I am studying the construction of the motivic stable homotopy category of schemes $\mathbf{SH}(S)$ following Riou's paper [Categorie homotopiquement stable d'un site suspendu avec intervalle](http://www.math.u-psud.fr/~riou/doc/sh.pdf) (click to see the paper) and I have a problem due to my weakness in homotopy theory. 


Riou proves that $\mathbf{SH}(S)$ is a triangulated category out of a result of Quillen in his book *Homotopical Algebra* . The result Riou uses is the following (*cf.* Corollaire 3.4 in Riou's paper)


**Corollary 3.4:** *Let $C$ be a pointed model category and let $H$ be its homotopy category. Assume that the suspension functor $\Sigma \colon H\rightarrow H$ is an autoquivalence of categories. Then $H$ is a triangulated category.*


Riou then proves that $\mathbf{SH}(S)$ is a triangulated category in the following result (I state it for the case of the stable homotopy category of schemes):


**Theorem 3.10:** *Recall there is an isomorphism $\mathbb{P}^1\simeq S^1\wedge \mathbb{G}\_m$ in $H\_\bullet (S)$. Then the functor $\underline{\phantom{a}}\wedge S^1\colon \mathbf{Spt}^{\mathbb{P}^1}\to \mathbf{Spt}^{\mathbb{P}^1}$ induces an equivalence of categories $\mathbf{SH}(S)\rightarrow \mathbf{SH}(S)$. The category $\mathbf{SH}(S)$ is therefore canonically triangulated.*


**Lemma 3.11:** *If the suspension functor $\mathbf{SH}^{S^1}(S)$ is an equivalence of categories then so is the suspension in $\mathbf{SH}^{S^1\wedge\mathbb{G}\_m}(S)=\mathbf{SH}(S)$.*


He assumes that the suspension functor in $\mathbf{SH}$ is $\underline{\phantom{a}}\wedge \mathbb{P}^1\colon \mathbf{Spt}^{\mathbb{P}^1}\to \mathbf{Spt}^{\mathbb{P}^1}$. My question is the following:


**Question:** Quillen's suspension on a model category $C$ is a very concrete one he defines in terms of the model structure (*cf.* I.2 Theorem 2 of *Homotopical algebra*) It is natural to expect that the suspension on $\mathbf{SH}$ is $\underline{\phantom{a}}\wedge \mathbb{P}^1$. But: **why is Quillen suspension functor $\underline{\phantom{a}}\wedge \mathbb{P}^1$ in $\mathbf{SH}(S)$? How does one prove such a thing?**


Thank you very much in advance for your time and help.



Triangulated structure on $\mathbf{SH}(S)$: $\mathbb{P}^1$-suspension versus classical suspension
This question is answered by Hoyois' first comment: suspension is always $\underline{\phantom{a}}\wedge S^1$. One should write $\mathbb{P}^1$-suspension otherwise. 


