Remark 15.71.7. Let us explain why the sign used in the direct construction in the proof of Lemma 15.71.6 agrees with the sign we get from the construction using the discussion in Remark 15.71.2 and Categories, Remark 4.43.12. Denote $- \otimes - = \text{Tot}(- \otimes _ R -)$ and $hom(-, -) = \mathop{\mathrm{Hom}}\nolimits ^\bullet (-, -)$. The construction using monoidal category language tells us to use the arrow

$hom(L^\bullet , M^\bullet ) \otimes K^\bullet \longrightarrow hom(hom(K^\bullet , L^\bullet ), M^\bullet )$

in $\text{Comp}(R)$ corresponding to the arrow

$hom(L^\bullet , M^\bullet ) \otimes K^\bullet \otimes hom(K^\bullet , L^\bullet ) \longrightarrow M^\bullet$

gotten by swapping the order of the last two tensor products and then using the evaluation maps $hom(K^\bullet , L^\bullet ) \otimes K^\bullet \to L^\bullet$ and $hom(L^\bullet , K^\bullet ) \otimes L^\bullet \to M^\bullet$. Only in swapping does a sign intervene. Namely, in the isomorphism

$K^\bullet \otimes hom(K^\bullet , L^\bullet ) \to hom(K^\bullet , L^\bullet ) \otimes K^\bullet$

there is a sign $(-1)^{r(q + r')}$ on $K^ r \otimes _ R \mathop{\mathrm{Hom}}\nolimits _ R(K^{-r'}, L^ q)$, see Section 15.72 item (9). The reader can convince themselves that, because of the correspondence we are using to describe maps into an internal hom, this sign only matters if $r = r'$ and in this case we obtain $(-1)^{r(q + r)} = (-1)^{r + qr}$ as in the direct proof.

Comment #4343 by Manuel Hoff on

In the sentence "The category of super vector spaces has an internal hom which we denote $V^\vee$.", i think "internal hom" should be replaced by something like "duals".

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).