CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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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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Tensor products are unique up to a unique isomorphism carrying elementary tensors to elementary tensors

Statement

Let $T, T^{'}$ be abelian groups equipped with balanced maps $τ : M \times N \to T$ and $τ^{'} : M \times N \to T^{'}$ , and suppose that each pair has the universal property of Universal property of the tensor product for balanced maps into abelian groups. Then there is a unique group isomorphism $u : T \to T^{'}$ such that $u \circ τ = τ^{'}$ . Its inverse is the unique map $v : T^{'} \to T$ with $v \circ τ^{'} = τ$ .

Facts & Assumptions

Given: Two representing pairs $(T, τ)$ and $(T^{'}, τ^{'})$ for balanced maps out of $M \times N$ .

[L1]

For any balanced map from $M \times N$ into an abelian group, a representing pair supplies a unique group homomorphism through which that map factors (Universal property of the tensor product for balanced maps into abelian groups).

Proof

technique · direct

1.1givenL1

Apply [L1] for $(T, τ)$ to the balanced map $τ^{'}$ and for $(T^{'}, τ^{'})$ to $τ$ ; this gives unique homomorphisms $u : T \to T^{'}$ and $v : T^{'} \to T$ with $u τ = τ^{'}$ and $v τ^{'} = τ$ .

2.1step 1.1L1

Both $v \circ u$ and ${id}_{T}$ compose with $τ$ to give $τ$ , so uniqueness in [L1] gives $v \circ u = {id}_{T}$ ; similarly $u \circ v = {id}_{T^{'}}$ .

3.1step 2.1L1∎

Thus $u$ is an isomorphism with inverse $v$ , and the same uniqueness clause shows that no other isomorphism carrying $τ$ to $τ^{'}$ exists.

Depends on

Universal property of the tensor product for balanced maps into abelian groups

Used by

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.

Sources

Stacks Project, Section 10.12: Tensor products (standard reference, not scraped)