Lemma 15.71.3. Let $R$ be a ring. Let $M^\bullet$ be a complex of $R$-modules. Let $N^\bullet , \eta , \epsilon$ be a left dual of $M^\bullet$ in the monoidal category of complexes of $R$-modules. Then

1. $M^\bullet$ and $N^\bullet$ are bounded,

2. $M^ n$ and $N^ n$ are finite projective $R$-modules,

3. writing $\epsilon = \sum \epsilon _ n$ with $\epsilon _ n : N^{-n} \otimes _ R M^ n \to R$ and $\eta = \sum \eta _ n$ with $\eta _ n : R \to M^ n \otimes _ R N^{-n}$ then $(N^{-n}, \eta _ n, \epsilon _ n)$ is the left dual of $M^ n$ as in Lemma 15.71.2,

4. the differential $d_ N^ n : N^ n \to N^{n + 1}$ is equal to $-(-1)^ n$ times the map

$N^ n = \mathop{\mathrm{Hom}}\nolimits _ R(M^{-n}, R) \xrightarrow {d_ M^{-n - 1}} \mathop{\mathrm{Hom}}\nolimits _ R(M^{-n - 1}, R) = N^{n + 1}$

where the equality signs are the identifications from Lemma 15.71.2 part (2).

Conversely, given a bounded complex $M^\bullet$ of finite projective $R$-modules, setting $N^ n = \mathop{\mathrm{Hom}}\nolimits _ R(M^{-n}, R)$ with differentials as above, setting $\epsilon = \sum \epsilon _ n$ with $\epsilon _ n : N^{-n} \otimes _ R M^ n \to R$ given by evaluation, and setting $\eta = \sum \eta _ n$ with $\eta _ n : R \to M^ n \otimes _ R N^{-n}$ mapping $1$ to $\text{id}_{M_ n}$ we obtain a left dual of $M^\bullet$ in the monoidal category of complexes of $R$-modules.

Proof. Since $(1 \otimes \epsilon ) \circ (\eta \otimes 1) = \text{id}_{M^\bullet }$ and $(\epsilon \otimes 1) \circ (1 \otimes \eta ) = \text{id}_{N^\bullet }$ by Categories, Definition 4.42.5 we see immediately that we have $(1 \otimes \epsilon _ n) \circ (\eta _ n \otimes 1) = \text{id}_{M^ n}$ and $(\epsilon _ n \otimes 1) \circ (1 \otimes \eta _ n) = \text{id}_{N^{-n}}$ which proves (3). By Lemma 15.71.2 we have (2). Since the sum $\eta = \sum \eta _ n$ is finite, we get (1). Since $\eta = \sum \eta _ n$ is a map of complexes $R \to \text{Tot}(M^\bullet \otimes _ R N^\bullet )$ we see that

$(d_ M^{-n - 1} \otimes 1) \circ \eta _{-n - 1} + (-1)^ n (1 \otimes d_ N^{-n}) \circ \eta _{-n} = 0$

by our choice of signs for the differential on $\text{Tot}(M^\bullet \otimes _ R N^\bullet )$. Unwinding definitions, this proves (4). To see the final statement of the lemma one reads the above backwards. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).