LemmaStatement: AI-generatedProof: AI-generated

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 right inverse makes a homomorphism surjective

PDF p. 9: Surjectivity of the determinant on K1

Statement

Let $d : G \to H$ and $s : H \to G$ be group homomorphisms with $d \circ s = {id}_{H}$ . Then $d$ is surjective and $s$ is injective. If $G$ is abelian, $G$ is the internal direct sum of $\ker (d)$ and $s (H)$ .

Here $\ker (d)$ is the subgroup of elements mapped to the identity, and an internal direct sum means that each element has a unique expression as a product of one element from each subgroup.

Facts & Assumptions

Given: Group homomorphisms $d, s$ with $d \circ s = {id}_{H}$ .

Goal: Prove surjectivity, injectivity and, when $G$ is abelian, the direct-sum decomposition.

Proof

technique · Apply the right-inverse identity and decompose an arbitrary element

1.1givenalgebra

Each $h \in H$ equals $d (s (h))$ , so $d$ is surjective. If $s (h) = s (h^{'})$ , applying $d$ gives $h = h^{'}$ , so $s$ is injective.

2.1givenstep 1.1algebra∎

Suppose $G$ is abelian. For $g \in G$ , put $k = g s (d (g))^{- 1}$ . Then $d (k) = 1$ , so $g = k s (d (g))$ lies in $\ker (d) s (H)$ . If $s (h) \in \ker (d)$ , then $h = d (s (h)) = 1$ , so the two subgroups intersect only in the identity. The expression is therefore unique, and commutativity gives the internal direct sum.

Used by · 1

Units split the K one determinant