What is a map of double complexes? What is a map of double complexes? I was reviewing exercise 1.2.8 in Weibel on mapping cones and it raised this question of what is a map of double complexes in general.
In general, the condition $d^h d^v + d^v d^ h =0$ implies that squares in a double complex do not commute (that is, the maps $d^h$ aren’t chain maps, for example). If we picture a double complex $C_{.,.}$ as living at height $1$ in the lattice $\mathbb{Z}^2 \times \mathbb{Z}$ and a double complex $D_{.,.}$ as living at height $0$, then (I’d imagine) that a map of double complexes from $C_{.,.} -> D_{.,.}$ is a collection of maps ${f_{p,q}}$ from $C_{p,q}-> D_{p,q}$, subject to some commutativity conditions I’m trying to understand:
A map of chain complexes is a collection of maps {f_.} such that squares which appear commute. It seems that the natural generalization for double complexes is a collection of maps such that cubes which appear commute, where these cubes have top face given by a square in C_{.,.} and a bottom face given by a square in D_{.,.} with the same indices (and side length 1). Do we just want all paths along the edges of this cube to commute from a given $C_{p,q}$ to $D_{p-1, q-1}$? We cannot hope for all faces of these cubes to commute, as the top and bottom faces to not commute as noted above. Do we want the ``side faces" (front, back, left, and right) to all commute? Or, just between paths of antipodal points?
To clarify: This is before we pass to total complexes, so I am not asking for the definition of a map of total complexes. Perhaps there is an interpretation of what a map of double complexes should be in terms of what if needed to get a map of total complexes, but this is not clear to me. This is also before using the `sign trick' which allows us to identify double complexes with $\textbf{Ch}(\textbf{Ch})$, where the definition of a map of chain complexes would prescribe what a map of double complexes is in this setting. Note also that if we think of a complex as a double complex by placing it as a row in $q=0$ then we recover the original definition of a map of chain complexes by asking for all ``side faces" (front, back, left, right) to commute.
 A: I see now how imposing the sign trick [I chose $d^h = (-1)^q d^h$ under Weibel's notation as a chain map between columns] and asking for a map in $\textbf{Ch}(\textbf{Ch})$ yields that all faces of cubes commute, and removing the sign trick will have:

*

*top and bottom faces anti-commute

*front and back faces still commute:
$$f_{p-1}\circ (-1)^q d_C^h = (-1)^q d_D^h \circ f_{p,q}$$ implies $$f_{p-1, q} \circ d_C^h = d_D^h \circ f_{p,q}$$ from multiplying through by $(-1)^q$ and linearity.

*left and right faces still commute, as $d^v$ has not been modified.

That is: a map of double complexes in the sign trick setting (so that all squares commute of double complexes from the start) is a collection $\{f_{p,q}\}$ of maps $C_{p,q} \rightarrow D_{p,q}$ such that all faces of all cubes commute, and a map of double complexes without the sign trick is a collection of maps $\{f_{p,q}\}: C_{p,q} \rightarrow D_{p,q}$ such that top and bottom faces of cubes anti-commute, and all other faces commute.
Note: A given cube that I refer to throughout for instance, has vertices:
$\{C_{p,q}, C_{p-1,q}, C_{p,q-1}, C_{p-1,q-1} \}$ on top face and vertices $\{D_{p,q}, D_{p-1,q}, D_{p,q-1}, D_{p-1,q-1} \}$ on bottom face.
