The universal property of the tensor product A ⊗ B of abelian groups states that Hom(A ⊗ B, C) ≅ Bilin(A × B, C). What does this characterization mean in practice?
AA ⊗ B is the direct product A × B equipped with an extra bilinear operation
BEvery bilinear map from A × B to any abelian group C factors uniquely through a group homomorphism out of A ⊗ B — the tensor product is the universal target for bilinear maps
CA ⊗ B always has strictly more elements than A × B because tensoring generates additional elements via bilinearity relations
DThe tensor product only exists when both A and B are free abelian groups
The universal property defines A ⊗ B as the object that 'represents' bilinear maps: there is a canonical bilinear map φ: A × B → A ⊗ B such that every bilinear map f: A × B → C factors uniquely as f = g ∘ φ for a unique group homomorphism g: A ⊗ B → C. This means studying bilinear maps out of A × B is exactly the same as studying linear (group homomorphism) maps out of A ⊗ B. The tensor product linearizes bilinearity. This universal property — not any particular construction of elements — is what defines the tensor product up to unique isomorphism.
Question 2 Multiple Choice
A short exact sequence 0 → A → B → C → 0 is tensored with a module M, yielding M ⊗ A → M ⊗ B → M ⊗ C → 0. This sequence is right-exact but the leftmost map M ⊗ A → M ⊗ B may fail to be injective. What does this failure measure?
AThe failure of M to be projective — projective modules fix the problem by making the sequence fully exact
BThe failure of the tensor product to preserve limits; the kernel of M ⊗ A → M ⊗ B is measured by Tor₁(M, A), the first derived functor of the tensor product
CA defect in the original exact sequence — if 0 → A → B → C → 0 were split exact, no failure could occur
DA computational artifact; the tensor product always preserves short exact sequences over commutative rings
The tensor product functor M ⊗ − is right-exact: it preserves the right part of exact sequences (surjectivity) but can destroy injectivity on the left. The precise measurement of this failure is Tor₁(M, A): if M is flat (in particular, if M is free or projective), Tor₁(M, A) = 0 and the sequence remains exact on the left. If Tor₁(M, A) ≠ 0, the leftmost map fails injectivity. This failure is the motivation for the Tor functor in homological algebra — Tor measures how far the tensor product is from being exact.
Question 3 True / False
Because the tensor product functor A ⊗ − is left adjoint to the internal hom Hom(A, −) in a closed monoidal category, it preserves all colimits, including coproducts and coequalizers.
TTrue
FFalse
Answer: True
Left adjoints preserve colimits — this is a general theorem of category theory. Since A ⊗ − is left adjoint to Hom(A, −), it preserves all colimits in the second argument. Concretely: A ⊗ (colim Bᵢ) ≅ colim (A ⊗ Bᵢ). In particular, tensor product distributes over direct sums (coproducts) and coequalizers. This is analogous to multiplication distributing over addition. The dual statement — that right adjoints preserve limits — explains why Hom(A, −) preserves products and equalizers.
Question 4 True / False
In a symmetric monoidal category, the symmetry isomorphism means A ⊗ B and B ⊗ A are the same (equal) object, not merely isomorphic.
TTrue
FFalse
Answer: False
Symmetry in a monoidal category provides a natural isomorphism σ_{A,B}: A ⊗ B ≅ B ⊗ A — a coherent family of invertible morphisms, not an equality of objects. The objects A ⊗ B and B ⊗ A are generally distinct (as constructions) but canonically isomorphic. Category theory carefully distinguishes equality from isomorphism: two objects are equal only when they are literally the same object, while isomorphism means there are invertible maps between them. This is a core example of why working 'up to isomorphism' is the right level of equivalence in category theory.
Question 5 Short Answer
Why is the tensor product defined by a universal property rather than by an explicit construction of its elements? What does this approach tell you about morphisms out of A ⊗ B?
Think about your answer, then reveal below.
Model answer: Defining the tensor product by a universal property means characterizing it by how it interacts with other objects via morphisms, rather than by describing its internal structure. The universal property — Hom(A ⊗ B, C) ≅ Bilin(A × B, C) — says exactly what a morphism out of A ⊗ B must look like: it corresponds to a bilinear map out of A × B. This characterization is preferred because (1) it defines the tensor product up to unique isomorphism, so any two objects satisfying the property are canonically identified; (2) it makes the relationship between tensor products and bilinear maps transparent and systematic; and (3) it generalizes immediately to any monoidal category without needing to specify elements. The construction via generators and relations (free group quotiented by bilinearity) is just one realization of the universal property, not the definition itself.
This question targets the central methodological insight of category theory: properties are more fundamental than constructions. Two different constructions of the tensor product (e.g., as a quotient of a free module, or via a basis in the finitely generated case) yield the same object up to unique isomorphism because they satisfy the same universal property. This is why the universal property is the definition — it captures everything that matters about the tensor product without fixing a particular set-theoretic representation.