# Waldhausen Cats 6: F1C Is a Category with Cofibrations

Let $\Ccl$ be a category with cofibrations. Recall that we have introduced a subcategory of the arrow category of $\Ccl$ as follows:

Definition. Define a new category $F_1\Ccl$ to be the full subcategory of ${\rm Ar}\Ccl$ whose objects are the cofibrations of $\Ccl$, and whose cofibrations $(A\to A')\to (B\to B')$ are those pairs $(\varphi:A\to B,\psi:A'\to B')$ that satisfy the following two properties:

1. $A\to B$ is a cofibration.
2. $A'\cup_A B\to B'$ is a cofibration.

We have already discussed this definition and hopefully motivated it. Now comes the hard part: we need to prove the next theorem! The tricky part of course will be the third axiom (isn't it always the third axiom?) because it requires more three-dimensional-ish diagram reasoning.

Theorem. The category $F_1\Ccl$ is a category with cofibrations.

We have to verify four things: that the cofibrations actually form a category, and the three axioms for cofibrations. We do it in two separate parts: first, we prove the three axioms, and then we prove that the cofibrations in $F_1\Ccl$ form a category.

Proof (Part 1, The Three Axioms). Axiom 1. For any object $A\to A'$ in $F_1\Ccl$, we ned to verfiy that $(*\to *)\to (A\to A')$ is a cofibration. This means that $*\to A$ and $A\to A'$ must be cofibrations in $\Ccl$, which is true by definition of $F_1\Ccl$.

Axiom 2. Next, we need to verify that isomorphisms are cofibrations. So suppose that $(A\to A')\to (B\to B')$ is an isomorphism. We need to verify that in $\Ccl$, the maps $A\to B$ and $A'\cup_AB\to B'$ are cofibrations. Since $A\to B$ is an isomorphism, it is a cofibration in $\Ccl$. Proving that $A'\cup_A B\to B'$ is a cofibration in $\Ccl$ goes as follows. First, the map $A'\to A'\cup_A B$ is the pushout along an isomorphism, and so it is also an isomorphism. So, the unique map $A'\cup_AB\to B'$ (indicate by a dotted arrow) making the diagram commute is just the composite of the two isomorphisms $A'\cup_AB\to A'\to B'$.

Axiom 3. The final axiom state that the pushout of a map along cofibrations exist, and that the pushout of such a map is also a cofibration. So, suppose that $(A\to A')\to (B\to B')$ is a cofibration and that $(A\to A')\to (C\to C')$ is any map in $F_1\Ccl$. We'll recall the pushout in the arrow category ${\rm Ar}\Ccl$ that we've already seen before: Let us describe first what we need to prove: in order to show that this is actually a pushout in, $F_1\Ccl$ we need to verify that $B\cup_A C\to B'\cup_A'C'$ is actually an object of $F_1\Ccl$, or in other words, that this map is a cofibration in $\Ccl$. Once we know that this is a pushout, we need to verify that $(C\to C')\to (B\cup_AC\to B'\cup_{A'}C')$ is a cofibration in $F_1\Ccl$. Indeed, for this to be so, we need to verify two things: firstly, that $C\to B\cup_A C$ is a cofibration in $\Ccl$. Secondly, we need to show that the canonical map $C'\cup_C (B\cup_A C)\to B'\cup_{A'}C'$ is also a cofibration in $\Ccl$. In essence, we have reduced checking Axiom 3 to the following three facts about maps in $\Ccl$:

1. The natural map $B\cup_A C\to B'\cup_{A'} C'$ is a cofibration.
2. The natural map $C\to B\cup_A C$ is a cofibration.
3. The natural map $C'\cup_{C}(B\cup_A C)\to B'\cup_A'C'$ is a cofibration.

We can check this in any order. The easiest in #2: it is true because the map $C\to B\cup_A C$ comes from the pushout along the cofibration $A\to B$. What about #3? Because $(A\to A')\to (B\to B')$ is a cofibration in $F_1\Ccl$, we know that $A'\cup_A B\to B'$ is a cofibration in $\Ccl$. And, the natural map $C'\cup_C (B\cup_A A')\to B'\cup_{A'}C'$ comes from the pushout This also shows that the natural map $B\cup_A C'\to B'\cup_{A'} C'$ is a cofibration (why?). Now we can tackle #1: The map $B\cup_A C\to B'\cup_{A'} C'$ can be written as the composition $B\cup_A C\to B\cup_A C'\to B'\cup_{A'} C'$. Hence, we need to show that $B\cup_AC\to B\cup_AC'$ is a cofibration: this is because it can be written as the pushout of the cofibration $C\to C'$ along $C\to B\cup_A C$.

Proof (Part 2, Cofibrations Form a Cat). Suppose that $(A\to A')\to (B\to B')$ and $(B\to B')\to (C\to C')$ are cofibrations in $F_1\Ccl$. We need to show that the composition $(A\to A')\to (C\to C')$ is a cofibration in $F_1\Ccl$. In $\Ccl$, the cofibrations form a category, so $A\to C$ in $\Ccl$ is a cofibration: this is one of the two requirements for $(A\to A')\to (C\to C')$ to be a cofibration in $F_1\Ccl$. What of the second? We need to show that the map $A'\cup_A C\to C'$ is a cofibration in $\Ccl$. To do this, it's best if we put all these maps in an awesome crazy diagram: Here, the important point is that all the squares are pushouts. Composing the isomorphism $A'\cup_AC\to (A'\cup_A B)\cup_B C$ with $\psi\circ\varphi$ gives the required map. But $\varphi$ is a cofibration in $\Ccl$ as it is the pushout along a cofibration. And $\psi$ is also a cofibration, being the composition of the isomorphism on the right together with the natural map $B'\cup_B C\to C'$ (why?). Hence the cofibrations as we have defined them in $F_1\Ccl$ form a category.