$\hom (M, \coprod_i N_i) \cong \bigoplus_i \hom (M, N_i)$ in abelian categories when $M$ is simple?

Question

For $M$ a simple $R$ module, and $N_i$ a family of $R$ modules, we have $$ \hom (M, \bigoplus_i N_i) \cong \bigoplus_i \hom (M, N_i) $$ as abelian groups. Since the direct sum is the coproduct in the category of $R$ modules, we could rewrite this as $$ \hom (M, \coprod_i N_i) \cong \bigoplus_i \hom (M, N_i). \tag{1} $$ This raises the question whether (1) is true for any abelian category, when $M$ is a simple object (defined I guess as an object such that any [regular?] monomorphism into it is an isomorphism). Does anyone know whether this is true?

Answer 1

It's not so much a question of simpleness as it is a question of finite generation. Indeed, in the case where $M$ is projective, the property in question is even equivalent to finite presentability.

Let $\mathcal{A}$ be an abelian category with filtered colimits. As was pointed out in the comments, infinite direct sums in $\mathcal{A}$ can be constructed using filtered colimits and finite direct sums. Thus, if $M$ is a finitely presentable object in $\mathcal{A}$, i.e. an object such that $\mathcal{A} (M, -) : \mathcal{A} \to \mathbf{Ab}$ preserves filtered colimits, then $\mathcal{A} (M, -) : \mathcal{A} \to \mathbf{Ab}$ also preserves infinite direct sums.

Now, suppose we have the following short exact sequence in $\mathcal{A}$: $$0 \longrightarrow K \longrightarrow P \longrightarrow M \longrightarrow 0$$ Then, for each object $N$ in $\mathcal{A}$, we have an induced exact sequence: $$0 \longrightarrow \mathcal{A} (M, N) \longrightarrow \mathcal{A} (P, N) \longrightarrow \mathcal{A} (K, N)$$ Suppose $P$ is finitely presentable. A little diagram chase then shows that the canonical comparison $$\bigoplus_{i \in I} \mathcal{A} (M, N_i) \to \mathcal{A} (M, \bigoplus_{i \in I} N_i)$$ is injective. In fact, it is bijective. For any morphism $M \to \bigoplus_{i \in I} N_i$ in $\mathcal{A}$, there is a finite subset $I' \subseteq I$ making the following diagram commute, $$\require{AMScd} \begin{CD} P @>>> \bigoplus_{i \in I'} N_i \\ @VVV @VVV \\ M @>>> \bigoplus_{i \in I} N_i \end{CD}$$ but $P \to M$ is an epimorphism and $\bigoplus_{i \in I'} N_i \to \bigoplus_{i \in I} N_i$ is a monomorphism, so there is a unique morphism $M \to \bigoplus_{i \in I'}$ making both triangles commute, i.e. $M \to \bigoplus_{i \in I} N_i$ factors through $\bigoplus_{i \in I'} N_i \to \bigoplus_{i \in I} N_i$.

If $\mathcal{A}$ is a locally finitely presentable abelian category, then any simple object in $\mathcal{A}$ is a quotient of a finitely presentable object. Indeed, by hypothesis, every object in $M$ is a filtered colimit of finitely presentable objects, so in particular there is an epimorphism $\bigoplus_{i \in I} N_i \to M$ where each $N_i$ is finitely presentable; but if $M$ is simple, that implies each $N_i \to M$ is either zero or an epimorphism, so there must be some $i \in I$ such that $N_i \to M$ is an epimorphism.

Thus, in view of Jeremy Rickard's counterexample, the conclusion must be that simple objects in abelian categories need not be finitely generated.

Answer 2

It's not true in the opposite category of the category of vector spaces over a field. If you take $M$ to be a one-dimensional space then it is the (false) statement that $$\left(\prod_i V_i\right)^\ast\cong\bigoplus_iV_i^\ast.$$