Lemma 15.72.1. Let $R$ be a ring. Let $M$ be an $R$-module. Let $N, \eta , \epsilon$ be a left dual of $M$ in the monoidal category of $R$-modules, see Categories, Definition 4.43.5. Then

1. $M$ and $N$ are finite projective $R$-modules,

2. the map $e : \mathop{\mathrm{Hom}}\nolimits _ R(M, R) \to N$, $\lambda \mapsto (\lambda \otimes 1)(\eta )$ is an isomorphism,

3. we have $\epsilon (n, m) = e^{-1}(n)(m)$ for $n \in N$ and $m \in M$.

Proof. The assumptions mean that

$M \xrightarrow {\eta \otimes 1} M \otimes _ R N \otimes _ R M \xrightarrow {1 \otimes \epsilon } M \quad \text{and}\quad N \xrightarrow {1 \otimes \eta } N \otimes _ R M \otimes _ R N \xrightarrow {\epsilon \otimes 1} N$

are the identity map. We can choose a finite free module $F$, an $R$-module map $F \to M$, and a lift $\tilde\eta : R \to F \otimes _ R N$ of $\eta$. We obtain a commutative diagram

$\xymatrix{ M \ar[rr]_-{\eta \otimes 1} \ar[rrd]_-{\tilde\eta \otimes 1} & & M \otimes _ R N \otimes _ R M \ar[r]_-{1 \otimes \epsilon } & M \\ & & F \otimes _ R N \otimes _ R M \ar[u] \ar[r]^-{1 \otimes \epsilon } & F \ar[u] }$

This shows that the identity on $M$ factors through a finite free module and hence $M$ is finite projective. By symmetry we see that $N$ is finite projective. This proves part (1). Part (2) follows from Categories, Lemma 4.43.6 and its proof. Part (3) follows from the first equality of the proof. $\square$

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