TheoremStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-09-29

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Short five lemma by pullback without members

Statement

For a morphism of short exact sequences in an abelian category, the three conclusions of the short five lemma hold without using members: monic outer maps force the middle map to be monic, epic outer maps force it to be epic, and isomorphic outer maps force it to be an isomorphism.

Facts & Assumptions

Given: A morphism of short exact sequences $0 \to A \overset{i}{\to} B \overset{p}{\to} C \to 0,$ $0 \to A^{'} \overset{i^{'}}{\to} B^{'} \overset{p^{'}}{\to} C^{'} \to 0,$ with vertical maps $f, g, h$ .

[L1]

In a short exact sequence, the left map is a kernel and the right map is a cokernel (A short exact sequence is a kernel-cokernel pair).

[L2]

Pullbacks of epimorphisms are epic, and the induced map on kernels in a pullback square is an isomorphism (The pullback of an epimorphism is an epimorphism, In a pullback square, the induced map on the kernels of the two parallel arrows is an isomorphism).

[L3]

A cartesian square over an epimorphism is also cocartesian (A cartesian square over an epimorphism is also cocartesian).

[L4]

In an abelian category, monic-plus-epic implies isomorphism (An abelian category is balanced).

[L5]

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

Pullback diagram

The proof uses the following pullback of $p^{'}$ along $h$ :

Proof

technique · direct

1.1L1L2L3construct

Form the pullback of $p^{'}$ along $h$ shown above. Because $p^{'} g = h p$ , there is a unique comparison map $u : B \to P$ with $α u = p$ and $β u = g$ . Since $p^{'}$ is the cokernel of $i^{'}$ by [L1], it is epic, so [L3] makes the square also cocartesian, and [L2] identifies $\ker (β)$ with $\ker (h)$ .

2.1L1step 1.1assume-hypalgebra

Assume $f$ and $h$ are monic. Then $\ker (h) = 0$ , so step 1.1 makes $β$ monic. If $u t = 0$ , then $p t = α u t = 0.$ Because $i = \ker (p)$ by [L1], there is $s$ with $t = i s$ . Now $0 = β u t = g i s = i^{'} f s .$ The map $i^{'}$ is monic by [L1], and $f$ is monic by hypothesis, so $s = 0$ and hence $t = 0$ . Thus $u$ is monic, and therefore $g = β u$ is monic.

2.2L1L2L5step 1.1assume-hypalgebra

Still with the pullback square of step 1.1, let $j : A^{'} \to P$ be a kernel of $α$ ; by [L2] this exists and agrees with the induced map from $\ker (p^{'}) = im (i^{'})$ to $P$ . Assume now that $f$ and $h$ are epic. To show that $u$ is epic, let $t : P \to T$ satisfy $t u = 0$ . Then $t j f = t u i = 0.$ Since $f$ is epic, $t j = 0$ . The map $α$ is epic as a pullback of the epic $p^{'}$ by [L2], and hence is the cokernel of its kernel $j$ by [L5]. Therefore $t = r α$ for a unique $r : C \to T$ . But then $0 = t u = r α u = r p,$ and $p$ is epic by [L1], so $r = 0$ and hence $t = 0$ . Therefore $u$ is epic.

3.1L3step 1.1step 2.2algebra

To show that $β$ is epic under the same hypotheses, let $s : B^{'} \to T$ satisfy $s β = 0$ . Since the square of step 1.1 is cocartesian by [L3], the compatible pair of maps $0 : C \to T$ and $s : B^{'} \to T$ induces a unique $v : C^{'} \to T$ with $v h = 0$ and $v p^{'} = s$ . Because $h$ is epic, $v = 0$ , so $s = 0$ . Thus $β$ is epic, and therefore $g = β u$ is epic.

4.1L4step 2.1step 3.1

If $f$ and $h$ are isomorphisms, steps 2.1 and 3.1 show that $g$ is both monic and epic. Therefore [L4] makes $g$ an isomorphism.

5.1step 2.1step 3.1step 4.1∎

This gives the short five lemma again, now by a pullback-and-pushout argument and without any use of members.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

25 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

David Mehrle, Category Theory, Part III, Lemma 7.23 (standard reference, not scraped)
The Stacks Project, Section 12.5, Lemmas 12.5.12 and 12.5.13 (standard reference, not scraped)