Suppose you have two $k$ algebras $A, B$ (say also finitely generated if this helps) and a functor $F: A-mod \to B-mod $ such that $| F(M) |= |M|$. Here $|U|$ denotes the underlying $k$ vector space.

Can you find sufficient conditions so that $F$ is the restriction functor relative to $f:B \to A$?

I suspect that a "nice" condition to start from is monoidality. Indeed, in the case in which the rings are the group rings of finite groups, by Tannaka reconstruction you could reconstruct the morphism.

My motivation: I had to find an $\mathbb{R}$-algebra injective morphism from $\mathbb{C} \to \mathbb{R}[u]/(u^2+1) ^m$. It's really easy with Jordan form theory to give a complex structure on a module for the right hand ring, but constructing the actual morphism it's not easy at all, and some non trivial ad-hoc argument is needed to make " reconstruction ".

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    $\begingroup$ Since restriction is exact and preserves all direct sums you should assume that (if it isn't already implied). Then by Eilenberg Watts your functor is tensor product with a bimodule. If your bimodule had to be A itself then you would get the functor comes from restriction. $\endgroup$ May 16, 2021 at 12:57
  • $\begingroup$ That's a very nice observation. Note that if the isomorphism with the fiber functor is functorial then your hypothesis are implied, since exactness and direct sums can be checked at the level of $k$ vector spaces. We still can't rule out that the functoe is restriction and then tensoring with a dimension 1 module! $\endgroup$ May 16, 2021 at 14:56
  • $\begingroup$ I think the one dimensional module issue is not a problem since you can change your homomorphism $\endgroup$ May 16, 2021 at 15:01
  • $\begingroup$ I think if your isomorphism with the fiber functor is natural then the bimodule I have above has to be isomorphic to A and hence if you are doing right modules your functor is given by restriction along a homomorphism $\endgroup$ May 16, 2021 at 15:35

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Let's assume that we are using right modules and that the isomorphism of underlying vector space functors is a natural isomorphism because we will just work up to isomorphism of functors. Note that $F$ is exact. To come from restriction $F$ should preserve all limits and colimits. Maybe this follows by functorially preserving underlying spaces but I'm not sure. If it doesn't you probably want to require that. It follows from Eilenberg-Watts that there is an $B$-$A$-bimodule $M$ such that $F\cong \hom_A(M,-)$. Moreover, since $A$ represents the underling vector space functor, we must have that $M\cong A$ as a right $A$-module. Since $End_A(A)\cong A$, it follows the left $B$-module structure comes from a homomorphism $F\colon B\to A$. So basically $F\cong \hom_A(A,-)$ via this bimodule structure. But then evaluating at $1$ gives an isomorphism of $F$ with the functor taking $N$ to $N$ with $B$-action $mb = mf(b)$. So $F$ is isomorphic to restriction along $f$.

  • $\begingroup$ Complete and well explained, thanks. For the sake of curiosity, It turned out that the hard passage in my concrete question was functoriality. "Taking the Jordan form" is definitely not functorial, and if one tries to reconstruct via generalized eigenspaces you don't know in advance how much the blocks are big. This causes the dimensions not to match, so that you lose the hypothesis on the fiber functor. Just to know, if you restrict to finitely generated modules, does Eilenberg Watts still works? $\endgroup$ May 16, 2021 at 22:45
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    $\begingroup$ That's outside my pay grade. It might be you can extend an exact functor defined in finitely generated modules to the whole category via direct limits. I know for Morita equivalence working with finitely generated but I don't know about Eilenberg Watts $\endgroup$ May 16, 2021 at 23:23

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