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.  
 A: 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. 
