying set of $A$; if $e_a,e_b$ are the generators corresponding to $a,b \in A$, then we define $e_a e_b = e_{a+b}$. This has the desired for the same reason as $Sym_{\mathbb Z}$: regarded as a functor $Ab \to CRing$, it is left adjoint to the "group of units" functor $R \mapsto R^\times$.
Again there are variants like $A \mapsto \mathbb Q[A]$.
- More generally, if $G$ is any affine abelian group scheme over $Spec \mathbb Z$, then $Hom(\Gamma(G),-): CRing \to Ab$ is a right adjoint functor, whose left adjoint has the desired property.
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3$\begingroup$ It might be worth mentioning that every right adjoint functor $\mathrm{CRing}→\mathrm{Ab}$ is of the form you describe (this is just because a right adjoint functor $\mathrm{CRing}→\mathrm{Set}$ is necessarily corepresentable plus some Yoneda) $\endgroup$ Commented Jun 3, 2019 at 12:05
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3$\begingroup$ Also, any functor $F: Ab \rightarrow Ab$ with the expected property is automatically a functor from $Ab$ to $CRing$, indeed due to the fact that the product in $Ab$ is also the coproduct, every object in $Ab$ has a unique monoid structure for the cartesian monoidal structure, and every morphism is a morphism of monoids, so for any functor $F$ from Ab to Ab which is monoidal from the cartesian to the tensor monoidal structure, $F(X)$ has a canonical ring structure for all $X$, a functoriality in $X$ is compatible to the ring structure. A similar argument show they are in fact Hopf algebras. $\endgroup$ Commented Jun 4, 2019 at 8:27