# When does a cofibration induce a cofibration on cylinder objects?

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.

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.