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In his Tohoku paper, section 1.5, Grothendieck states the following axioms that an abelian category might satisfy:

AB4)Infinite sums exist, and the direct sum of monomorphisms is a monomorphism.

AB5)Infinite sums exist, and the and if $A_i$ (indices in some possibly infinite set $I$) is a filtrated family of subsets of some object A in the category, and B another subset of A, then $(\sum A_i)\cap B = \sum (A_i\cap B)$

(A subset is what I am translating sous-truc as meaning... I am not sure if this is the correct English notation for this notion.)

Grothendieck states that AB5 is stronger than AB4, without proof. I cannot prove it myself; can someone enlighten me as to why this is true?

tceps
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2 Answers2

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So let us suppose we have a family of monomorphisms $A_i \to B_i$, where $i$ varies over an indexing set $I$. Let $A = \bigoplus_i A_i$ and $B = \bigoplus_i B_i$; we want to show that $A \to B$ is also a monomorphism. Let $K$ be the kernel, so that we have an exact sequence $$0 \longrightarrow K \longrightarrow A \longrightarrow B$$ Let $\mathcal{J}$ be the system of all finite subsets of $I$: this is a filtered poset, and if we define $A_j = \bigoplus_{i \in j} A_i $ and $B_j = \bigoplus_{i \in j} B_i$ for each finite subset $j$ of $I$, we get a filtered system. In any abelian category, given a diagram $$\begin{alignedat}{3} 0 \longrightarrow \mathord{} & K_j & \mathord{} \longrightarrow \mathord{} & A_j & \mathord{} \longrightarrow \mathord{} & B_j \\ & \downarrow && \downarrow && \downarrow \\ 0 \longrightarrow \mathord{} & K & \mathord{} \longrightarrow \mathord{} & A & \mathord{} \longrightarrow \mathord{} & B \\ \end{alignedat}$$ if the rightmost vertical arrow is a monomorphism and both two rows are exact, then the left square is a pullback square; since $A_j \to A$ is also a monomorphism, this amounts to saying that $K_j = A_j \cap K$. Since $j$ is a finite set, $A_j \to B_j$ is automatically a monomorphism, so $K_j = 0$. Taking the filtered colimit over $\mathcal{J}$, we obtain $$\sum_j K_j = \left( \sum_j A_j \right) \cap K = A \cap K = K$$ so $K = 0$. Thus, $A \to B$ is a monomorphism.

In fact, AB5 is strictly stronger than AB4. Take $\mathcal{A} = \textbf{Ab}^\textrm{op}$. The opposite of an abelian category is an abelian category, and it is not hard to show that $\mathcal{A}$ satisfies AB3 and AB4; but $\mathcal{A}$ does not satisfy AB5. This is the same example used by Grothendieck in his paper.

Community
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Zhen Lin
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$\def\colim{\operatorname{colim}}$For me, AB5 means "all coproducts exist and filtered colimits are exact." A proof of the equivalence with OP's definition is to be found in N. Popescu, Abelian categories with applications to rings and modules, Sect. 2.8, Theorem 8.6.

If our category is AB3 (i.e., all coproducts exist; thus it is cocomplete), then we can write [ref] $$ \label{1}\tag{1} \coprod_{i\in I}X_i=\underset{F\in T}{\colim}\coprod_{j\in F}X_j, $$ where $T$ is the set of finite subsets of $I$, which is a directed set. On the one hand, since finite coproducts are also products in a preadditive category [Tag 0101], they are exact (i.e., they commute with all finite limits and colimits). Thus, if our category is AB5, then the expression \eqref{1} is exact in $\{X_i\}_{i\in I}$, i.e., $A_i\xrightarrow{f_i} B_i\xrightarrow{g_i} C_i$ exact for all $i$ implies $\coprod A_i\to\coprod B_i\to\coprod C_i$ exact.

Elías Guisado Villalgordo
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