Recall that a (combinatorial) simplicial complex $X$ is said to be $n$-dimensional if it contains at least one face of dimension $n$ and no faces of dimension $n+1$. Further, an $n$-dimensional finite simplicial complex $X$ is said to be *pure* if every face $\delta$ of dimension $k\lt n$ is a face of some $n$-dimensional face of $X$.


For simplicial sets, we have a different definition for dimension given as:


A simplicial set $S$ is said to be $n$-skeletal if the inclusion $\operatorname{Sk}^n S \subseteq S$ of the $n$-skeleton of $S$ is an isomorphism. We say that $S$ is $m$-dimensional if $m$ is the smallest number for which $S$ is $m$-skeletal. 


Given a simplicial complex $X$ as a simplicial set, we see that the associated simplicial set $\Delta[X]$ is $n$-dimensional if and only if $X$ is $n$-dimensional. 


Is there a useful notion extending the definition of purity to those simplicial sets with only finitely many nondegenerate simplices?


The property that I'm hoping for in such an extension is sort of a generalization of the "prism decomposition" for products of simplices. That is, for the product of two simplicial sets $S$ and $T$, pure of dimension $s$ and $t$ respectively, I would like the product $S\times T$ to be pure of dimension $s+t$. It is certainly the case that the product has dimension $s+t$ (this can be seen working simplex by simplex), but without a proper definition of purity, we can't give a real generalization.



Extending the definition of "pure of dimension n" from simplicial complexes to simplicial sets?
I [EDIT: almost] would follow the suggestion you made in your comment. 


[EDIT: below the line was my original answer, which is wrong, as pointed out by Karol in the comments. Here is an answer which is less functorial, but I hope more correct.]


* It would be nice to define pure dimension internal to categories other than $Set$. Unfortunately the definition so far need the internal logic to satisfy excluded middle, and I'm not sure how to get around this. My intuition is that a finitely extensive regular category should be the minimum requirement.


Assume $S$ is $n$-dimensional (i.e. $n$-skeletal). Let $ND(S\_n) \subset S\_n$ be the non-degenerate $n$-simplices (in Harry's original example this is a finite set), that is, the compliment of the joint image of 
$$
s\_{i\_1}\ldots s\_{i\_{n-k}} : \coprod\_{k = 0}^{n-1} S\_k \to S\_n
$$
(we can do this step for simplicial objects in a Boolean topos, or more generally in a regular category with complements of all subobjects and finite coproducts). This defines a presheaf on $\tilde\Delta\_n$, the subcategory of $\Delta$ consisting of objects $\le n$ and only the coface maps. The $n$-simplices are $ND(S\_n)$ and all lower dimensional simplices are the same as for $S$. Unfortunately this construction is *not* functorial, because a map of simplicial sets may send non-degenerate simplices to degenerate ones.


**Definition**: A presheaf $S'$ on $\tilde\Delta\_n$ is of *pure dimension* if the collection of maps


$$
d\_{i\_1}\ldots d\_{i\_{n-k}} : S'\_n \to \coprod\_{k = 0}^{n-1} S'\_k
$$


is jointly regular epimorphic. This step works in any regular category with finite coproducts. This differs from Harry's suggestion in that we ask that *all* $k$-simplices are a face of a non-degenerate $n$-simplex.


I'll have a think about the proof that pure dimension, defined this way, is additive. It is here that I think the extensivity should be used.




---


I believe there is a functor
$$
R:Set^{\Delta^{op}} \to Set^{\tilde\Delta^{op}}
$$
(where $\tilde\Delta$ is the subcategory of $\Delta$ with the same objects but only coface maps) that takes a simplicial set and removes all degenerate simplices. If I am not mistaken, this should be adjoint to one of the adjoints to the restriction along the inclusion $\tilde\Delta \to \Delta$, the one which adds the smallest number of degenerate simplices. (Check this, I am in a rush to leave on an interstate drive)


There are analogues $R\_n$ for the finite subcategories of $\Delta$ consisting of all objects $\le n$.


Assume $S$ is $n$-dimensional. Define it to be of *pure dimension* if $S' := R\_n sk\_n S$ satisfies:


* for all $m \le n$, the face maps $d\_i : S'\_m \to S'\_{m+1}$ are jointly surjective.


I think the functors $R,R\_n$ preserve products, as do the skeleton functors, so showing that dimension is additive should be ok.


