I was wondering, what a $N$-dimensional simplicial space $X$ should be. Of course the degeneracy maps force the spaces to be nonempty in high dimensions. Currently I have two different versions and i am wondering whether they are equivalent:


1) There are no nondegenerate simplexes above dimension $N$. 


2) Let $\Delta|\_N$ be the full subcategory of $\Delta$ consisting of the objects $[0],\ldots ,[N]$. The inclusion induces a restriction functor $R$ from $\Delta-$spaces to $\Delta|\_N$-spaces, which has a left adjoint $L$. There is a canonical map $L(R(X))\rightarrow X$. $X$ should be called $N$-dimensional, iff this map is an isomorphism,i.e. a homeomorphism on each object of $\Delta$.


To make the question more precise: $2)\Rightarrow 1)$ can be easily seen, as the map $L(R(X))\rightarrow X$ cannot hit any nondegenerate simplex above dimension $N$ by construction. The other way round:
Given $1)$, then all the maps occuring in the natural transformation are surjective and continuous. Injectivity should follow from the relations in $\Delta$. So why is the inverse map of sets continuous?


(In the category of simplicial sets (and not spaces) both notions should be equivalent using the same argumentation.)



Notion of finite dimensional simplicial space
Interesting question!


I'm cautiously optimistic that 1) $\Rightarrow$ 2). As you say, if $X$ satisfies 1), then $L(R(X))\to X$ is a continuous bijection. (Because it's a statement about point-sets, and the thing is true for simplicial sets.)


If $N=0$, it should be easy: $L(R(X))$ is the constant simplicial space, whose value at each [k] is $X\_0=X([0])$. The canonical map $L(R(X))\to X$ is the one which at degree $k$ is given by the map $s:X\_0\to X\_k$ defined by the composite of degeneracy operators. But we know that the composite $X\_0\to X\_k\to X\_0$ is the identity, where $d:X\_k\to X\_0$ is a composite of face operators. So if $s$ is a bijection, $d$ is its continuous inverse.


For general $N$, let $Y=L(R(X))$. The functor $R$ also has a right adjoint, which I'll call $M$. Let $Z=M(R(X))$. Just as the space $Y\_k$ looks like a colimit of a certain diagram of the spaces $X\_0,\dots,X\_N$, the space $Z\_k$ looks like a limit of a certain diagram of these spaces.


There are canonical maps $Y=L(R(X)) \to X \to M(R(X))=Z$. I would like to claim that the composite $f:Y\to Z$ gives a homeomorphism of $Y$ onto its image. If you can prove this, that will give your result, since a continuous inverse to $Y\to X$ will be given by $X\to f(Y)\approx Y$.


