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The pullback of an epimorphism is an epimorphism

Statement

In a pullback square in an abelian category

if $e$ is epic, then $α$ is epic.

Facts & Assumptions

Given: The displayed pullback square in an abelian category, with $e$ epic.

[L1]

The pullback is the kernel of the difference map on $A \oplus B$ (A pullback is the kernel of the difference of the two legs, and dually for pushouts).

[L2]

Abelian categories are preadditive with zero morphisms (Abelian category, A preadditive category with a zero object has zero morphisms in the published sense).

[L3]

In an abelian category, a morphism is epic exactly when its cokernel is zero (In an abelian category, monic means zero kernel and epic means zero cokernel).

[L4]

In an abelian category every epic morphism is the cokernel of its kernel (Every monomorphism is the kernel of its cokernel, and dually every epimorphism is the cokernel of its kernel).

Proof

technique · direct

1.1L1L2

By [L1], there is a monomorphism $k : P \to A \oplus B$ with $α = p_{A} k$ , $β = p_{B} k$ , and $k$ the kernel of $d : = f p_{A} - e p_{B}$ . Since $d i_{B} = - e$ , the epicity of $e$ implies that $d$ is epic as well.

2.1L1L2L3L4step 1.1∎

Let $q : A \to Q$ be a cokernel of $α$ . Because $d$ is epic by step 1.1, [L4] says that $d$ is a cokernel of its kernel $k$ . Since $q α = q p_{A} k = 0$ , the map $q p_{A} : A \oplus B \to Q$ kills $k$ , so it factors through $d$ as $\overline{q} d = q p_{A}$ . Composing with $i_{B}$ gives $\overline{q} (- e) = 0$ , and since $e$ is epic, $\overline{q} = 0$ . Thus $q p_{A} = 0$ , and composing with $i_{A}$ yields $q = 0$ . So $α$ has zero cokernel and is epic by [L3].

Depends on

Dependency tree · 2 levels

Sources

The Stacks Project, Section 12.5, Lemma 12.5.13