$X$ is smooth Poisson. Kontsevich formality theorem says that there is a $L_\infty$ quasi-isomorphism $$T_{\text{poly}}\xrightarrow{L_\infty}D_{\text{poly}},$$ where $T_{\text{poly}}:=(\bigwedge^\bullet_{\mathcal{O}_X}T^1(X))[1]$ is the dgla of (shifted) polyvector fields on $X$ and $D_{\text{poly}}:=C^\bullet(\mathcal{O}_X)[1]$ is the dgla of (shifted) Hochschild chains on $X$ (computing Hochschild cohomology of $\mathcal{O}_X$). The Poisson structure $\pi$ is an MC element on the LHS, we can form the MC twisting and get an $L_\infty$ quasi-isomorphism $$(T_{\text{poly}},d_\pi)\xrightarrow{L_\infty}(D_{\text{poly}},d_{\text{Hoch}_\hbar}),$$ where $d_{\text{Hoch}_\hbar}=d_{\text{Hoch}}+[\mu_\hbar-\mu,-]$, $\mu_\hbar$ is the Kontsevich quantisation.

My question is if this map is $\mathcal{O}_X$-linear. The question makes sense because $\mathcal{O}_X$ acts on both sides. It is clear it is $\mathbb{C}$-linear. It is not an algebra map, but I guess $\mathcal{O}_X$-linear sits somewhere in-between.

I am sorry if this is a trivial question.