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I have a proper model category and in it two coequalisers, $A_i \rightrightarrows B_i \to C_i$, $i=1,2$. I have a map of diagrams arising from maps $A_1 \to A_2$, $B_1 \to B_2$ where these two arrows are cofibrations. Can I reasonably expect the canonical arrow $C_1 \to C_2$ to be a cofibration? Are there additional properties on the model category I can assume such that it is?

It may help that the maps $A_i \to B_i$ could also be cofibrations, if I am not mistaken.

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  • $\begingroup$ I wonder if you could do something similar to what Hovey does at the bottom of page 126 of his book on Model Categories, namely consider the category of diagrams of shape $A_i \rightrightarrows B_i$ and make it Reedy by forcing one of the $A_i \to B_i$ to be positive degree and the other negative. Then prove the colimit functor is left Quillen as Hovey does. You'd need all objects to be cofibrant and all to be cofibrations. If this sketch works then it would tell you that $C_1\to C_2$ is a cofibration if the vertical arrows form a Reedy cofibration $\endgroup$ Nov 10, 2013 at 22:09
  • $\begingroup$ (continued) So you can extract conditions on the arrows rather than on the category. I'm not sure how this compares to Tom Goodwillie's "no" answer, but it seems worth a shot. At the very least, forcing one map to be positive degree and the other negative would probably rule out Tom's counterexample. $\endgroup$ Nov 10, 2013 at 22:09
  • $\begingroup$ Thanks, @DavidWhite - unfortunately I think I needed a pretty general input, but this suggestion may be useful for others. $\endgroup$ Nov 11, 2013 at 0:54

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No. Think of examples in which $A_1$ is initial (so $B_1=C_1$). There's no reason why $B_1\to C_2$ should be a cofibration just because $B_1\to B_2$ is.

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  • $\begingroup$ Thanks Tom. I think there must be something more at work in my example, then... $\endgroup$ Apr 6, 2011 at 2:44

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