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(Inspired by an answer given here: does monomorphism of sheaves of abelian groups imply it's injective on the level of open sets? )

I wonder if, in a general (elementary) topos $\mathcal{T}$, if you have a ring object $R$, then the underlying set functor $R{-}\mathbf{Mod} \to \mathcal{T}$ has a left adjoint.

Out of the general constructions I can think of:

  • The construction as a submodule of $R^S$ requires decidability of equality on $S$, in order to get the coprojections $S \to R^S$.
  • The construction in terms of formal linear combinations $\sum_{i=1}^n r_i e_{s_i}$ requires a natural numbers object.
  • The construction a la Freyd's adjoint theorem (take the collection of all possible module structures on a subset of a "large enough" base set, with maps $S$ to the subset; take the product of all these modules; then take the submodule generated by $S$) typically requires some cardinality arguments to ensure the base set is indeed "large enough". These arguments might not be valid in a general topos, as they might require excluded middle and/or axiom of choice. On top of that, the cardinality argument that immediately comes to my mind also involves a natural numbers object.

So, to try to come up with a counterexample, I looked at the case of the topos of finite sets (to avoid the topos having a natural numbers object). However, in this topos, free modules do exist, since this topos has decidable equality on every object. To avoid this, I considered passing to the topos of "pointwise finite" presheaves on a finite category. However, in this topos, you can still form free modules "pointwise".

(Of course, if a general topos does have free modules over ring objects, then it would have to be something special to this case. For example, certainly free abelian groups don't exist in the topos of finite sets.)

Daniel Schepler
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  • An idea could be to try to produce $R^{(A)}$ as the submodule of $R^{R^A}$ generated by the image of $A$ (evaluation map). But I don't know how to prove this is the free one. In the classical case, we still use decidability to show that. At least it is free relatively to $R$ since given $f:A→R$, the extension $R^{(A)}→R$ is given by the composite $R^{(A)}↪R^{R^A}→R$ where the second map is evaluation at $f$. Is it true in non-Boolean topoi but with a natural number object? – Dabouliplop Oct 11 '22 at 07:43
  • So it is also free relatively to submodules of modules of the form $R^X$... If there is a cogenerator $G$, we can do the same trick with $G$ instead of $R$ and get a free module over $A$. This is close to what we would do with the argument with an adjoint functor theorem, probably. – Dabouliplop Oct 11 '22 at 07:48
  • I did not intend to get your hopes up with my answer, but I’ll be interested to see whether anyone can come up with a topos without free modules. It would be even more exciting if someone came up with a free module construction which didn’t require LEM or an NNO. – Mark Saving Oct 11 '22 at 19:21

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