Hochschild chains are bar tensor chains

For $n\ge 0$, define on pure tensors $f_n(a_0\otimes \cdots \otimes a_{n+1},m)=(a_{n+1}m{a}_0)\otimes a_1\otimes \cdots \otimes a_n,$ where at $n=0$ the value is $a_{1}m{a}_0\in M$. The formula is $k$-multilinear in the $n+2$ algebra slots and $m$, so [F7] gives a bilinear map $f_n:{\mathrm{Bar}}_n(A)\times M\to C_n(A,M)$. It is balanced over $A^e$: for $z=a_0\otimes \cdots \otimes a_{n+1}$, $f_n(z\cdot (c\otimes d^{\mathrm{o}\mathrm{p}}),m)=(a_{n+1}c)md{a}_0\otimes a_1\otimes \cdots \otimes a_n=a_{n+1}(cmd)a_0\otimes a_1\otimes \cdots \otimes a_n=f_n(z,(c\otimes d^{\mathrm{o}\mathrm{p}})m),$ using [F2], [F3], and associativity. By [F6] it induces ${\mathrm{\Phi }}_n$ with the stated formula. In degree zero this is precisely ${\mathrm{\Phi }}_0((a_0\otimes a_1)\otimes m)=a_{1}m{a}_0$.

Define ${\mathrm{\Psi }}_n$ on pure Hochschild tensors by ${\mathrm{\Psi }}_n(m\otimes a_1\otimes \cdots \otimes a_n)=(1\otimes a_1\otimes \cdots \otimes a_n\otimes 1){\otimes }_{A^e}m\mathrm{.}$ This prescription is $k$-multilinear and hence defines a linear map by [F7]. At $n=0$, set ${\mathrm{\Psi }}_0(m)=(1\otimes 1){\otimes }_{A^e}m$, consistent with $C_0(A,M)=M$.

For $n\ge 1$, the bar face $r=0$ becomes the first Hochschild face, because $a_{n+1}m(a_0{a}_1)=(a_{n+1}m{a}_0)a_1$. Each internal face $1\le r<n$ keeps the coefficient $a_{n+1}m{a}_0$ and multiplies the same adjacent pair $a_r,a_{r+1}$. The last bar face $r=n$ becomes the cyclic face because $(a_n{a}_{n+1})m{a}_0=a_n(a_{n+1}m{a}_0)$. These faces have the same alternating sign $(-1{)}^r$, so ${\mathrm{\Phi }}_{n-1}(d_n\otimes 1_M)=b_n{\mathrm{\Phi }}_n$. When $n=1$, there are no internal faces: applying ${\mathrm{\Phi }}_0$ to $d_1\otimes 1_M$ gives $a_{2}m(a_0{a}_1)-(a_1{a}_2)m{a}_0$, equal to $(a_{2}m{a}_0)a_1-a_1(a_{2}m{a}_0)=b_1{\mathrm{\Phi }}_1$ by associativity. At $n=0$ both outgoing differentials are zero.

On a pure Hochschild tensor, ${\mathrm{\Phi }}_n{\mathrm{\Psi }}_n$ is the identity because the outer units act trivially on $M$. Conversely, for $z=a_0\otimes \cdots \otimes a_{n+1}$, [F2] gives $(1\otimes a_1\otimes \cdots \otimes a_n\otimes 1)\cdot (a_{n+1}\otimes a_0^{\mathrm{o}\mathrm{p}})=z$, and [F3] gives $(a_{n+1}\otimes a_0^{\mathrm{o}\mathrm{p}})m=a_{n+1}m{a}_0$. The balancing relation in ${\otimes }_{A^e}$ therefore gives ${\mathrm{\Psi }}_n{\mathrm{\Phi }}_n(z\otimes m)=z\otimes m$. The same calculation at $n=0$ uses the empty middle tensor. Thus ${\mathrm{\Phi }}_n$ and ${\mathrm{\Psi }}_n$ are inverse in every degree.

If $g:M\to N$ is an $A$-bimodule map, then ${\mathrm{\Phi }}_n^N(z\otimes g(m))=a_{n+1}g(m)a_0\otimes a_1\otimes \cdots \otimes a_n=g(a_{n+1}m{a}_0)\otimes a_1\otimes \cdots \otimes a_n$, so the isomorphisms are natural in the coefficient bimodule. If $A=k$, the multiplication map $k{\otimes }_k{k}^{\mathrm{o}\mathrm{p}}\to k$ has inverse $\lambda \mapsto \lambda \otimes 1$, since $c\otimes d=cd\otimes 1$ in the tensor product; [F10] identifies $A^e=k{\otimes }_k{k}^{\mathrm{o}\mathrm{p}}$ with $k$. Under this identification, the right action in [F2] on ${\mathrm{Bar}}_n(k)\cong k$ is scalar multiplication by $cd$, and the left action in [F3] on $M$ is also multiplication by $cd$. Thus ${\mathrm{Bar}}_n(k){\otimes }_{k^e}M\cong k{\otimes }_{k}M\cong M$ by [F8]. The Hochschild term $C_n(k,M)\cong M$ by the tensor-unit maps, since the $n$ scalar factors multiply into the coefficient. Every bar and Hochschild face preserves this total scalar, so each face identifies with ${\mathrm{id}}_M$, and the formula for ${\mathrm{\Phi }}_n$ also identifies with ${\mathrm{id}}_M$. This checks the degenerate ground-field case directly. The displayed isomorphisms use no projectivity or choice. [step 1.1, step 2.1, step 1.3, F1, F2, F3, F4, F5, F8, F10, given, algebra] $\mathrm{\square }$