A minus sign when rotating two odd cochain factors

The complexes $M$ and $N$ are each concentrated in cochain degree $1$ with zero differential, so each is bounded with terms that are free of rank one over $k$; the only nonzero summand of $\mathrm{Tot}(M{\otimes }_{k}N{)}^r$ is at $r=1+1=2$, where it is $k{\otimes }_{k}k\cong k$ by [F3], and the differential is zero because both input differentials vanish. Similarly $\mathrm{Tot}(N{\otimes }_{k}M{)}^2\cong k$ with zero differential. The terms are finite projective over the opposite algebra $k^{\mathrm{o}\mathrm{p}}=k$, so the hypotheses of [F1] and [F2] hold.

In the notation of [F1] the first block is ${\mathrm{Bar}}_0(k){\otimes }_{k}M$ in bar degree $p=0$ and cochain degree $i=1$, and the second block is ${\mathrm{Bar}}_0(k){\otimes }_{k}N$ in bar degree $q=0$ and cochain degree $l=1$. The rotation formula $(-1{)}^{(i-p)(l-q)}$ of [F1] therefore reads $(-1{)}^{(1-0)(1-0)}=(-1{)}^1=-1$ on the unique summand; the termwise map of [F2] reads $(-1{)}^{il}=(-1{)}^{1\cdot 1}=-1$ on the same summand, so the two formulations of the sign agree. Applying the rotation twice multiplies $(-1{)}^{il}(-1{)}^{li}=(-1{)}^2=1$, so the square of the rotation is the identity here.

The coefficient differentials vanish. The sign is evaluated on the degree-zero Hochschild class, which is a cycle because $b_0=0$. This does not make the chain-level compatibility checks in higher bar degrees vacuous; those are part of [F1], and this example uses only the induced map on $H{H}_0$. The only surviving Hochschild degree is $j=0$: by [F4] (equivalently [F6]) the Hochschild complex of the ground field has $H{H}_0(k,k)=k$ and $H{H}_j(k,k)=0$ for $j\ge 1$, because the alternating boundary is $0$ for odd $j$ and an isomorphism for even $j$. Total degrees are computed by $n=i-j$: with $j=0$ and the tensor factor concentrated in cochain degree $2$, the only nonzero hyperhomology is in total cochain degree $2$.

Collecting: the derived cyclicity isomorphism and the termwise cyclicity isomorphism both carry the class of the unique summand by the factor $-1$, so the rotation is a nontrivial automorphism of the one-dimensional vector space in total degree $2$, not merely a sign that could be removed by choosing a different basis; and applying it twice is the identity. This verifies the sign instance of the cyclic comparison and exhibits the necessity of the Koszul sign in the convention $D=d_{\mathrm{c}\mathrm{o}\mathrm{m}\mathrm{p}\mathrm{l}\mathrm{e}\mathrm{x}}+(-1{)}^{i}b$; it does not reprove the general comparison theorem.