CorollaryStatement: AI-generatedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-11

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A pushout along an isomorphism is isomorphic to the other factor

Statement

If $f:K\to G$ is an isomorphism and $h:K\to H$ is any homomorphism, then the pushout is isomorphic to $H$ , compatibly with the canonical maps.

Facts & Assumptions

Given: The objects and hypotheses in the statement.

[L1]

Given homomorphisms $f:K\to G$ and $h:K\to H$ as in def-group-homomorphism, a pushout is a group $P$ with homomorphisms $i_G:G\to P$ and $i_H:H\to P$ such that $i_G\circ f=i_H\circ h$ , and such that every compatible pair $u:G\to Q$ , $v:H\to Q$ factors through a unique $w:P\to Q$ with $w\circ i_G=u$ and $w\circ i_H=v$ . The maps $f,h$ need not be injective. (Pushouts of group homomorphisms).

[L2]

Group isomorphisms, automorphisms and the set $\operatorname{Aut}(G)$ . An isomorphism $f:G\to H$ is a bijective group homomorphism (def-group-homomorphism, def-injection-surjection-bijection). When $G=H$ , it is an automorphism of $G$ . Write $\operatorname{Aut}(G):=\{f:G\to G:f\text{ is an automorphism}\}.$ (Group isomorphisms, automorphisms and the set $\operatorname{Aut}(G)$ ).

[L3]

The inverse of a bijective group homomorphism is a group homomorphism. If $f:G\to H$ is a bijective group homomorphism, then its set-theoretic inverse $f^{-1}:H\to G$ is a group homomorphism. (The inverse of a bijective group homomorphism is a group homomorphism).

Proof

technique · direct

1.1

The maps $h\circ f^{-1}:G\to H$ and $\mathrm{id}_H:H\to H$ are compatible because $(h\circ f^{-1})\circ f=h$ .

givenL1L2L3

2.1

For any other compatible pair $u:G\to Q$ , $v:H\to Q$ , compatibility forces $u=v\circ h\circ f^{-1}$ , so the unique mediator from $H$ is $v$ .

step 1.1

3.1

Thus $H$ satisfies the pushout universal property and is uniquely isomorphic to the given pushout. The proof also covers trivial groups.

step 2.1algebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 24 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.