I have two very concrete and simple question. Just in case I write downwards what led me into this.

My questions: Let $\mathbf{SH}(X)$ be the stable homotopy category of Voevodsky. Denote $S^n$ the simplicial $n$-sphere and also the spectrum it defines by infinite suspension.

1.- **Is the following diagram commutative?**
$$
\begin{matrix}
S^a\wedge S^{b}&\buildrel{\sim}\over{\to} & S^{a+b}\\
\sigma \downarrow & &\downarrow ^{(-1)^{a+b}}\\
S^b\wedge S^a &\buildrel{\sim}\over{\to} & S^{a+b}
\end{matrix}
$$
where $\sigma$ denotes the permutation.

2.- If so, **can you give a proof or a concrete reference?**

Thanks in advance.

What led me to this question? Recall from Cisinki and Deglise that the algebraic de Rham cohomology is represented by a **commutative** ring spectrum $E_{\mathrm{dR}}$, in other words:
$$
H^p(X,\Omega_X^\bullet)=\mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p]).
$$
Note that commutative spectrum means that there is a product $\mu \colon E\wedge E\to E$ satisfying the commutative diagram if we switch the source. We know that the algebraic de Rham cohomology is **anticommutative** or **graded commutative**, (because the product comes from the wedge product on differential forms) In other words, $\alpha \in H^p(X,\Omega_X^\bullet)$ and $\beta \in H^q(X,\Omega_X^\bullet)$ then $\alpha \cdot \beta =(-1)^{p+q}\beta\cdot \alpha$.

Let me recall the definition of product in the $E$-cohomology. Let $\alpha\in \mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p])$ and $\beta \in \mathrm{Hom}_{\mathbf{SH(X)_\mathbb{Q}}}(1_X,E_{\mathrm{dR}}[p'])$, then $$ \alpha\cdot \beta \colon 1_X\to E_{\mathrm{dR}}[p]\wedge E_{\mathrm{dR}}[p']\simeq E_{\mathrm{dR}}\wedge E_{\mathrm{dR}}\wedge 1_X[p+p']\buildrel{\mu}\over{\longrightarrow}E_{\mathrm{dR}}\wedge 1_X[p+p']\simeq E_{\mathrm{dR}}[p+p'] $$ Recall that $1_X[p]=S^p\wedge 1_X$, in order for this product to be anticommutative or graded commutative the diagram $$ \begin{matrix} 1_X[p]\wedge 1_X[p']&\buildrel{\sim}\over{\to} & 1_X[p+p']\\ \sigma \downarrow & &\downarrow (-1)^{p+p'}\\ 1_X[p']\wedge 1_X[p]&\buildrel{\sim}\over{\to} & 1_X[p+p'], \end{matrix} $$ where $\sigma$ denotes the permutation, must commute. Right?