Bourbaki

Lemma 2

Tag 00RS, permanent. Cite it as [Bourbaki, Tag 00RS].

Algebra, VIII, p. 389, § 20. Linear Representations of Algebras

Let c:Θ(A)→Θ(A)⊗Θ(A)c : \Theta(A) \to \Theta(A) \otimes \Theta(A) be a K-linear mapping. Then η\eta is a homomorphism from AA to the dual algebra of the cogebra (Θ(A),c)(\Theta(A), c) if and only if the following diagram commutes:

Θ(A)→cΘ(A)⊗Θ(A)∣∣A∗→tmA(A⊗A)∗ \begin{array}{ccc} \Theta(A) & \xrightarrow{c} & \Theta(A) \otimes \Theta(A) \\ | & & | \\ A^* & \xrightarrow{t m_A} & (A \otimes A)^* \end{array}

Indeed, η\eta is a homomorphism from AA to the dual algebra of the cogebra (Θ(A),c)(\Theta(A), c) if and only if we have η(ab)=mK∘(η(a)⊗η(b))∘c\eta(ab) = m_K \circ (\eta(a) \otimes \eta(b)) \circ c for all a,b∈Aa, b \in A, that is

η(ab)(f)=mK∘(η(a)⊗η(b))(c(f)) \eta(ab)(f) = m_K \circ (\eta(a) \otimes \eta(b))(c(f))

for a,b∈Aa, b \in A and f∈Θ(A)f \in \Theta(A). Now, we have

η(ab)(f)=f(ab)=⟨tmA(j1(f)),a⊗b⟩ \eta(ab)(f) = f(ab) = \langle t m_A(j_1(f)), a \otimes b \rangle

and, by (14),

mK∘(η(a)⊗η(b))(c(f))=⟨j2(c(f)),a⊗b⟩ m_K \circ (\eta(a) \otimes \eta(b))(c(f)) = \langle j_2(c(f)), a \otimes b \rangle

for all a,b∈Aa, b \in A and every f∈Θ(A)f \in \Theta(A). The lemma follows.

Since j2j_2 is injective, there exists at most one linear mapping cc that makes the diagram above commute. To prove its existence, we must prove that the image of tm∘j1t m \circ j_1 is contained in that of j2j_2. In other words, we must prove that there exist, for every element ff of Θ(A)\Theta(A), a natural number nn and elements f1′,…,fn′,f1′′,…,fn′′f'_1, \ldots, f'_n, f''_1, \ldots, f''_n of Θ(A)\Theta(A) satisfying the relations

f(ab)=∑i=1nfi′(a)fi′′(b) f(ab) = \sum_{i=1}^n f'_i(a) f''_i(b)

for a,b∈Aa, b \in A. We will then have

c(f)=∑i=1nfi′⊗fi′′. c(f) = \sum_{i=1}^n f'_i \otimes f''_i.

By the corollary of VIII, p. 379, there exists a left AA-module EE of finite dimension over KK with ff as a coefficient. Let (e1,…,en)(e_1, \ldots, e_n) be a basis of EE, (e1∗,…,en∗)(e_1^*, \ldots, e_n^*) the dual basis, xx an element of EE, and x∗x^* an element of E∗E^* such that f=cE(x,x∗)f = c_E(x, x^*). Set fi′=cE(ei,x∗)f'_i = c_E(e_i, x^*) and fi′′=cE(x,ei∗)f''_i = c_E(x, e_i^*) for i∈[1,n]i \in [1, n]; for a, b in A, we have

f(ab)=⟨x∗,abx⟩=⟨x∗a,bx⟩=⟨∑i⟨x∗a,ei⟩ei∗,bx⟩ f(ab) = \langle x^*, abx \rangle = \langle x^* a, bx \rangle = \left\langle \sum_i \langle x^* a, e_i \rangle e_i^*, bx \right\rangle
=∑i⟨x∗a,ei⟩⟨ei∗,bx⟩=∑i⟨x∗,aei⟩⟨ei∗,bx⟩=∑ifi′(a)fi′′(b), = \sum_i \langle x^* a, e_i \rangle \langle e_i^*, bx \rangle = \sum_i \langle x^*, ae_i \rangle \langle e_i^*, bx \rangle = \sum_i f_i'(a) f_i''(b),
and therefore (16).

Let us prove the coassociativity of c. For this, consider the K-linear mappings

c′=(c⊗1Θ(A))∘candc′′=(1Θ(A)⊗c)∘c c' = (c \otimes 1_{\Theta(A)}) \circ c \quad \text{and} \quad c'' = (1_{\Theta(A)} \otimes c) \circ c
from Θ(A)\Theta(A) to Θ(A)⊗3\Theta(A)^{\otimes 3}. We have the relations
⟨j3(f⊗c(g)),a⊗b⊗c⟩=⟨f,a⟩⟨j2∘c(g),b⊗c⟩ \langle j_3(f \otimes c(g)), a \otimes b \otimes c \rangle = \langle f, a \rangle \langle j_2 \circ c(g), b \otimes c \rangle
=⟨f,a⟩⟨g,bc⟩ = \langle f, a \rangle \langle g, bc \rangle
=⟨j2(f⊗g),a⊗bc⟩ = \langle j_2(f \otimes g), a \otimes bc \rangle
=⟨t(IdA⊗mA)∘j2(f⊗g),a⊗b⊗c⟩ = \langle t(\mathrm{Id}_A \otimes m_A) \circ j_2(f \otimes g), a \otimes b \otimes c \rangle
for f,g∈Θ(A)f, g \in \Theta(A) and a,b,c∈Aa, b, c \in A. From this, we deduce that the following diagram commutes:

Θ(A)⊗Θ(A)→IdΘ(A)⊗cΘ(A)⊗Θ(A)⊗Θ(A)↓j2↓j3(A⊗A)∗→t(IdA⊗mA)(A⊗A⊗A)∗. \begin{array}{ccc} \Theta(A) \otimes \Theta(A) & \xrightarrow{\mathrm{Id}_{\Theta(A)} \otimes c} & \Theta(A) \otimes \Theta(A) \otimes \Theta(A) \\ \downarrow j_2 & & \downarrow j_3 \\ (A \otimes A)^* & \xrightarrow{t(\mathrm{Id}_A \otimes m_A)} & (A \otimes A \otimes A)^*. \end{array}

Because of the commutativity of this diagram and that of (15), for f∈Θ(A)f \in \Theta(A) and a,a′,a′′∈Aa, a', a'' \in A, we have

⟨j3∘c′(f),a⊗a′⊗a′′⟩=⟨f,(aa′)a′′⟩; \langle j_3 \circ c'(f), a \otimes a' \otimes a'' \rangle = \langle f, (aa')a'' \rangle;
we can show the relation
⟨j3∘c′′(f),a⊗a′⊗a′′⟩=⟨f,a(a′a′′)⟩ \langle j_3 \circ c''(f), a \otimes a' \otimes a'' \rangle = \langle f, a(a'a'') \rangle
likewise. Since multiplication in A is associative, we have j3∘c′=j3∘c′′j_3 \circ c' = j_3 \circ c'', and therefore c′=c′′c' = c'' because j3j_3 is injective.

Finally, formulas (16) and (17) imply that Θ(A)\Theta(A) admits ε\varepsilon as a counit.

Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.

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Transcribed from the printed edition, 2023, Springer Nature. Not translated.