I have asked the following question on MSE a few days ago, but without any success.

I am interested in proving the following statement:

Let $\mathcal{A}$ be a tensor category. Then the category of modules over a commutative monoid $A$ in $\text{Ind}\mathcal{A}$ is abelian and symmetric monoidal, and every such module is the quotient of a flat module.

Here by 'tensor category' I mean a $k$-linear abelian and rigid symmetric monoidal category whose tensor product is $k$-bilinear and exact in both variables. An $A$-module $M$ is flat if the funtor $-\otimes_A M$ is exact.

I am able to show that this category is symmetric monoidal, but I am having some trouble showing that it is abelian and that every module is the quotient of a flat one. What I found out so far:

  • The zero object of $\mathcal{A}$ also is a zero object in $A\textbf{Mod}$ together with the trivial action
  • As biproduct of $(M,\rho_M)$ and $(N,\rho_N)$ one can just take the biproduct $M\oplus N$ in $\text{Ind}\mathcal{A}$ with the action $\rho_M \oplus \rho_N$
  • I think the following diagrams make the kernel and cokernel of a morphism of $A$-modules (taken in $\text{Ind}(\mathcal{A})$) into $A$-modules, hence showing that $A\textbf{Mod}$ has all kernels and cokernels (I think $A\otimes \text{coker} f = \text{coker}(\text{id}\otimes f)$):

enter image description here

  • Since the forgetful functor from $A\textbf{Mod}$ to $\text{Ind}(\mathcal{A})$ is a right adjoint of the free functor $X\rightsquigarrow A\otimes X$, it preserves limits and in particular monomorphisms. Therefore a monomorphism in $A\textbf{Mod}$ also is a monomorphism in $\text{Ind}(\mathcal{A})$ and since this is abelian, it is a kernel there. Then since the underlying objects of kernels in $A\textbf{Mod}$ are kernels in $\text{Ind}(\mathcal{A})$, the morphism also is a kernel there.

I have no idea though how to show that every epimorphism is a cokernel in $A\textbf{Mod}$. Also, I do not know how to go about that property with the flatness. I think this is a quite common statement, and yet I could not find a proof anywhere. Any help would be appreciated!

  • $\begingroup$ Can you mimic the usual proof that free modules are flat? Maybe you need to restrict to free modules on projective objects? Do you not need some kind of “enough projectives” assumption in your definition of tensor category? $\endgroup$ Jun 21, 2021 at 14:56
  • $\begingroup$ @NoahSnyder Yes, all free modules are flat here. That follows directly from the fact that the tensor product in $\text{Ind}(\mathcal{A})$ is exact. That in turn is a consequence of $\mathcal{A}$ being rigid. $\endgroup$
    – S.Farr
    Jun 21, 2021 at 15:10
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    $\begingroup$ Sorry, why can't we just use $\rho_M$ as the map $A \otimes M \rightarrow M$ to realize $M$ as a quotient? $\endgroup$ Jun 21, 2021 at 15:53
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    $\begingroup$ Epi in Amod implies that the cokernel in Amod is 0, which then means that the cokernel in Ind(A) is 0, which then means it's Epi in Ind(A)? $\endgroup$ Jun 21, 2021 at 16:33
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    $\begingroup$ No problem, when you get a chance maybe write up a short summary of this as an answer and I’ll upvote it. $\endgroup$ Jun 21, 2021 at 16:58

1 Answer 1


This answer is a summary of the discussion in the comments.

To the question why $A\textbf{Mod}$ is abelian: A morphism in an abelian category is monic (resp. epic) if and only if its kernel (resp. cokernel) is zero. Now suppose we are given an epimorphism $f$ in $A\textbf{Mod}$; then its cokernel is zero, and since the underlying object in $\text{Ind}(\mathcal{A})$ of $\text{coker}(f)$ is the cokernel of the underlying morphism of $f$, we have $\text{coker}(f)= 0$ in $\text{Ind}(\mathcal{A})$ also. Thus, $f$ is epic in $\text{Ind}(\mathcal{A})$ and hence a cokernel. By the construction of cokernels in $A\textbf{Mod}$, this shows that $f$ is a cokernel in this category also. The same works for monomorphisms and kernels.

To the question why every module is the quotient of a flat module: Because the tensor product in $\text{Ind}(\mathcal{A})$ is exact, every free $A$-module is flat. By the unit axiom of modules, the composition

$$ M\simeq I\otimes M \xrightarrow{e\otimes \text{id}_M} A\otimes M \xrightarrow{\rho} M$$

must be the identity on $M$ ($e$ is the unit of the monoid $A$), so in particular $\rho$ is split epic, and hence $M$ is a quotient of $A\otimes M$ viewed as free $A$-module.


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