TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-29

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A cartesian square over an epimorphism is also cocartesian

Statement

In an abelian category, if

is a cartesian square and $e$ is epic, then the square is also cocartesian.

Facts & Assumptions

Given: The displayed cartesian square, with $e$ epic.

[L1]

The square is a pullback exactly when $0 \to P \overset{(\binom{α}{β})}{\to} X \oplus Y \overset{(f, - e)}{\to} Z$ is exact (A square is cartesian exactly when a short sequence is exact).

[L2]

In an abelian category, an exact sequence $A \to B \to C \to 0$ is exactly one in which the last map is a cokernel of the first (Degenerate exactness criteria).

[L3]

Pushouts are defined by the usual colimit universal property (Pullbacks and pushouts as limits and colimits of cospans and spans).

[L4]

The biproduct $X \oplus Y$ carries the usual injections from $X$ and $Y$ (Biproduct).

Proof

technique · direct

1.1L1givenL4algebra

By [L1], the cartesian hypothesis makes $0 \to P \overset{(\binom{α}{β})}{\to} X \oplus Y \overset{v}{\to} Z$ exact, where $v : = (f, - e)$ . If $r, s : Z \to T$ satisfy $r v = s v$ , then $r e = - r v i_{Y} = - s v i_{Y} = s e$ , so the epicity of $e$ gives $r = s$ . Thus $v$ is epic, and the sequence $P \overset{(\binom{α}{β})}{\to} X \oplus Y \overset{v}{\to} Z \to 0$ is exact.

2.1L2L3L4step 1.1constructalgebra

By [L2], the map $v$ is a cokernel of $(\binom{α}{β})$ . Now let $x : X \to T$ and $y : Y \to T$ satisfy $x α = y β$ . By the coproduct side of [L4], there is a unique morphism $w : X \oplus Y \to T$ with $w i_{X} = x$ and $w i_{Y} = - y$ . Then $w (\binom{α}{β}) = x α - y β = 0,$ so the cokernel property gives a unique $u : Z \to T$ with $u v = w$ . Composing with $i_{X}$ and $i_{Y}$ yields $u f = u v i_{X} = w i_{X} = x, u e = - u v i_{Y} = - w i_{Y} = y .$ This is exactly the pushout universal property from [L3].

3.1step 2.1∎

Therefore every pullback square over an epimorphism is also a pushout square.

Depends on

Used by

Short five lemma by pullback without members Theorem
The pullback and pushout theorems Theorem

Dependency tree · two levels

13 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)
David Mehrle, Category Theory, Part III, Corollary 7.18 (standard reference, not scraped)