# Cube of cofibrations II

Let $\mathcal{C}$ be a category with cofibrations in the sense of (Waldhausen, Algebraic K-Theory of Spaces) and denote by $F_n(\mathcal{C})$ the category with cofibrations consisting of sequences of $n$ cofibrations $A_0 \rightarrowtail A_1 \rightarrowtail \dotsc \rightarrowtail A_n$ in $C$. A cofibration $A \to B$ in $F_n(\mathcal{C})$ is a commutative ladder consisting of "lattices"

$$\begin{matrix} A_i & \rightarrowtail & A_{i+1} \\\\ \downarrow & & \downarrow \\\\ B_i & \rightarrowtail & B_{i+1} \end{matrix}$$

which means that $A_i \rightarrowtail B_i$ and $B_i \sqcup_{A_i} A_{i+1} \rightarrowtail B_{i+1}$ (which already implies $A_{i+1} \rightarrowtail B_{i+1}$).

There is a notion when a commutative cube of cofibrations is a lattice: We require that 1) each one of the six squares is a lattice, and 2) that the map from the pushout of (the cube without its tip) to the tip is a cofibration. This appears naturally in the proof of $F_n F_m \mathcal{C} \cong F_m F_n \mathcal{C}$, see my former question.

Question. Do we really have to require 2), or does it already follow from 1)?

In the category of (finite) pointed sets one can check that 2) directly follows from 1). In that situation a diagram of cofibrations above is a lattice iff $B_i \cap A_{i+1} = A_i$, the intersection taken inside of $B_{i+1}$. Then one can make a diagram chase and verify 2). Similarly, it is true for the category of finite pointed CW-complexes.

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