Let $C$ be a category with cofibrations in the sense of (Waldhausen, Algebraic K-Theory of Spaces) and denote by $F_n(C)$ the category with cofibrations consisting of sequences of $n$ cofibrations $A_0 \rightarrowtail A_1 \rightarrowtail \dotsc \rightarrowtail A_n$ in $C$. A cofibration in $F_n(C)$ is a commutative ladder consisting of "lattices". See the attached PDF for a precise definition.

Waldhausen gives a beautiful graphical proof that there is an equivalence $F_n F_m C \cong F_m F_n C$ (Lemma 1.1.5). The reason is, basically, that an object in both categories is given by a rectangular array of cofibrations, such that each square is a lattice, which is a symmetric condition. Now, Waldhausen only indicates why this equivalence $F_n F_m \cong F_m F_n$ also preserves cofibrations, and is therefore exact. Namely, he claims that a cofibration in $F_n F_m C$ is a $3$-dimensional diagram satisfying similarly some symmetric condition, but he does not name it explicitly. So what is this condition? So basically the question is: When should we call a commutative cube of cofibrations a lattice?

The full question contains many diagrams (also a 3-dimensional one), which cannot be properly displayed with MO-Latex. Therefore I've decided to write the full question, also with some of my attempts so far, as a PDF. I hope that this is not unappropriate for MO.

EDIT: Eva Höning pointed out to me that a cube of fibrations is a lattice if every one of the six squares is a lattice, and the map from the pushout of the cube without its tip to the tip is a cofibration. And it is rather straight forward to see that both $F_n F_m$ and $F_m F_n$ have these cubes as cofibrations. In the first place I was not careful enough with the definitions to see that, but it is really easy.

  • $\begingroup$ @MartinBradenburg, the links appear to be broken, though I realize that might be by design after this many years. $\endgroup$ Mar 12, 2016 at 16:29

1 Answer 1


The notion you are looking for is well-known in homotopy theory under the name Reedy cofibration, but for some reason this name doesn't show up in papers about Waldhausen categories, even though the concept is used all the time.

To keep things close to your question let's say that $J$ is a finite poset (in general it can be any direct category). For a diagram $X : J \to \mathcal{C}$ and $j \in J$ we define a latching object $L_j X$ as the colimit of the restriction of $X$ to the subposet $\lbrace i \in J \mid i < j \rbrace$. If $L_j X$ exists, then it comes with a canonical map $L_j X \to X_j$. We say that a diagram $X$ is Reedy cofibrant if all $L_j X$s exist and the canonical maps $L_j X \to X_j$ are cofibrations. More generally, a map of Reedy cofibrant diagrams $X \to Y$ is a Reedy cofibration if all the induced maps $X_j \sqcup_{L_j X} L_j Y \to Y_j$ are cofibrations.

You can easily verify that if $\mathcal{C}$ is a category with cofibrations, then so is the category of Reedy cofibrant diagrams $J \to \mathcal{C}$ (and Reedy cofibrations as cofibrations). The same holds for Waldhausen categories.

Now, $F_n \mathcal{C}$ is nothing else but the category of Reedy cofibrant diagrams $[n] \to \mathcal{C}$ and you can verify that both categories $F_n F_m \mathcal{C}$ and $F_m F_n \mathcal{C}$ can be identified with the category of Reedy cofibrant diagrams $[m] \times [n] \to \mathcal{C}$ (as categories with cofibrations), which fills in the gap in the proof.

Edit: I realized that to answer all your concerns raised in the comments, I would have to go through the basic theory of Reedy cofibrations and an MO answer is not a good place for this. Instead I will try to indicate what needs to be done and point you to a reference (unfortunately, I don't know a reference that discusses exactly what you need).

Initially, I thought that I would simplify things by restricting to finite posets, but this only resulted in obscuring an important part of the story. So let's go to general direct categories (by the way, "direct" is very different from "directed"). A small category $J$ is direct if there is a functor $\mathrm{deg} : J \to \mathbb{N}$ such that for any non-identity morphism $i \to j$ in $J$ we have $\mathrm{deg} i < \mathrm{deg} j$. Everything I said above goes through, you just have to modify the definition of latching objects, $L_j X$ is defined to be the colimit over $\partial J \downarrow j$ i.e. the full subcategory of the slice $J \downarrow j$ spanned by non-identity morphisms.

