2
$\begingroup$

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.

$\endgroup$

1 Answer 1

8
$\begingroup$

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.

$\endgroup$
3
  • $\begingroup$ Oh, I thought that it is true in every abelian category ... $\endgroup$ Apr 20, 2012 at 11:29
  • $\begingroup$ But of course you're right. The pushout of the cube without its tip is just the direct sum of the three lines, which does not embed into the given space. Thank you Tom! $\endgroup$ Apr 20, 2012 at 11:37
  • $\begingroup$ The condition given on the cube reminds of the dual notion of fibrant cube discussed and used in Steiner, R., Resolutions of spaces by cubes of fibrations, J. London Math. Soc. (2), 34, 1986, 1, 169--176. $\endgroup$ May 3, 2012 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.