\Pi {\{s : S\}}.\, \Pi p : P\, s.\, P\, (s {\mathrel{\downarrow}}p) \to P\,
  s$ (translation of subshape positions into positions in the global shape).

satisfying the following two shape equations and three position equations:

$\forall {\{s\}}.\, s {\mathrel{\downarrow}}{\mathsf{o}}= s$,

$\forall {\{s, p, p'\}}.\,  s {\mathrel{\downarrow}}(p {\mathrel{\oplus}}p') = (s {\mathrel{\downarrow}}p) {\mathrel{\downarrow}}p'$,

$\forall {\{s, p\}}.\, p {\mathrel{\oplus}}{\{s\}}\, {\mathsf{o}}= p$,

$\forall {\{s, p\}}.\, {\mathsf{o}}\, {\{s\}} {\mathrel{\oplus}}p = p$,

$\forall {\{s, p, p', p''\}}.\, (p {\mathrel{\oplus}}{\{s\}}\, p') {\mathrel{\oplus}}p'' = p {\mathrel{\oplus}}(p' {\mathrel{\oplus}}p'')$.

(Using ${\mathrel{\oplus}}$ as an infix operation, we write the first, implicit, argument next to the operation symbol when we want to give it explicitly.) Modulo the fact that the positions involved come from different sets, laws 3–5 are the laws of a monoid. In the special case $S = 1$, we have exactly one set of positions, namely $P\, \ast$, and that is a monoid. If $S$ is general, but $s {\mathrel{\downarrow}}p$ does not depend on $p$ (in this case $s {\mathrel{\downarrow}}p = s$ thanks to law 1), then each $P\, s$ is a monoid. (One might also notice that laws 1–2 bear similarity to the laws of a monoid action. If none of $P\,
s$, ${\mathsf{o}}\, {\{s\}}$, $p \oplus\, {\{s\}}\, 