# Infinite Sum Axioms in Tohoku

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?

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.