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In the excellent "A Handbook of Model Categories" (2021), cofibrant and fibrant homotopies are defined exactly as it seemed natural to me: immediately through functorial cylindrical objects / path space objects (hereinafter I will only talk about cylindrical objects), and the concept of non-functorial cylindrical objects in general not mentioned. But for proofs of elementary statements such as "cofibrate homotopy is an equivalence relation if dom is a cofibrant" Scott Balchin refers to Hovey, who proves symmetry and transitivity by constructing new cylindrical objects.

I would like to dispense with the notion of a non-functorial cylindrical object altogether. Instead, I am attracted by the search for natural operations on functorial cylindrical objects. For example, applying the cofibrate functorial factorization to the automorphism $\rm{swap}$ (of the codiagonal morphism) defines a natural automorphism of a cylinder, which gives a natural proof of the symmetry of cofibrate homotopy.

To prove transitivity, it is natural to glue two cylinders along the upper and lower bases and construct a morphism $\alpha \colon A \times I \to A \times 2I$ such that $i_0 \circ \alpha = i_0 \circ s_0$ and $i_1 \circ \alpha = i_1 \circ s_1$. Here $i_0, i_1 \colon A \to A \times I$ are compositions of inclusions $A \to A \amalg A$ with canonical embedding $A \amalg A \to A \times I$, and $s_0, s_1$ are push-out arrows.

Is there a natural wide class of model categories (possibly with an additional structure) in which it is possible to construct such morphism $A \times I \to A \times 2I$ (at least for cofibrates $A$)? Or does the desire to limit ourselves to such categories select convenient model categories for some interesting homotopy categories? (I'm just starting to learn model categories)

P.S. Of course, I write function composition in direct order (less common).

UPD. The question has been slightly edited, see version history for the context of Mike Shulman's answer.

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    $\begingroup$ It seems to me your real objection is not non-functoriality but rather non-uniqueness. Put it another way, it sounds to me you want to find a way to choose just one cylinder object (for each object) once and for all. I think this is not always possible. Certainly standard practice in model category theory is to allow oneself to choose ad hoc cylinder objects as required for the argument. $\endgroup$
    – Zhen Lin
    Commented Aug 7, 2022 at 14:53
  • $\begingroup$ By functorial cylindrical object I mean the cylindrical object resulting from the factorization of the codiagonal map. I want to fix it as the meaning of the word cylinder, in particular, it is unique (and functorial), yes. But no other choice (once and for all) of a cylindrical object also seems natural to me. $\endgroup$ Commented Aug 7, 2022 at 16:10

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You might be interested in looking into enriched model categories. If $\mathcal{V}$ is a monoidal model category (with cofibrant unit, for simplicity) and $\mathcal{C}$ is a $\mathcal{V}$-model category, then for any interval object (i.e. cylinder object for the unit object) $\mathbb{1}+\mathbb{1} \to \mathsf{I} \to \mathbb{1}$ in $\mathcal{V}$, the copower (tensor) $(-\odot \mathsf{I})$ is a well-behaved "cylinder functor" on $\mathcal{C}$. I put it in quotes because in general $A\odot \mathsf{I}$ may not be a "cylinder object" in the model categorical sense, but it is whenever $A$ is cofibrant, which is usually enough.

In this case, there really is an object $2\mathsf{I} = \mathsf{I} \sqcup_{\mathbb{1}} \mathsf{I}$ such that your "long cylinder" is also an enriched copower $A \odot 2\mathsf{I}$. So what you want is an isomorphism $\mathsf{I} \cong 2\mathsf{I}$ in $\mathcal{V}$, or at least a morphism $\mathsf{I} \to 2\mathsf{I}$ that you can compose with to implement transitivity.

Whether such a thing exists depends of course on $\mathcal{V}$. The most common choice of $\mathcal{V}$ is simplicial sets, in which case such an (iso)morphism does not exist. But another traditionally important choice of $\mathcal{V}$ (though it's fallen somewhat out of favor recently) is topological spaces, and in this case there really is an isomorphism $\mathsf{I} \cong 2\mathsf{I}$. So you may be interested in looking into topologically enriched model categories.

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  • $\begingroup$ Interesting, thanks! But topologically enriched model categories are a rather narrow class, right? For example, the model categories of chain complexes are probably not such? $\endgroup$ Commented Aug 9, 2022 at 16:38
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    $\begingroup$ It depends on what you mean by "narrow". It's true that many natural examples are not topologically enriched. But it's known that any combinatorial model category is Quillen equivalent to a simplicially enriched one, and I expect that that could be transferred across the Qullen equivalence between simplicial sets and topological spaces to show an analogous result for topologically enriched model categories. $\endgroup$ Commented Aug 9, 2022 at 17:42
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    $\begingroup$ Model categories of chain complexes, on the other hand, are generally enriched over a monoidal model category of chain complexes, and I believe the latter does have a morphism (though not an isomorphism) $\mathsf{I} \to 2\mathsf{I}$ that can implement transitivity. $\endgroup$ Commented Aug 9, 2022 at 17:43
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    $\begingroup$ Indeed, what is really needed is only a morphism $\alpha A \times I \to A \times 2I$ such that $i_0 \circ \alpha i_0 \circ s_0$ and $i_1 \circ \alpha = i_1 \circ s_1$. Here $i_0, i_1 \colon A \to A \times I$ compositions of inclusions $A \to A \amalg A$ with canonical embedding $A \amalg A \to A \times I$, and $s_0, s_1$ push- out arrows. In my example with a trivial model structure, such a morphism is $A \amalg A \to A \amalg A \amalg A$ (embeddings $1 \mapsto 1, 2 \mapsto 3$). I'll edit the question accordingly, thank you! $\endgroup$ Commented Aug 9, 2022 at 20:20
  • $\begingroup$ Taking the opportunity: of course, I have been reading you for a long time and a lot on MO, n-cafe, I have incredibly pleasant associations with your name, it's nice to chat with you :) $\endgroup$ Commented Aug 9, 2022 at 20:20

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