The key fact is that the colimit of a Reedy cofibrant diagram always exists, the proof can be found in http://arxiv.org/abs/math/0610009v4 (Theorem 9.3.5). If you go through the proof you will see that this means in particular that if you know that some diagram is Reedy cofibrant below certain degree $m$, then the latching objects in degree $m$ exist. This should help you with seeing why $L_j (X_{i,\bullet})$ from your comment exists.

If $I$ and $J$ are direct, then so is $I \times J$ (just take the sum of degrees). Using the remarks above you should be able to verify that a diagram $I \to \mathcal{C}^J$ is Reedy cofibrant if and only if the corresponding diagram $I \times J \to \mathcal{C}$ is Reedy cofibrant. Moreover a morphism $X \to Y$ in $\mathcal{C}^J$ is a Reedy cofibration between Reedy cofibrant diagrams if and only if the corresponding diagram in $\mathcal{C}^{J \times [1]}$ is Reedy cofibrant. Those things are hopefully sufficient to answer your questions.

  • $\begingroup$ Ah, this makes everything clear as crystal :). Thanks a lot! I will try to fill in the gaps. $\endgroup$ Feb 22, 2012 at 18:05
  • $\begingroup$ Hm, it didn't work out. Let's call $R(I,C)$ the category of Reedy cofibrant diagrams $I \to C$. The cofibrations in $R([n] \times [m],C)$ are the three-dimensional diagrams which are built up out of cubes such that every face is a lattice and the morphism from the pushout of the cube without its tip to the tip is a cofibration. The latter condition is equivalent to (D1) as well as to (D2) (notation from the PDF above). Every cofibration in $R([n] \times [m],C)$ is a cofibration in $R([n],R([m],C))$. $\endgroup$ Feb 23, 2012 at 14:03
  • $\begingroup$ But in order to show the converse, I'm in the same situation as before: Assume that the front and back faces of a cube of cofibrations are lattices and that (D1) is a lattice. Why are then also the left and the right faces lattices? This is not obvious to me. I hoped that the more general question makes it easier: Let $I,J$ be directed posets, do we then have $R(I,R(J,C)) \cong R(I \times J,C)$ as categories with cofibrations? Even on objects this is not clear to me: If $X : I \times J \to C$ is a diagram such that $L_{i,j}(X)$ exists, why does for every $i,j$ then $L_j(X_{i,*})$ exist? $\endgroup$ Feb 23, 2012 at 14:08
  • $\begingroup$ Finally: If (D3) denotes the condition for a cube above which is equivalent to (D1) and (D2), do we need this condition at all, when all faces are assumed to be lattices? In the case $C=\mathrm{Set}$ with injections as cofibrations this answer is no: Every cube of injections such that every face is a pullback is a pullback as a whole. I tried to generalize this to the given setting (where no pullbacks are defined), but it didn't work out. Thus, my initial question "when is a cube a lattice" is still open. $\endgroup$ Feb 23, 2012 at 14:14
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    $\begingroup$ The case $I = J = [1]$ is indeed easy. If $X \to Y$ is a morphism in $\mathcal{C}^{[1]}$ and $Z$ is the corresponding object of $\mathcal{C}^{[1] \times [1]}$, then we have $L_{1,1} Z = Z_{1,0} \sqcup_{Z_{0,0}} Z_{0,1} = X_1 \sqcup_{X_0} Y_0 = X_1 \sqcup_{L_1 X} L_1 Y$. Thus the canonical morphism $L_{1,1} Z \to Z_{1,1}$ coincides with $X_1 \sqcup_{L_1 X} L_1 Y \to Y_1$ and this morphism being a cofibration is equivalent to both $Z$ being Reedy cofibrant and $X \to Y$ being a Reedy cofibration (assuming that we already verified this at $(0,0)$, $(1,0)$ and $(0,1)$, which is easier). $\endgroup$ Feb 24, 2012 at 7:07

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