In Keven Walker's answer to the question, [Cubical vs. simplicial singular homology](https://mathoverflow.net/questions/3656/cubical-vs-simplicial-singular-homology)  it is written:



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This raises a few questions. I see how simplices are the "linear trees", and I see how a map of two simplices will become a map between two such trees. I also see how to send the n-cube to the appropriate star-shaped tree. It remains mysterious to me what we send a map between two cubes. To make this a bit more precise:



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> Is their a small category whose objects are rooted trees, such that the morphisms give the right morphisms of prod simplices (or more aptly name prisms). The sort of prisms I have in mind are the objects in Gugenhiem's paper, "On Supercomplexes".
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In the comments of said answer, it was said that the combinatorics are related to blob homology and dendroidal sets. Any references either on point or even slightly off point will be appreciated.



What is the precise relationship between "prodsimplicial sets" and rooted trees?
There are a short list of operations described as generating the desired polyhedra: 


* $ X : \mathrm{Prism} \vdash C X : \mathrm{Prism} $
* $ l : \mathrm{list}\ \mathrm{Prism} \vdash \Pi l : \mathrm{Prism} $


There are a short list of operations needed to generate the family of rooted trees:


* $ T : \mathrm{Tree}\ \vdash \mathrm{Stem}\ T : \mathrm{Tree} $
* $ F : \mathrm{list}\ \mathrm{Tree}\ \vdash \mathrm{graft}\ F : \mathrm{Tree}$


That makes the correspondence obvious; the distinction is in the semantics of prisms vs. trees.


Now, the generators I've given would seem to distinguish between $ \Pi (\Pi (A, B), C) $ and $ \Pi (A, B, C) $, just as they seem to distinguish between ... , well, I'd have to draw pictures, and I don't really want to. I think that's quite alright, I don't mind many isomorphic polyhedra having distinct descriptions. But if that worries you, you can argue that the strange plurality of trees is isomorphic to the strange plurality of products in a natural way. And that should be enough.


