Connected targets have repetition

Suppose $A\in {\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^k(H)$, and take any $k$-colouring of $qA$. Restrict it to the first tagged part and transport it to $A$ by $v\mapsto (1,v)$. Membership of $A$ gives a monochromatic induced embedding of $H$ in $A$. Composing with that tagged-copy map gives a monochromatic induced copy in $qA$. Since the original colouring was arbitrary, $qA\in {\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^k(H)$.

Suppose instead that $A\mathrm{\notin }{\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^k(H)$. Choose a $k$-colouring $c$ of $A$ with no monochromatic induced copy of $H$, and colour every vertex $(i,v)$ of $qA$ by $c(v)$. If this colouring had a monochromatic induced copy of $H$, connectivity would put all its vertices in one tagged part. Projecting that part by $(i,v)\mapsto v$ would give a monochromatic induced copy of $H$ in $A$, contradicting the choice of $c$. Therefore the repeated colouring avoids such a copy and $qA\mathrm{\notin }{\mathcal{R}}_{\mathrm{i}\mathrm{n}\mathrm{d}}^k(H)$.

Step 1.2 proves the contrapositive of the implication from membership of $qA$ to membership of $A$; step 1.1 proves the converse. Since $A$ and $q$ were arbitrary, their membership predicates are equivalent for every graph $A$ and positive integer $q$. This is the Ramsey repetition property at $k$.