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Let $M$ be a model category, consider a very good cylinder object $X \coprod X \to X \times I \overset{\operatorname{pr}}{\to} X$ (here $X \times I$ is just a notation, no object $I$ is implied), that is, the first arrow $\operatorname{in}_1$ is cofibration, and the second is trivial fibration.

Now for $f\colon X \to Y$, the mapping cylinder is defined as the pushout of the natural morphism $\operatorname{in}_1\colon X \to X \times I$ along $f\colon X \to Y$. The mapping cylinder is supplied with canonical arrows $X \to M_f \to Y$, where the first one is $\operatorname{in}_0$ and the second one is induced by:

  • composition $\operatorname{pr} \colon X \times I \to X$ and $f$ on the cylinder of $X$
  • $\rm{id}$ on $Y$

Let us assume that both $X$ and $Y$ are bifibrant (fibrant and cofibrant). It is easy to show (using only that $X$ is cofibrant) that the retraction $M_f \to Y$ is a weak equivalence. Is it true that the natural morphism $\operatorname{in}_0:X \to M_f$ is a cofibration? If not, what should be required from the model category to make this true?

enter image description here

Dual, I'm interested in the same question about mapping cocylinder / path fibration (of course, the answers will be dual, I'm just mentioning them mainly for web searches)

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2 Answers 2

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The short answer is "yes," it is true that the induced map $X\to M_f$ is a cofibration. I refer you to Section IX of Williamson's thesis Cylindrical model structures, page 114 of the pdf. He says that this observation goes back to a 1974 paper by Kamps.

The long answer is: if you are interested in developing abstract homotopy starting from a cylinder, there has already been a LOT written on this topic. First, I advise you to read Williamson's paper in full. I worry slightly from the wording of the question that you're hoping too much for everything to work exactly the same way as it does for spaces (e.g., the notation $X\times I$, the "inclusion" $X\to X\times I$, etc.). Actually, cylinder objects (and the associated mapping cylinders) can look very different from what you might expect based on intuition from topological spaces.

It might also be wise to read the nLab article on cylinders. I want to point out that what you're requiring in your post is actually a "very good" cylinder not only a "good cylinder." I also think it would be wise to ready the nLab article on mapping cylinders and cocylinders. Lastly, this post of Emily Riehl explains a bit about mapping cylinders in topology and in chain complexes, and points to Chapter 18 of the book by May and Ponto and to two papers by Riehl (along with Barthel and May). Also, this MO post might help.

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  • $\begingroup$ Thank you very much for the answer and all these links! $\endgroup$ Commented Feb 6 at 15:51
  • $\begingroup$ I used quotation marks around the word "inclusion" precisely because I do not rely (in this aspect) on the intuition from Top (but only use it terminologically, as well as the words fibration/cofibration). $\endgroup$ Commented Feb 6 at 15:51
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    $\begingroup$ I am familiar with the terminology on nlab, but in the books by Hirschorn, Hovey, Balchin there is a different (more familiar to me) terminology (when a cylindrical object (without epithets) requires cofibration by default). This condition is essentially used in demonstrating the properties of (left and right) homotopy and constructing a homotopy category. Due to the presence of different terminologies, I clarified mine. $\endgroup$ Commented Feb 6 at 15:54
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    $\begingroup$ Great! I always support more clarity! $\endgroup$ Commented Feb 6 at 16:03
  • $\begingroup$ Hmm, looking at "Cylindrical Model Structures" I did not find there (as well as in Kamps) proof in the model category context. The setting in which he (and Kamps) work (and prove this statement) induces a model structure (see the nice picture on page 6), but not in the opposite direction. On the other hand, after spending some more time searching, I found out that what I needed was called the factorization lemma, the trivial proof of which is fully presented on nlab . This link is the perfect answer for me :) $\endgroup$ Commented Feb 8 at 8:49
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This statement is called the "factorization lemma", the trivial proof of which is fully presented on nlab. Moreover, the arrow from the mapping cylinder in $Y$ is not just a weak equivalence, but a section to an acyclic fibration, which is often useful.

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