of rigid spaces $X_p(\overline\rho_x)\longrightarrow {\mathfrak X}_\infty:=({\mathrm{Spf}}\, R_\infty)^{\rm rig}$. Let $L(\lambda)$ be the finite dimensional algebraic representation of $G(F^+\otimes_{{\mathbb{Q}}}{\mathbb{Q}}_p)$ associated (via the usual shift) to the Hodge-Tate weights $({\bf k}_v)_{v\in S_p}$. Proving classicality of $x$ turns out to be equivalent to proving that the image of $x$ in ${\mathfrak X}_\infty$ is in the support of the $R_\infty$-module $\Pi_\infty(\lambda)'$ which is the continuous dual of: $$\Pi_\infty(\lambda):={\mathrm{Hom}}_{{\prod}_{v\in S_p}{\mathrm{GL}}_n(\mathcal{O}_{F_{\tilde{v}}})}\big(L(\lambda),\Pi_\infty\big).$$ By [@CEGGPS Lem.4.17], the $R_\infty$-module $\Pi_\infty(\lambda)'$ is essentially a Taylor-Wiles-Kisin “usual” patched module for the trivial inertial type and the Hodge-Tate weights $({\bf k}_v)_{v\in S_p}$. Forgetting the factors ${\mathfrak{X}}_{{\overline{\rho}}_x^p}$ and ${\mathbb{U}}^g$ which appear in ${\mathfrak X}_\infty$, its support is a union of irreducible components of the smooth rigid space $\prod_{v\in S_p}{\mathfrak{X}}_{\overline \rho_{x,v}}^{\square,{\bf k}_v{\rm -cr}}$. It is thus enough to prove that the unique irreducible component $Z_{\rm cris}(\rho_x)$ of $\prod_{v\in S_p}{\mathfrak{X}}_{\overline \rho_{x,v}}^{\square,{\bf k}_v{\rm -cr}}$ passing through $(\rho_{x,v})_{v\in S_p}$ contains a point which is in the support of $\Pi_\infty(\lambda)'$. But it is easy to find a point $y$ in $Z_{\rm tri}(x)$ sufficiently close to $x$ such that $(\rho_{y,v})_{v\in S_p}\in Z_{\rm cris}(\rho_x)$ (in particular $\rho_{y,v}$ is crystalline of the same Hodge-Tate weights as $\rho_{x,v}$) [*and*]{} moreover $\rho_{y,v}$ is [*generic*]{} in the sense of [@BHS Def.2.8] for all $v\in 