gn.general topology - When is the realization of a simplicial space compact ?

Ok I checked the idea mentioned in the comments above.

As $X_n$ is a closed subspace of $X_M$ (use any degeneracy; it has a left inverse; composing both the other way round yields a projection and projections have closed images in the Hausdorff setting), it is enough to show, that $X_M$ is compact.

So consider the "generalized midpoint map"

$i: X_Mrightarrow coprod_{nle M} X_ntimes Delta^nrightarrow |X|qquad xmapsto [x,c]$, where $cinDelta^M$ denotes the barycenter.

For example, if $M=2$ and $x$ is a nondegenerate one simplex in $X$, there are two degeneracies $s_0,s_1:X_1rightarrow X_2$. Then one gets in |X|:

$i(s_0(x))=[s_0(x),frac{1}{3},frac{1}{3},frac{1}{3}]=[x,frac{2}{3},frac{1}{3}]$

$i(s_1(x))=[s_1(x),frac{1}{3},frac{1}{3},frac{1}{3}]=[x,frac{1}{3},frac{2}{3}]$

So in the realisation one really picks several midpoints of a simplex, that doesn't have maximal dimension.

But the map $i$ is still injective. Any simplex $xin X_M$ might be written in a unique way as $x=s(y)$, where $yin X_n$ is nondegenerate and $s$ is a degeneracy map (it corresponds to a surjective map $[M]rightarrow [n]$ in $Delta^{Op}$, $Delta^{Op}$ can be identified with the category whose objects are the sets ${0,ldots,n}$ and whose morphisms are nondecreasing maps).

Then $i(x)=[y,s(c)]=[y,frac{s^{-1}(1)}{3},ldots,frac{s^{-1}(1)}{3}]$. Now the right side has normal form (There is a normal form for points in $|X|$. This can be found in every book about simplicial sets and it works the same way for simplicial spaces). Given any other $x'=s'(y')$ with $i(x)=i(x')$, we get

$[y,frac{|s^{-1}(1)|}{M+1},ldots,|frac{s^{-1}(1)|}{M+1}]=[y',frac{|s'^{-1}(1)|}{M+1},ldots,frac{|s'^{-1}(n)|}{M+1}]$. As this is in normal form, we get $y=y'$ and $|s'^{-1}(i)|=|s^{-1}(i)|$. But such nondecreasing maps are characterized by the size of the preimages, so $s=s'$ and hence $x=x'$. So the map $i$ is injective.

We still have to show, that $i$ is a homeo onto its image and that the image of $i$ is closed. Then $X_M$ is a closed subspace of a compact space and hence compact. Both will follow from the next claim:

For any closed subset $Asubset X_M$ the saturation $A'$ of $Atimes {c}$ is still a closed subset of $coprod_{nle M} X_ntimes Delta^n$. This is an even worse calculation than the last one. One has to consider for each degeneracy map $s:X_Nrightarrow X_M$ the preimage $s^{-1}(A)times {s(c)}$ and the set $B_s$ of all points that can be simplified to one of those points. Then one can show, that $B_s$ is closed and $A'=bigcup_s B_s$. As there are only finitely many degeneracy maps (below dim $M$) this is a finite union and hence closed.