One representative from each matching attachment reconstructs the target

Define $f:V(J)\to V(H)$, where $H$ is the displayed induced subgraph, by $f(v)=(0,\phi (v))$ for $v\in V(F)=V(J)\setminus I$ and $f(u)=x_u$ for $u\in I$. The core vertices are distinct because $\phi$ is injective; the representatives are distinct because their indexed parts are disjoint, and none lies in the core. Thus $f$ is injective, and it is surjective onto $V(H)$ by the definition of $H$. This also covers an empty $I$ or empty $V(F)$.

For distinct vertices of $F$, their images are adjacent exactly when their images under $\phi$ are adjacent in $A$, equivalently when the original vertices are adjacent in $J$, since $F$ is induced. For $v\in V(F)$ and $u\in I$, the attachment rule says $f(v)$ and $f(u)$ are adjacent exactly when some $w\in V(F)$ satisfies $\phi (w)=\phi (v)$ and $uw\in E(J)$; injectivity forces $w=v$. For distinct $u,u^{\mathrm{\prime }}\in I$, their images lie in distinct parts and are nonadjacent, as are $u,u^{\mathrm{\prime }}$ because $I$ is independent. These cases exhaust distinct pairs, so $f$ preserves and reflects adjacency and is an isomorphism from $J$ onto $H$. If all vertices of $H$ have one common colour, its induced copy is monochromatic.