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](https://mathoverflow.net/questions/89213) 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.



Cube of cofibrations II
For vector spaces 1 does not imply 2. Make a cube by taking a vector space, three subspaces, their pairwise intersections, and the triple intersection. If the big one is two-dimensional and the other three are three distinct lines you have a counterexample.


