Posted by Jason Polak on 29. January 2016 · Write a comment · Categories: math

In Waldhausen Cats 4, we gained some intuition on the arrow category. Recall that given any category $\Ccl$, we can define the arrow category ${\rm Ar}\Ccl$ to be the category whose objects are the morphisms of $\Ccl$. A morphism $(A\to A’)\to (B\to B’)$ in ${\rm Ar}\Ccl$ is a pair of morphisms $(\varphi:A\to B,\psi:A’\to B’)$ making the resulting square commute. If $\Ccl$ is a category with cofibrations, then ${\rm Ar}\Ccl$ can be made into a category with cofibrations by declaring $(\varphi,\psi)$ a cofibration exactly when $\varphi$ and $\psi$ are cofibrations back in $\Ccl$. The content of Waldhausen Cats 4 was a proof that this is so.

Now we level up, and do something a little cooler and more complicated. We start with a category of cofibrations $\Ccl$, and make ${\rm Ar}\Ccl$ into a category with cofibrations like we just did.

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. The natural map $A’\cup_A B\to B’$ is a cofibration.

In the next post, we shall prove that $F_1\Ccl$ is indeed a category with cofibrations. In this post, we’ll just play around with this definition a little to get some intuition for it.

The Morphism $A’\to B’$

Well, what about the morphism $A’\to B’$? Why didn’t we require that this map be a cofibration?! Because it follows from the axioms! Indeed, since $A\to B$ is a cofibration in $\Ccl$, so is $A’\to A’\cup_A B$. In our definition, #2 requires that $A’\cup_AB\to B$ is a cofibration. So the composite of these two is just $A’\to B’$, and hence $A’\to B’$ is also a cofibration as the cofibrations in $\Ccl$ form a subcategory.

A Morphism in ${\rm Ar}\Ccl$

Let’s first look at an example of a map $(A\to A’)\to (B\to B’)$ in ${\rm Ar}\Ccl$ that is not a cofibration, but that the two maps $A\to B$ and $A’\to B’$ are both cofibrations in $\Ccl$.

Consider the category of abelian groups, where the cofibrations are the injections. Let $A$ be any nonzero abelian group.

Then the zero map $0\to A$ and the identity $A\to A$ are both cofibrations, so they are objects of $F_1\Ccl$. But, the morphism in ${\rm Ar}\Ccl$ given by the commutative square

is not a cofibration in $F_1\Ccl$. This is because condition two would require that the map $A\oplus A\to A$ given by $(x,y)\mapsto x + y$ is an injection, which it clearly is not.

In The Next Post

The next post will contain a proof that $F_1\Ccl$ is in fact a category with cofibrations. The reader may wish to try and prove this first, and consult the paper for hints.

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