TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-14

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Universal property of the tensor algebra

Statement

If $A$ is a unital associative $k$ -algebra, every linear map $f : V \to A$ extends uniquely to a unital $k$ -algebra homomorphism $\hat{f} : T (V) \to A$ .

Facts & Assumptions

Given: A vector space $V$ , a unital associative $k$ -algebra $A$ , and a linear map $f : V \to A$ .

[L1]

Multilinear maps on $V^{n}$ factor uniquely through $V^{\otimes n}$ (Finite iterated tensor products represent multilinear maps independently of parenthesization).

[L2]

Maps from a direct sum are determined uniquely by their restrictions to the summands (Universal property of a direct sum of modules).

[L3]

The grading and concatenation multiplication are those of Tensor algebra of a vector space.

Proof

technique · direct

1.1

For $n \geq 1$ , the map $(v_{1}, \dots, v_{n}) \mapsto f (v_{1}) \dots f (v_{n})$ is multilinear, so [L1] gives a linear map $f_{n} : V^{\otimes n} \to A$ with $f_{n} (v_{1} \otimes \dots \otimes v_{n}) = f (v_{1}) \dots f (v_{n})$ . Put $f_{0} (a) = a 1_{A}$ .

L1construct

2.1

By [L2], the maps $f_{n}$ combine uniquely to a linear map $\hat{f} : T (V) \to A$ . On pure homogeneous tensors, concatenation gives $\hat{f} (u v) = \hat{f} (u) \hat{f} (v)$ , and bilinearity extends this to all finite sums; also $\hat{f} (1) = 1_{A}$ and $\hat{f} \circ j = f$ .

step 1.1L2L3algebra

3.1

If $F : T (V) \to A$ is any unital algebra homomorphism with $F j = f$ , then $F (v_{1} \otimes \dots \otimes v_{n}) = F (j v_{1}) \dots F (j v_{n}) = f (v_{1}) \dots f (v_{n})$ and $F (1) = 1_{A}$ . Pure tensors span every homogeneous summand, so [L2] gives $F = \hat{f}$ .

step 2.1L2L3algebra

4.1

The map constructed in step 2.1 is therefore the unique unital algebra extension of $f$ , including the boundary case $V = 0$ , where $T (V) = k$ .

step 2.1step 3.1∎

Depends on

Used by

Universal property of the enveloping algebra Theorem
Universal property of the symmetric algebra Theorem

Dependency tree · two levels

11 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

Etingof, MIT 18.745 notes, §12.1, printed pp. 69–70 (standard reference, not scraped)
Kirillov, An Introduction to Lie Groups and Lie Algebras, §5.1, printed pp. 71–72 (standard reference, not scraped)