a diagram $\delta_s: D_s\rightarrow \ugraph{\mathcal{A}}{\Sigma}$. We can therefore build a diagram $\bar{\delta}$ over $\ugraph{\mathcal{A}\cup\mathcal{B}}{\Sigma}$ by *blowing up* $\delta$ along the diagrams $\delta_s$. More precisely, we construct the graph $\bar{D}_\delta$ obtained by replacing each vertex $s$ by the graph $D_s$; this is well-defined as each ray in $\delta(s)$ comes from a unique ray from a star $\delta_s(s')$ for some $s'\in V^{D_s}$, and therefore each edge $e$ in $D_\delta$ of source $s$ becomes an edge of source the unique star $s'\in V^{D_s}$. The morphism $\delta$ then extends uniquely to a morphism $\bar{\delta}: \bar{D}_\delta\rightarrow \ugraph{\mathcal{A}\cup\mathcal{B}}{\Sigma}$ whose action on the subgraphs $D_s$ coincides with that of $\delta_s$ (as a morphism into $\ugraph{\mathcal{A}\cup\mathcal{B}}{\Sigma}$).

To check that this mapping from $\mathcal{B}$-diagrams on $\exec^*_\mathcal{A}(\Sigma)$ to $(\mathcal{A\cup B})$-diagrams on $\Sigma$ is indeed an isomorphism, one can directly define an inverse mapping. For this purpose, the essential remark is that given a $(\mathcal{A\cup B})$-diagram $\bar{\delta}$ on $\Sigma$, one can recover the underlying $\mathcal{B}$-diagrams on $\exec^*_\mathcal{A}(\Sigma)$ as the restriction of $\bar{\delta}$ to the connected components of the graph obtained from $D_{\bar{\delta}}$ by removing the edges mapped to $\mathcal{A}$-coloured edges in $\ugraph{\mathcal{A}\cup\mathcal{B}}{\Sigma}$. The underlying graph of the corresponding $\mathcal{B}$-diagram on $\exec^*_\mathcal{A}(\Sigma)$ is then defined from $D_{\bar{\delta}}$ by contracting each of these 