# Opposite of a category of modules is not a category of modules

So the answer in this question: https://mathoverflow.net/questions/29442/what-is-the-opposite-category-of-the-category-of-modules-or-hopf-algebra-repres

refers to the book Abelian Categories by Freyd. I think the idea of the proof is to show using exercise B on page 116 in that book (ie. if a category $\bf{A}$ and its opposite are Grothendieck then for all $A \in \bf{A}$ the natural map $\bigoplus A \rightarrow \prod A$ is an isomorphism and $A=0$) that if $\bf{R-Mod}^{op}$ is also a module category it is Grothendieck and so $\bf{R-Mod} = 0$.

So I wanna know how to do that exercise. I've been looking at other sources, for example here: http://therisingsea.org/notes/AbelianCategories.pdf in Proposition 51 we have that $\bigoplus A_i \rightarrow \prod A_i$ is a monomorphism for $\bf{A}$ Grothendieck. However the proof of that Proposition uses results from Theory of Categories by Mitchell and basically the problem is that the results being used are getting more and more advanced and I would like to keep this as simple as possible. Also I'm not sure that if we have a monomorphism from $\bigoplus A_i \rightarrow \prod A_i$ and another one from $\prod A_i \rightarrow \bigoplus A_i$ then this means that they are isomorphic as required by the exercise.

Is there an easier way to do the exercise or even show that $\bf{R-Mod}^{op}$ can't be a module category?

Thinking about Grothendieck categories is making things way more complicated than necessary to prove just that $\bf{R-Mod}^{op}$ is not a module category. You can prove this by simply thinking about the natural map $f:\bigoplus A_i\to \prod A_i$ in very concrete terms for modules. To be clear, this map $f$ is defined as the unique map such that $p_kfe_j=Id_{A_j}$ if $j=k$ and $0$ otherwise, where the maps $e_j:A_j\to\bigoplus A_i$ and $p_k:\prod A_i\to A_k$ are the inclusion and projection maps of the coproduct and product.
For a module category, $\prod A_i$ is just the cartesian product, $\bigoplus A_i$ is the submodule of $\prod A_i$ consisting of elements that are nonzero on only finitely many coordinates, and this map $f:\bigoplus A_i\to \prod A_i$ is the inclusion map. In particular, $f$ is always a monomorphism.
Now consider the opposite of a module category. Note that our module homomorphism $f:\bigoplus A_i\to \prod A_i$ is still the natural morphism from a coproduct to a product when considered as a morphism of $\bf{R-Mod}^{op}$: the direction of $f$ as a morphism is reversed, but coproducts and products are also swapped, and the condition on $p_kfe_j$ ends up being self-dual. So, if $\bf{R-Mod}^{op}$ were equivalent to a module category, then $f$ would be a monomorphism as a morphism of $\bf{R-Mod}^{op}$. That just means $f$ would be an epimorphism in $\bf{R-Mod}$.
So, if $\bf{R-Mod}^{op}$ is equivalent to a module category, then the map $f:\bigoplus A_i\to \prod A_i$ is both a monomorphism and an epimorphism and hence an isomorphism for any collection of objects $A_i$. But this is false as long as infinitely many of the $A_i$ are nonzero, since there are then elements of $\prod A_i$ which have infinitely many nonzero coordinates. So assuming $R$ is not the zero ring (so there exists a nonzero module over it), $\bf{R-Mod}^{op}$ cannot be equivalent to a module category.
• Ok cool I think I get it; so in a category of modules we can show that the induced map from a coproduct to a product is mono. Then if the opposite category was a category of modules too this would imply the same thing but due to the nature of opposite categories the induced map from coproducts to products in there is exactly the opposite of the induced map in the normal category. Hence $f$ and $f^{op}$ are both mono hence $f$ is mono and epi hence an iso since we're in an abelian category (I'm gonna try and prove this). – Fromage Nov 26 '17 at 10:34