CorollaryStatement: AI-generatedProof: AI-generatedprecheck 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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

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 $Aut (G)$ . 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 $Aut (G) : = {f : G \to G : f is an automorphism} .$ (Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[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

1.1givenL1L2L3

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

2.1step 1.1

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 $v$ .

3.1step 2.1algebra∎

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

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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.