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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?

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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 $$0 \longrightarrow K \longrightarrow A \longrightarrow B \longrightarrow 0$$ 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.

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