# Why does this construction give a (homotopy-invariant) suspension (resp. homotopy cofiber) in an arbitrary pointed model category?

In their text Foundations of Stable Homotopy Theory, Barnes and Roitzheim define the suspension of a cofibrant object X of a pointed model category to be the pushout of the diagram $$*\leftarrow X\coprod X\to Cyl(X)$$, where the second map is the structure map of the cylinder object. By contrast, there is a more manifestly homotopy-invariant definition of suspension given in e.g. Dwyer and Spalinski, which is the homotopy pushout of the diagram $$*\leftarrow X\to *$$. It is not clear to me why these definitions agree; if we don't assume properness, I don't even see why the first is homotopy-invariant! (If we assume the model category is proper, then the first diagram's pushout is equal to its homotopy pushout.) There is a similar issue with the cofiber, which they define for a cofibration of cofibrant objects $$f:A\to X$$ as the pushout of $$*\leftarrow A\to X$$: again, it is not clear why this is homotopy-invariant (with respect to maps between such $$f$$ in the comma category) unless the model category is proper. Can we drop the properness assumption and still get homotopy colimits or at least homotopy invariance? Even if so, why are the definitions of suspension equivalent?

• Shouldn't the first of these diagrams be $\ast \amalg \ast \leftarrow X \amalg X \to \text{Cyl}(X)$? Jun 16, 2020 at 2:46
• @JohnKlein In a category with a zero object, we have $*\coprod *\cong *$, so the diagrams are naturally isomorphic. Jun 16, 2020 at 5:03
• Ah...I didn't notice that you were assuming a pointed model structure. I wrote my comment and answer without that assumption. Jun 16, 2020 at 12:12
• I changed my answer to reflect that we are in the pointed case. Jun 16, 2020 at 14:04

if we don't assume properness, I don't even see why the first is homotopy-invariant!

The pushout of a diagram A←B→C in which all objects are cofibrant and one of the maps is a cofibration is always its homotopy pushout in any model category, see Proposition A.2.4.4 in Lurie's Higher Topos Theory.

This is the case for both of your examples, since the initial object is cofibrant.

An argument showing that the two models of suspension are equivalent will probably be based on something like the following:

Assertion: Suppose we are given a commutative diagram of the form $$\require{AMScd}$$ $$\begin{CD} \ast @<<< C @= C \\ @VVV @VVV @VV V \\ Y @<<< A @>g>> X \\ @| @VVV @VVV\\ Y @<<< A/C @>>h > X/C \end{CD}$$ in which the vertical directions form cofibration sequences (when I write $$A/C$$, I mean $$A \amalg_C \ast$$, where $$\ast$$ is the zero object), and the maps $$g$$ and $$h$$ are cofibrations.

Then the map of pushouts $$Y \cup_A X \to Y \cup_{A/C} X/C$$ is a weak equivalence, or better still, it is an isomorphism.

It seems to me that this is true by the assumption of properness, since we have a cofibration sequence given by the pushouts $$\ast\cup_C C \to Y \cup_A X \to Y \cup_{A/C} X/C$$ in which the first term is isomorphic to $$\ast$$.

Let's call the first suspension $$SX$$ and the second one $$\Sigma X$$.

Given the assertion, we can show that the two models for suspension are weakly equivalent as follows:

Apply the assertion to the diagram $$\begin{CD} \ast @<<< \ast\amalg X @= X \\ @VVV @VVV @VVV \\ \ast @<<< X\amalg X @>g >> \text{Cyl}(X) \\ @| @VVV @VVV\\ \ast @<<< X @>>h > CX \end{CD}$$ (where $$CX = \text{Cyl}(X)/X$$) to get that the map $$SX\to \Sigma X$$ is a weak equivalence.