Bar transfer is a chain map independent of the transversal

The retraction satisfies $r(hx)=hr(x)$ for $h\in H$, since $Hx=Hhx$. For $x\in G$ and $t\in T$ write $tx=h_t{t}_x$. Right multiplication by x permutes the right cosets, so $t\mapsto t_x$ permutes T. The t-summand on $[(x{g}_0,\dots ,x{g}_n)\otimes xm]$ is $[(h_{t}r(t_x{g}_0),\dots ,h_{t}r(t_x{g}_n))\otimes h_t{t}_{x}m]$, equal in H-coinvariants to the $t_x$-summand of the original chain. This proves diagonal invariance and well-definedness.

For two H-equivariant vertex maps f,u define the prism $P_n(v_0,\dots ,v_n)={\sum }_{i=0}^n(-1{)}^i(f{v}_0,\dots ,f{v}_i,u{v}_i,\dots ,u{v}_n)$. Expansion gives $dP+Pd=u_{\ast }-f_{\ast }$: deletions away from the switch cancel the corresponding terms of Pd; the two switch faces at consecutive values of i cancel, leaving only deletion of the first f-vertex at i=0 and of the last u-vertex at i=n, with signs + and -. For n=0 this reads $d(f{v}_0,u{v}_0)=(u{v}_0)-(f{v}_0)$. Equivariance makes this descend to diagonal coinvariants. If $v_j=v_{j+1}$, each prism summand has either an adjacent equal f-pair or an adjacent equal u-pair, so it also descends to normalization.

For each deletion index $0\le j\le n$, deleting the jth vertex of $(r(t{g}_0),\dots ,r(t{g}_n))$ is exactly applying r to the tuple with $g_j$ deleted; the coefficient remains tm. Thus all faces, including j=0,n, commute with the sum and so does their alternating differential. Adjacent equal input vertices stay equal after r, so degeneracies map to degeneracies. The formula descends to normalized chains.

Let U be another transversal with retraction u. For each right coset write its representative in U as $a_{t}t$, $a_t\in H$. The corresponding U-summand becomes $[(u(t{g}_0),\dots ,u(t{g}_n))\otimes tm]$ after translating by $a_t^{-1}$ in the H-coinvariants. Thus compare r and u on the same inputs $(t{g}_0,\dots ,t{g}_n)$ and same coefficient tm. Sum the prism of step 1.2 over T. The permutation calculation of step 1.1 applies to the prism too, because both vertex maps are H-equivariant. It gives a well-defined normalized homotopy K with $dK+Kd={\mathrm{Tr}}_U-{\mathrm{Tr}}_T$.

For completeness, in the inhomogeneous tuple $(1,g_1,g_1{g}_2,\dots ,g_1\cdots g_n)$, the recursion gives $r(t{g}_1\cdots g_i)=h_1\cdots h_i$ and $r(t)=1$. Consecutive vertex ratios are therefore the printed h_i, proving the inhomogeneous formula. All sums and choices are finite; for H=G use T={1}.