UPDATE: One way to write this is in degree a group homology style. Then we can view elements of $1$ \Gamma(C)_n$ as decorated with these$n$-tuple paths, the width and height determine which group we have land in, and the boundary map takes an alternating sum of deleting commas with the understanding that$(C_1 NN = EE = 0$. So, for example, taking the boundary of this element in $C_3 \otimes D_0) D_2$:$$d(a \oplus otimes b)_{(N,E,N,NE)} = (C_1 da) \otimes D_1) b_{(E,N,NE)} - a \oplus (C_0 otimes b_{(NE,N,NE)} + a \otimes D_1), corresponding to N, NE, and E respectively. consider the tuple (N,E) in degree 2, which comes from b_{(N,NE,NE)} - a factor C_1 \otimes D_1b_{(N,E,NNE)}.The face map d_0 induces d \otimes id to the "E" factor; the face map (The last term involving d_1 induces the projection NNE is dropped because it is zero. The first term had a boundary map applied to the "NE" first factor ; the face map because it was d_2 induces zero.N.) 2 added 1281 characters in body The face maps have two characters. The map d_i for i > 0 simply deletes the element i from the ordered set [n], and reindexes; if the resulting map [n-1] \to [k] is no longer surjective, the corresponding factor maps to zero. By contrast, the map d_0 deletes 0 and reindexes, but if the corresponding map \phi is no longer surjective its image is isomorphic to [k-1], and we apply the boundary map. Unfortunately, a chain complex isn't very useful without its differential, and that's more complicated to describe. The boundary map is the alternating sum of face maps; each face map deletes i from the ordered set 0 < \cdots < n and reindexes. If i > 0 and one of the resulting projections to [p] or [q] is no longer surjective, the corresponding factor maps to zero; if i = 0 then one or both of the maps to [p] or [q] misses zero, the appropriate image(s) are isomorphic to [p-1] or [q-1], and we apply the boundary map on those factors. For example, in degree 1 we have (C_1 \otimes D_0) \oplus (C_1 \otimes D_1) \oplus (C_0 \otimes D_1), corresponding to N, NE, and E respectively. consider the tuple (N,E) in degree 2, which comes from a factor C_1 \otimes D_1. The face map d_0 induces d \otimes id to the "E" factor; the face map d_1 induces the projection to the "NE" factor; the face map d_2 induces zero. 1 The bad news is that in degree n, this tensor product has 3^n terms. The functor \Gamma can be roughly described as follows. If we write [n] for the ordered set 0 < 1 < \cdots < n, then $$ \Gamma(C)_n = \bigoplus_{k} \bigoplus_{\phi\colon [n] \twoheadrightarrow [k]} C_k. $$ When you take the tensor product of \Gamma(C) and \Gamma(D) levelwise, you get a direct sum indexed by pairs of surjections [n] \twoheadrightarrow [p] and [n] \twoheadrightarrow [q]. The functor N then will take the quotient of this by the subcomplex of degenerate ones; those where the maps [n] \twoheadrightarrow [p] and [n] \twoheadrightarrow [q] factors through a surjection [n] \twoheadrightarrow [m]. In practice, the pairs which are not degenerate are those for which the map [n] \to [p] \times [q] is injective. As a result, we have that $$ N(\Gamma(C) \otimes \Gamma(D))_n = \bigoplus_\phi C_p \otimes D_q$$ where the sum is indexed by injections$[n] \to [p] \times [q]$where composing with either projection is surjective. In practice, you can index this direct sum by$n$-tuples of strings of elements from${N, NE, E}$, representing a path of length$n$;$p$and$q\$ are determined by the height and width of the path.