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Since about the time I asked the question, What is the precise relationship between "prodsimplicial sets" and rooted trees? I have been playing with these rooted trees and their correspondence to certain regular cell complexes. If $T$ is a rooted tree, let $X_T$ be the regular cell complex associated with $T$. Then if $X_T$, and $X_{T'}$ are isomorphic as regular cell complexes (not as spaces), then are the trees, $T$ and $T'$ isomorphic?

Some Backround

Let us sketch the association between rooted trees and this class of regular cell complexes.

The tree that consists of one element and no edges will be sent to the one point space.

If $T$ is a tree, then we may form the join $pnt*T$ thinking of the tree as a partial order with some properties, and adding a disjoint point as a new mininmal element.) Will give us the cone on $X_T$. Where we cone the cell complex in the standard way.

If $T_1, T_2, \ldots T_n$ are trees, the the tree $graft(T_1,T_2,\ldots T_n)$ (which is obtained by identifying all the roots) gets sent to $T_1\times T_2\times\ldots T_n$.

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up vote 2 down vote accepted

Here I will sketch a proof, leaving many of the details out. The proof will be by induction on the number of edges of the rooted tree, which is the same as the dimension of the PS-complex (prod-simplicial complex). Here are some lemmas that are certainly not too difficult. By trees, we mean rooted trees, and by cell complexes, we mean regular cell complexes. The product of cell complexes is the standard one. Namely that the cells of $X\times Y$ are of the form, $cell\times cell$. For the cone on $X$, we give a cell complex on $CX$ by adding a disjoint point and adding new cell which are cones on the old, with the attaching maps being the base inclusions.

  1. Suppose that $CX\sim CY$, where $C$ is the cone on the cell complexes $X,Y$, then $X\sim Y$

  2. Suppose that $CX\sim Y_1\times Y_2$, where everything is taken in cell complexes. Then $CX\sim \{pnt\}$.

  3. All prod-simplicial complexes have one top cell.

  4. All prod-simplicial complexes may be written as a product $CX_1\times CX_2\times\cdots CX_n$.

  5. Their is exactly one prod simplicial complex of dimension one.

Now for the hard part.

Proposition: Let $CX_1\times CX_2\times \cdots CX_n\sim CY_1\times CY_2\times\cdots CY_m$. Then $m=n$ their is $ \sigma\in S_n$ such that $CX_i\sim CX_{\sigma i} $.

proof:

We first make some assumptions. Each of the cell complexes, $X_i,Y_j$ are non-empty. This means that $CX_i,CY_j$ are not the one point space. Furthermore, we may assume that $n\leq m$, and we also pick an isomorphism $CX_1\times CX_2\times \cdots CX_n\rightarrow CY_1\times CY_2\times\cdots CY_m$. We also pick a one point cell $\{pnt\}\rightarrow CX_1\times CX_2\times \cdots CX_n$. Then any projection $ CX_1\times CX_2\times \cdots CX_n\rightarrow CX_{i_1}\times CX_{i_2}\times \cdots CX_{i_k}$ has a canonical section with respect to the basepoint. We define this by sending $CX_{i_l}\rightarrow CX_{i_l}$ via the identity for $C_{i_l}$ one of the factors of $CX_{i_1}\times CX_{i_2}\times \cdots CX_{i_k}$ and $pnt\rightarrow CX_{1}\times CX_{2}\times \cdots CX_{n}\rightarrow CX_j$ and by if $CX_j$ is not one of the factors of $CX_{i_1}\times CX_{i_2}\times \cdots CX_{i_k}$. Furthermore, the section associated to the composition of two projections is canonically the composite of two sections.

Now, we consider maps of the form, $CX_i\rightarrow CX_1\times CX_2\times \cdots CX_n\rightarrow CY_1\times CY_2\times\cdots CY_m$, where the first map is our cannonical section, and the second is the isomorphism in question. Since each of the $CX_i$ has one top cell, the subcell complex that is the isomorphic to the image of this map has one top cell. Furthermore, the cell complex in the image is given by the cells below the top cell. But this top cell is a product of cells from each factor. From this, we get the image of the map
$CX_i\rightarrow CX_1\times CX_2\times \cdots CX_n\rightarrow CY_1\times CY_2\times\cdots CY_m$ is the product of the images of the maps, $CX_i\rightarrow CX_1\times CX_2\times \cdots CX_n\rightarrow CY_1\times CY_2\times\cdots CY_m\rightarrow CY_p$. But as each $CX_i$ is cone, and by (2) above, their is a unique such $p$ such that the map $CX_i\rightarrow CX_1\times CX_2\times \cdots CX_n\rightarrow CY_1\times CY_2\times\cdots CY_q\rightarrow CY_p$ is an inclusion. This defines a function between the sets {1,2,...n}$\rightarrow${1,2,...,m}, defined by $q\mapsto p$. Furthermore, one can show that if $q_1,q_2,...,q_r\mapsto p$, then the map, $CX_{q_1}\times CX_{q_2}\times...CX_{q_r}\rightarrow CX_1\times CX_2\times \cdots CX_n\rightarrow CY_1\times CY_2\times\cdots CY_m\rightarrow CY_p$ is also injective. But by simple cell counting arguments this cannot happen. In particular our set function is injective. By further cell counting arguments, $m=n$, and all of the injections of the form $CX_i\rightarrow CY_k$ are isomorphisms.

QED

We are now in a position to prove that if two trees $T,S$ give rise to the same prod-simplicial complex, then they are isomorphic trees. Note that the dimension of the prod-simplicial complex is the same as the number of edges. So for zero and one dimensional prod-simplicial complexes, this is clear. Now suppose that the proposition is true for prod-simplicial complexes of dimension less than $n$. Then $S,T$ both have $n$ edges and both may be written as the $k_S$, and $k_T$ fold grafting of the join with a point of smaller trees. These two trees give rise to two different products of cones. But since the associated pro-simplicial complexes are isomorphic, the two products of cones must be the same by the above argument. Now apply the induction to each factor to see that each cone has a unique tree.

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