PropositionStatement: 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.
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Module homomorphisms induce tensor-product homomorphisms functorially

Statement

Let $f : M \to M^{'}$ be a homomorphism of right $R$ -modules and $g : N \to N^{'}$ a homomorphism of left $R$ -modules. There is a unique group homomorphism

$f \otimes g : M \otimes_{R} N ⟶ M^{'} \otimes_{R} N^{'}$

such that $(f \otimes g) (m \otimes n) = f (m) \otimes g (n)$ . These maps satisfy

${id}_{M} \otimes {id}_{N} = {id}_{M \otimes_{R} N}$

and

$(f^{'} \circ f) \otimes (g^{'} \circ g) = (f^{'} \otimes g^{'}) \circ (f \otimes g) .$

Facts & Assumptions

Given: Homomorphisms $f : M \to M^{'}$ of right $R$ -modules and $g : N \to N^{'}$ of left $R$ -modules.

[L1]

A balanced map $M \times N \to A$ into an abelian group extends uniquely to a group homomorphism $M \otimes_{R} N \to A$ (Universal property of the tensor product for balanced maps into abelian groups).

[L2]

A module homomorphism preserves addition and the relevant scalar action (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct

1.1givenL2algebra

The pairing $(m, n) \mapsto f (m) \otimes g (n)$ is additive in both variables by [L2], and $(f (m r)) \otimes g (n) = (f (m) r) \otimes g (n) = f (m) \otimes (r g (n)) = f (m) \otimes g (r n)$ , so it is balanced.

2.1step 1.1L1

By [L1] the pairing of step 1.1 induces a unique homomorphism $f \otimes g$ with the stated formula.

3.1step 2.1L1

The maps ${id}_{M} \otimes {id}_{N}$ and ${id}_{M \otimes_{R} N}$ agree on every elementary tensor, so uniqueness in [L1] makes them equal.

3.2step 2.1L1

The two sides of the composition formula both send $m \otimes n$ to $f^{'} (f (m)) \otimes g^{'} (g (n))$ , so uniqueness in [L1] makes them equal.

4.1step 2.1step 3.1step 3.2∎

Steps 2.1, 3.1 and 3.2 prove existence, uniqueness, identity preservation, and composition preservation.

Depends on

Used by

Dependency tree · two levels

10 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

H. Miller, Lectures on Algebraic Topology I, Sections 20-21 (standard reference, not scraped)