TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-28

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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

Used by

The pushout of a monomorphism is a monomorphism Corollary
FALSE: member equivalence is transitive in any pointed category with pullbacks False statement
FALSE: pullbacks preserve epimorphisms in every category with pullbacks False statement
Bounded above complexes admit projective replacements Lemma
The cost of the member calculus Remark
A chain map is a quasi-isomorphism exactly when its cone is acyclic Theorem
Epimorphisms in an abelian category are universal Theorem
Epimorphy is detected by members Theorem
Exactness is detected by members Theorem
In a pullback square, the induced map on the kernels of the two parallel arrows is an isomorphism Theorem
Member equivalence is transitive Theorem
Projective object characterisations Theorem
Schanuel's lemma in an abelian category Theorem
Short five lemma by pullback without members Theorem
Simple classes freely generate the Grothendieck group of a length category Theorem
Snake lemma in an abelian category Theorem
Snake lemma under the weaker Stacks hypotheses Theorem
The connecting morphism exists and is unique Theorem
The pullback and pushout theorems Theorem
The Yoneda product is associative and unital Theorem

Dependency tree · two levels

20 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

The Stacks Project, Section 12.5, Lemma 12.5.13 (standard reference, not scraped)