Double bar comparison for cyclic bimodule tensor products

Since $M$ is finite projective as a right $B$-module, [F11] gives a finite split retraction $M\to B^r\to M$. Tensoring it over $B$ with $N$ exhibits $X:=M{\otimes }_{B}N$ as a direct summand of $N^r$ as a right $A$-module. Thus $X$ is finite projective as a right $A$-module. Symmetrically, $Y:=N{\otimes }_{A}M$ is finite projective as a right $B$-module. Only the stated right-module structures are used.

For projectivity, the balanced tensor products identify $P_{A,p}$ with $A{\otimes }_k{A}^{\otimes p}{\otimes }_{k}X$ and $Q_{A,p,q}$ with $A{\otimes }_k{V}_{p,q}{\otimes }_{k}N$, where $V_{p,q}=A^{\otimes p}{\otimes }_{k}M{\otimes }_k{B}^{\otimes q}$. On either module the right $A^e$-action is $(a\otimes v\otimes z)\cdot (c\otimes d^{\mathrm{o}\mathrm{p}})=da\otimes v\otimes zc$. Since $X$ and $N$ are finite projective right $A$-modules, each is a retract of some $A^m$. Under AC, choose a $k$-basis of the middle vector space $V$; then $A{\otimes }_{k}V{\otimes }_{k}Z$ is a retract of a direct sum of copies of $A{\otimes }_{k}A$. The map $A^e\to A{\otimes }_{k}A$, $c\otimes d^{\mathrm{o}\mathrm{p}}\mapsto d\otimes c$, identifies the latter with a free right $A^e$-module of rank one. Thus each $P_{A,p}$ and $Q_{A,p,q}$ is projective as a right $A^e$-module; the same proof with $A,B$ exchanged handles the other side. This proves projectivity from the actual outer action and retains the middle factor $M$, without an unsupported split through the $A$-tensor.

The augmented complex $P_{A,\bullet }\to X$ is the standard bar resolution of the left $A$-module $X$; its underlying augmented complex is contractible by the bar extra-degeneracy that inserts $1_A$, so it is exact. For the double bar, write $U_p:={\mathrm{Bar}}_p(A){\otimes }_{A}M\cong A{\otimes }_k{A}^{\otimes p}{\otimes }_{k}M\mathrm{.}$ As a right $B$-module, $U_p$ is a direct sum of copies of $M$ (choose a $k$-basis of $A{\otimes }_k{A}^{\otimes p}$), hence is flat by the right $B$-projectivity of $M$ and [F6]. The augmented left $B$-bar complex ${\mathrm{Bar}}_{\bullet }(B){\otimes }_{B}N\to N$ is the standard bar resolution of the left $B$-module $N$; its terms are free left $B$-modules under AC, and its augmentation is a quasi-isomorphism by the extra-degeneracy contraction. Thus $U_p{\otimes }_B{\mathrm{Bar}}_{\bullet }(B){\otimes }_{B}N\to U_p{\otimes }_{B}N$ is a quasi-isomorphism for each $p$, by [F7]. The first-quadrant assembly lemma [F13] now gives a quasi-isomorphism $Q_A\to P_A$. Each total homological degree $s$ contains only the $s+1$ pairs $p+q=s$, so the direct-sum total has finite diagonals; no boundedness of the entire vertical bar complex is claimed. Together with 1.2 this proves that $Q_A$ is a projective right $A^e$-resolution of $X$. The symmetric proof gives the asserted resolutions $P_B,Q_B$ of $Y$.

Put $U_{\bullet }={\mathrm{Bar}}_{\bullet }(A){\otimes }_{A}M$ and $V_{\bullet }={\mathrm{Bar}}_{\bullet }(B){\otimes }_{B}N$, with homological bar degrees $p,q$. Then $Q_A=\mathrm{Tot}(U_{\bullet }{\otimes }_B{V}_{\bullet })$ and $Q_B=\mathrm{Tot}(V_{\bullet }{\otimes }_A{U}_{\bullet })$. The class map $[u\otimes v]⟼[v\otimes u]$ is well defined from ${\mathrm{coInv}}_A(U{\otimes }_{B}V)$ to ${\mathrm{coInv}}_B(V{\otimes }_{A}U)$. Indeed, a $B$-balance relation $ub\otimes v=u\otimes bv$ maps to classes $[v\otimes ub]$ and $[bv\otimes u]$, which agree in $B$-coinvariants; an $A$-coinvariant relation $au\otimes v=u\otimes va$ maps to $[v\otimes au]$ and $[va\otimes u]$, which agree by $A$-balance. The same construction in reverse is its inverse. On bidegree $(p,q)$ multiply this map by $(-1{)}^{pq}$. With source differential $D=d_A+(-1{)}^p{d}_B$ and target $D^{\mathrm{\prime }}=d_B^{\mathrm{\prime }}+(-1{)}^q{d}_A^{\mathrm{\prime }}$, the $d_A$ terms agree because $(-1{)}^{(p-1)q}=(-1{)}^q(-1{)}^{pq}$; the $d_B$ terms agree because $(-1{)}^p(-1{)}^{p(q-1)}=(-1{)}^{pq}$. Thus it is an isomorphism of chain complexes, and applying it twice gives sign $(-1{)}^{pq+qp}=1$. This rotation is asserted only after taking enveloping coinvariants.

By [F4], ${\mathrm{coInv}}_A(P_A)=\mathrm{Bar}(A){\otimes }_{A^e}X$ identifies with $C_{\bullet }(A,X)$, and similarly on the $B$-side. The degree-zero case also agrees with the coinvariant description [F5]. By 1.1, 1.2 and 2.1, $P_A,Q_A$ are projective resolutions of the same right $A^e$-module $X$; AC implies DC by [F10], so [F9] supplies comparison maps in both directions whose composites are chain-homotopic to the identities. The same holds for $P_B,Q_B$. In a graded instance, these comparisons and homotopies can be chosen of internal degree zero: take an ungraded lift and then its degree-zero homogeneous component. Because the lifted map and epimorphism have degree zero, that component still lifts the map; the same argument applies at each stage of the comparison and homotopy constructions. The additive coinvariant functors preserve these homotopies. Composing these comparison zigzags with the rotation of 2.2 gives the claimed chain-homotopy equivalence of Hochschild complexes. By [F15], it induces the asserted isomorphism on every $H{H}_j$. By [F16], comparison maps lifting the same object morphism are unique up to homotopy. For a morphism of bimodule pairs, the two composites around each comparison square lift the same induced morphism of $X$ (or $Y$), so [F16] makes that square commute up to homotopy. Thus the equivalence is natural up to homotopy; since the rotation itself squares to the identity, the resulting equivalence is involutive up to homotopy. The argument uses only right projectivity of $M_B$ and $N_A$.