2
$\begingroup$

Suppose we have a model category $C$ and a cofibration $A\hookrightarrow B$. I want to know under what reasonable assumptions we can conclude that $A\wedge I\to B\wedge I$ is a cofibration (where $\wedge I$ denotes taking a cylinder object).

A first remark is that $A\wedge I\to B\wedge I$ need not be well-defined, indeed there can be many different cylinder objects. However if $B\wedge I$ is a very good cylinder object, meaning that $B\coprod B \to B\wedge I$ is a cofibration and $B\wedge I\to B$ is a (acyclic) fibration, and $A\wedge I$ is a good path object (meaning only the cofibration part of "very good"), then we can get a map $A\wedge I\to B\wedge I$ making the following diagram commute :

$\require{AMScd} \begin{CD} A\coprod A @>>> A\wedge I @>>> A \\ @VVV @VVV @VVV\\ B\coprod B @>>> B\wedge I @>>> B \end{CD}$

The question is : can we choose this map to be a cofibration ? If not, are there reasonable hypotheses one can add to make it a cofibration ?

Of course, what one might want to do is factor $A\wedge I\to B\wedge I$ as a cofibration followed by an acyclic fibration. The middle object is then automatically a cylinder object for $B$ (we can see that $A\coprod A\to B\coprod B$ is a cofibration, so that we can lift the map from $B\coprod B$ to $B\wedge I$ to this new thing, making it a cylinder object), which unfortunately need not be very good. Then if we factor the map from $B\coprod B\to B\wedge I$ as a cofibration followed by an acyclic fibration, we get a very good cylinder object, but the lift $A\wedge I\to B\wedge I'$ that we obtain need not be a cofibration anymore (or as far as I can see, at least)

Is there any way to make this work ? That is, to get a (very) good cylinder object $B\wedge I$ and a cofibration $A\wedge I \hookrightarrow B\wedge I$ making the above commute ?

Let me also add that I have examples in mind that suggest that we have to change the first $B\wedge I$ in general.

$\endgroup$
0
$\begingroup$

Turns out it's easier than I thought, but in this solution I don't use $B\wedge I$ to construct my new cylinder object.

So suppose I have a very good cylinder object $A\wedge I$ for $A$, and construct the pushout

$\require{AMScd} \begin{CD} A\coprod A @>>> A\wedge I \\ @VVV @VVV \\ B\coprod B @>>> Z \end{CD}$

Then automatically, $A\wedge I\to Z$ is a cofibration (as a pushout of a cofibration), and similarly for $B\coprod B\to Z$ ($A\wedge I$ is a good cylinder object). Then by the pushout property and on the one hand $B\coprod B\to B$, on the other hand $A\wedge I\to A \to B$, we get a map $Z\to B$. Then we factor this as $Z\to W\to B$, a cofibration followed by an acyclic fibration. Then we get a diagram

$\require{AMScd} \begin{CD} A\coprod A @>>> A\wedge I @>>> A \\ @VVV @VVV @VVV\\ B\coprod B @>>> W @>>> B \end{CD}$

The map $B\coprod B\to W$ is a cofibration as a composite of two cofibrations, same for $A\wedge I\to W$, and the map $W\to B$ is an acyclic fibration, by definition. It follows that $W$ is a very good cylinder object for $B$ : we have our desired property.

I think that's as good as one can hope, because as I said it seems unlikely that we can keep our $B\wedge I$ that we started with, so building a new one is about as good as it gets.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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