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Aug 9, 2022 at 20:20 comment added Arshak Aivazian 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 :)
Aug 9, 2022 at 20:20 comment added Arshak Aivazian 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!
Aug 9, 2022 at 17:43 comment added Mike Shulman 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.
Aug 9, 2022 at 17:42 comment added Mike Shulman 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.
Aug 9, 2022 at 16:38 comment added Arshak Aivazian 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?
Aug 7, 2022 at 18:05 history answered Mike Shulman CC BY-SA 4.0