Remark 15.67.4. In the yoga of super vector spaces the sign used in the proof of Lemma 15.67.3 above can be explained as follows. A super vector space is just a vector space $V$ which comes with a direct sum decomposition $V = V^+ \oplus V^-$. Here we think of the elements of $V^+$ as the even elements and the elements of $V^-$ as the odd ones. Given two super vector spaces $V$ and $W$ we set

$(V \otimes W)^+ = (V^+ \otimes W^+) \oplus (V^- \otimes W^-)$

and similarly for the odd part. In the category of super vector spaces the isomorphism

$V \otimes W \longrightarrow W \otimes V$

is defined to be the usual one, except that on the summand $V^- \otimes W^-$ we use the negative of the usual identification. In this way we obtain a tensor category (where $\otimes$ is symmetric and associative with $1$). The category of super vector spaces has an internal hom which we denote $V^\vee$. One checks that the canonical isomorphisms $\mathop{\mathrm{Hom}}\nolimits (V, W) = W \otimes V^\vee$ and $\mathop{\mathrm{Hom}}\nolimits (V, W)^\vee = V \otimes W^\vee$ do not involve signs. Finally, given three super vector spaces $U$, $V$, $W$ we can consider the analogue

$c : \mathop{\mathrm{Hom}}\nolimits (V, W) \otimes U \longrightarrow \mathop{\mathrm{Hom}}\nolimits (\mathop{\mathrm{Hom}}\nolimits (U, V), W)$

of the maps $c_{p, r, s}$ which occur in the lemma above. Using the formulae given above (which do not involve signs) this becomes a map

$W \otimes V^\vee \otimes U \longrightarrow W \otimes U \otimes V^\vee$

which involves a $(-1)$ on elements $w \otimes v^\vee \otimes u$ if $v^\vee$ and $u$ are odd.

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