In the last post we proved that $F_1\Ccl$ is a category. Consider now the category $F_1^+\Ccl$ whose objects are pairs $(A\to A’, A’/A)$ where $A\to A’$ is an object of $F_1\Ccl$ and $A’/A$ is a choice of pushout of $*\leftarrow A\to A’$. In other words, we can think of $F_1^+\Ccl$ to be the category of sequences $A\to A’\to A’/A$ where $A’/A$ is the pushout. Perhaps even better, we can think of $F_1^+\Ccl$ as the category whose objects are all commutative squares
that are pushout squares. Here I used the symbol $B$ to denote an arbitrary pushout as there may be more than one choice of pushout. In general I’ll just write $A’/A$ as I’ll just be talking about one choice at a time.
Notice that a morphism $(A\to A’)\to (B\to B’)$ in $F_1\Ccl$ determines a morphism of pushout squares an vice-versa: indeed, given such a morphism, there is a unique morphism $A’/A\to B’/B$. Hence, the functor $F_1^+\Ccl\to F_1\Ccl$ given by sending $A\to A’\to A’/A$ to $A\to A’$ is an equivalence of categories because it is fully faithful and essentially surjective. This equivalence naturally gives $F_1^+\Ccl$ the structure of a category with cofibrations.
There are also functors $s,t,q:F_1^+\Ccl\to\Ccl$ given by:
- $s(A\to A’\to A’/A) = A$
- $t(A\to A’\to A’/A) = A’$
- $q(A\to A’\to A’/A) = A’/A$
We will now see in the rest of the post that these three functors $s,t,q$ are exact functors; in other words, they preserve the structure of a category with cofibrations. In the following proofs, we won’t say anything about the zero object, since that part is trivial. Also, that $s$ is exact is pretty straightforward, but it will be a good warmup for the harder ones.
Next, let us show that $s$ preserves the pushout square in $F_1^+\Ccl$. What is the pushout square? Once we choose a pushout for the diagram $C’/C\leftarrow A’/A\to B’/B$, say $(B’/B)\cup_{A’/A}C’/C$ we get a pushout square in $F_1^+\Ccl$:
You should convince yourself that this is actually a pushout in $F_1^+\Ccl$. Of course, we didn’t really need to know the entire pushout square to see that applying $s$ to the diagram gives the familiar pushout square with pushout $B\cup_A C$.
Now, we need to show that $t$ takes the big pushout square to a pushout square that satisfies the third axiom of a category with cofibrations. From the way the big pushout was constructed, this is again clear.
Finally, the nontrivial and tricky part!

The three inner squares are pushouts, so the two big rectangles give the isomorphisms and the required map, being a pushout along $A’\cup_AB\to B’$. (One should compare this proof to the diagram-chasing proof for abelian groups and cofibrations=injections.)
Now we have to show that $q$ takes pushouts to pushouts. Let us again recall the big pushout diagram
Now we already mentioned that this is actually a pushout square, though we didn’t prove it. This comes down to proving that $B’/B\cup_{A’/A}C’/C$ is actually a choice for the quotient $(B’\cup_{A’}C’)/(B\cup_AC)$. To do so, first assume we have a commutative diagram
We would like to show that there exists a unique arrow $(B’\cup_A’ C’)/(B\cup_A C)\to X$ making the entire diagram commute. Do do so, first note that this diagram gives a commutative diagram
which in turn, together with the first diagram gives the commutative diagram
Which gives the desired arrow (I’ve left out a few steps, here, but these are straightforward diagram manipulations. Check them!)
In a few more posts we will be coming to the point where we can define a Waldhausen category, or a category with cofibrations and weak equivalences. It’s probably helpful to look at a rough roadmap of our immediate goals:
- Define a Waldhausen category
- Define the $K$-theory of a Waldhausen category
- Show that this agrees with the “obvious” definition for $K_0$.
- State the additivity theorem and explain its importance
- Prove the additivity theorem
- Deduce various corollaries from the additivity theorem
After these goals, I shall probably go through the rest of Section 1, including explaining how $K$-theory from Waldhausen’s construction gives the same thing as Quillen’s Q-construction for exact categories.
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Math is fun!