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The question is regarding the following axioms

AB$4$. If $(f_i\colon A_i\to B_i)_{i \in I}$ is a family of monomorphisms in an abelian category, then the induced $\bigoplus_{i \in I} f_i\colon \bigoplus_{i \in I} A_i\to \bigoplus_{i \in I} B_i$ is also a monomorphism.

AB$5$. In an abelian category, for a directed family $(A_i)_{i \in I}$ of subobjects of $A$ and a subobject $B$ of $A$, we have $(\sum_{i \in I} A_i)\cap B = \sum_{i \in I} (A_i\cap B)$.

I'm trying to follow the following proof from N.Popescu's book Abelian categories with applications to rings and modules:

Corollary 8.9. An AB5-category $\mathscr{C}$ is AB4.

Proof. Let $\left\{f_i: X_i^{\prime} \rightarrow X_i\right\}_{i \in I}$ be a set of monomorphisms of $\mathscr{C}$ and $$ f: \coprod_i X_i{ }^{\prime} \rightarrow \coprod_i X_i $$ the sum morphism. Let $T$ be the set of finite subsets of $I$; for any $F \in T$ we denote by $X_F{ }^{\prime}$ the canonical image of $\coprod_{j \in F} X_j{ }^{\prime}$ in $\coprod_i X_i{ }^{\prime}$. Then $\left\{X_F{ }^{\prime}\right\}_{F \in T}$ is a direct set of subobjects and $\sum_F X_F^{\prime}=\coprod_i X_i^{\prime}$. Let $K=\operatorname{ker} f$. Then $\left.K=\left(\sum_F X_F{ }^{\prime}\right) \cap K=\sum_F{ }^{\prime} \cap K\right)$. If $K \neq 0$, then $X_F{ }^{\prime} \cap K \neq 0$ for some $F$. Now, let $u_F: \coprod_{j \in F} X_j^{\prime} \rightarrow \coprod_i X_i^{\prime}$ be the canonical monomorphism. Obviously, $u_F^{-1}\left(X_F^{\prime} \cap K\right) \neq 0$. Also, let $v_F: \coprod_{j \in F} X_j \rightarrow \coprod_i X_i$ and $$ f_F: \coprod_{j \in F} X_j{ }^{\prime} \rightarrow \coprod_{j \in F} X_j $$ be canonically constructed. Then $f u_F=v_F f_F$ and $u_F^{-1}\left(X_F^{\prime} \cap K\right)=$ $\operatorname{ker}\left(f u_F\right)=\operatorname{ker}\left(v_F f_F\right)=0, v_F$ and $f_F$ being monomorphisms. This is a contradiction, hence $f$ is a monomorphism. $\square$

However, I get stuck at the end, not seeing why $v_F$ is a monomorphism. The way I understand it, if $(l_F)_i\colon X_i \to \coprod_{i \in F} X_i$ and $l_i\colon X_i\to \coprod_{i \in I} X_i$ are canonical injections, then $v_F$ is the unique morphism such that $v_F\circ (l_F)_i = l_i$ for all $i \in F$. Why would it be monic?

I understand, however, why $f_F$ is monic. It is because direct sums of finite families of monomorphisms coincides their products, hence they are monomorphisms as limits commute with limits. But this logic doesn't apply to $v_F$ since it has possibly infinite set $I$ and the colimit of the constant discrete category is not the same as the limit of the constant connected category, so we can't use the trick of reducing the universal property morphism to the morphism induced by colimits, as in connected case.

Elías Guisado Villalgordo
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Jxt921
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2 Answers2

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Special thanks to Captain Lama in the comment section!

Let $(l_{I\setminus F})_i\colon X_i\to \coprod_{i \in I\setminus F} X_i$. Form a coproduct $(\coprod_{i \in F} X_i)\sqcup (\coprod_{i \in F\setminus I} X_i)$ together with canonical injections $q_1\colon \coprod_{i \in F} X_i \to (\coprod_{i \in F} X_i)\sqcup (\coprod_{i \in F\setminus I} X_i)$ and $q_2 \colon \coprod_{i \in F} X_i \to (\coprod_{i \in F} X_i)\sqcup (\coprod_{i \in F\setminus I} X_i)$. Then, by the associativity theorem for coproducts, $(\coprod_{i \in F} X_i)\sqcup (\coprod_{i \in F\setminus I} X_i)$ is a coproduct of the family $(X_i)_{i \in I}$ together with the canonical injections $q_i\colon X_i \to (\coprod_{i \in F} X_i)\sqcup (\coprod_{i \in F\setminus I} X_i)$ where $q_i = q_1\circ (l_F)_i$ if $i \in F$ and $q_i = q_2\circ (l_{F\setminus I})_i$ otherwise. By uniqueness of coproduct, there is a unique isomorphism $u$ such that $l_i = u\circ q_i = u\circ q_1\circ (l_F)_i$ for all $i \in F$ and $l_i = u\circ q_2 \circ (l_{F\setminus I})_i$ for all $i \in F\setminus I$. In particular, $v_F\circ (l_F)_i = l_i = u\circ q_1 \circ (l_F)_i$ for all $i \in F$, so $v_F = u\circ q_1$ is a monomorphism as coproduct injections are monomorphisms in abelian categories.

Jxt921
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The maps $u_F$, $v_F$ are monomorphisms because they have a retraction. Denote $v_j:X_j\to\coprod_{i\in I}X_i$ to the inclusion, $j\in I$. Define $p_F:\coprod_iX_i\to\coprod_{j\in F}X_j$ as the unique map such that $p_F\circ v_i$ is the inclusion $X_i\to\coprod_{j\in F}X_j$ if $i\in F$ and such that $p_F\circ v_i=0$ if $i\not\in F$. You can check that $p_F\circ v_F$ is the identity of $\coprod_{j\in F}X_j$.

Elías Guisado Villalgordo
  • 4,758