Linear realisations preserve induced pieces

Fix $e\in \mathcal{E}$. Every edge of $J_e$ belongs to $E(R)$ and has both endpoints in $e$; hence $E_e\subseteq E(R[e])$.

Conversely, let $\{u,v\}\in E(R[e])$. The endpoints are distinct and lie in $e$, and the union defining $E(R)$ supplies some $h\in \mathcal{E}$ with $\{u,v\}\in E_h$. Both vertices also lie in $h$. If $h\mathrm{\ne }e$, then $e\cap h$ contains the two distinct vertices $u,v$, contrary to linearity. Thus $h=e$ and $\{u,v\}\in E_e$. The two graphs have the same vertex set $e$, so $R[e]=J_e$.

In the template construction, fix $e$ and distinct $x,y\in W$. The transport formula gives the forward implication from $\{x,y\}\in F$ to $\{f_e(x),f_e(y)\}\in E_e$. Conversely, if the image pair belongs to $E_e$, the formula supplies an edge $\{a,b\}\in F$ with the same image pair; injectivity of $f_e$ gives $\{a,b\}=\{x,y\}$, so $\{x,y\}\in F$. Since $f_e$ is injective, its image pairs have distinct endpoints; by the equality just proved, this equivalence is also adjacency equivalence in $R$. Surjectivity gives $f_e(W)=e$, so $f_e$ is an induced embedding and $R[e]$ is its induced copy of $T$.