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3. A closed oriented manifold M of dimension n can be regarded as a bordism from the
empty (n − 1)-manifold to itself, M : ∅ → ∅. Thus
Z(M) : K ∼ = Z(∅) → Z(∅) ∼ = K
and thus Z(M) ∈ Hom K (K, K) ∼ = K is a number: an invariant assigned to every closed
oriented manifold of dimension n.
Observation 4.1.7.
Let Z be an n-dimensional topological field theory. For any closed oriented (n − 1)-dimensional
manifold M, Z(M) is a vector space. The cylinder on M gives a bordism d M : M ∐ M → ∅
which is a right evaluation. Similarly, we get a right coevaluation b M : ∅ → M ∐ M.
Applying the functor Z, we get a vector space Z(M) together with another vector space
Z(M) that is a right dual.
We need two lemmas:
Lemma 4.1.8.
Let V be an object in a tensor category. Let (V ∨ , d V , b V ) and (Ṽ ∨ , ˜d V , ˜b V ) be two right duals.
Then V ∨ and Ṽ ∨ are canonically isomorphic: there is a unique isomorphism ϕ : V ∨ → Ṽ ∨ , such
that the two diagrams
V ∨ ⊗ V
ϕ⊗id V
Ṽ ∨ ⊗ V
V ⊗ V ∨
id V ⊗ϕ
V ⊗ Ṽ ∨
d V
˜d
V
I
b V
˜bV
I
commute.
Proof.
For simplicity, we assume that the tensor category is strict. The axioms of a duality imply that
α : V ∨ id V ∨ ⊗˜b V
−→
V ∨ ⊗ V ⊗ Ṽ ∨ d V ⊗idṼ ∨
−→
Ṽ ∨
and
β : Ṽ ∨ id Ṽ ∨ ⊗b V
−→
are inverse to each other. Uniqueness is easy to see.
Ṽ ∨ ⊗ V ⊗ V ∨ ˜d V ⊗id V ∨
−→
V ∨
✷
Lemma 4.1.9.
A K-vector space V has a right dual, if and only if it is finite-dimensional.
Proof.
Consider the element
b V (1) =
N∑
b i ⊗ β i with b i ∈ V and β i ∈ V ∗
i=1
which is necessarily a finite linear combination. Then by the axioms of a duality
v = (id V ⊗ d V )(b V (1) ⊗ id V )(v) =
N∑
b i β i (v) .
i=1
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