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$\DeclareMathOperator{\Hom}{Hom}$ Dear all,

The question is for teaching purposes and rather basic, so I hope that it also allows (relatively) easy answer.

By abstract homotopy theory we know that if the object $A$ in a model category $\mathcal{C}$ is cofibrant, then for any object $X$ the left homotopy relation on $\Hom(A,X)$ is equivalence relation. The only thing that can go wrong if $A$ is not cofibrant, then the relation might not be transitive.

So the question is: in the Quillen model structure on topological spaces, what is an example of two spaces $A,X$ with $A$ non-cofibrant and three maps $f,g,h$ such that the transitivity of left homotopy is violated?

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  • $\begingroup$ I'm just asking for my own education... it seems you have to find some non-cell complex A in Top which is not cofibrant, then study the l.h.r. on Hom(A,X) and find your counter example there? $\endgroup$
    – cheyne
    Commented Mar 16, 2023 at 12:58
  • $\begingroup$ When looking at non-cofibrant objects the left homotopy relation become ill-defined in the sense that it depends on the choice of the cylinder. Do you want an example of a single cylinder for $A$ such that the homotopy relation defined relative to this cylinder isn't transitive, or do you want an exemple in the case where le left homotopy relation is defined as "there exists a cylinder $IA$ and a map $IA \to X$? I think these are fairly different questions. (though I don't have an example for any - so maybe there is an answer that works for both). $\endgroup$
    – Simon Henry
    Commented Mar 16, 2023 at 16:07
  • $\begingroup$ ... and In the second case, the precise definition of cylinder you are using might be relevant as well. $\endgroup$
    – Simon Henry
    Commented Mar 16, 2023 at 16:09

1 Answer 1

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This is not possible in the Quillen model structure on topological spaces. (Here I am interpreting in the weak sense: "$f$ and $g$ are left homotopic if there exists some cylinder object $A \amalg A \to C \xrightarrow{\sim} A$ and a map $C \to Y$ restricting to $f$ and $g$". If you restrict to using one fixed cylinder object, like $A \times [0,1]$ or $A$, then the answer depends on exactly which.)

Suppose we have three maps $f,g,h: X \to Y$, $C$ and $D$ are cylinder objects for $A$, $H: C \to Y$ is a left homotopy from $f$ to $g$, and $K: D \to Y$ is a left homotopy from $g$ to $h$. Without loss of generality we may assume that $C$ is a good cylinder object (the map $A \amalg A \to C$ is a cofibration) by factoring $A \amalg A \rightarrowtail C' \xrightarrow{\sim} C$ . We get a map $C \amalg_A D \to Y$ which restricts to $f$ and $h$, and so it is sufficient to prove that $C \amalg_A D$ is a cylinder object: equivalently, that the map $C \amalg_A D \to A$ is an equivalence.

Consider the following diagram of pushouts: $$ \require{AMScd} \begin{CD} \emptyset \amalg A \amalg \emptyset @>>> \emptyset \amalg A \amalg A @>>> \emptyset \amalg D\\ @VVV @VVV @VVV\\ A \amalg A \amalg \emptyset @>>> A \amalg A \amalg A @>>> A \amalg D\\ @VVV @VVV @VVV\\ C \amalg \emptyset @>>> C \amalg A @>>> C \amalg_A D\\ \end{CD} $$ The bottom-left vertical map is a cofibration, hence so are all the bottom vertical maps. The middle composite is a weak equivalence (disjoint unions of spaces preserve weak equivalences, cofibrant or not), and hence the bottom composite is a weak equivalence (topological spaces are a left proper model category -- pushouts along cofibrations preserve weak equivalences). However, this means that in the composite $C \to C \amalg_A D \to A$, two out of the three maps are equivalences, and hence so is the third: $C \amalg_A D$ is a cylinder object for $A$.

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  • $\begingroup$ Thank you! Given the answer, I have two questions: $\endgroup$
    – Igor Sikora
    Commented Mar 17, 2023 at 8:10
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    $\begingroup$ 1. What is special about the Quillen model category on topological spaces, that it is impossible? Is it some property of the model structure? $\endgroup$
    – Igor Sikora
    Commented Mar 17, 2023 at 8:11
  • $\begingroup$ 2. And where would you look for an example that the cofibrancy assumption matters in the left homotopy relation being equivalence relation? I.e., what would be a counterexample when transitivity is violated, in the weak or the stronger sense? (Of course, dual example about fibrancy and right homotopy would also be good) $\endgroup$
    – Igor Sikora
    Commented Mar 17, 2023 at 8:14
  • $\begingroup$ @IgorSikora In this specific proof that I've put down, we use two special properties: (a) coproducts of weak equivalences are weak equivalences, and (b) that this model structure is left proper. It also doesn't help that (c) all objects in Spaces are fibrant. The easiest technique that makes (a) fail involves taking an undercategory. The example I remember most where (b) fails is a model structure on differential graded algebras over the integers, lifted from the projective model structure on chain complexes. (Sorry that I don't have an example in my pocket that I know works!) $\endgroup$
    – Tyler Lawson
    Commented Mar 17, 2023 at 13:54

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