# Simplicial space whose all face/degeneracy maps are homotopy equivalences

I believe that the following is true, but I cannot find a proof. Let $X_\bullet$ be a simplicial topological space (I can add that my $X_\bullet$ comes from a bisimplicial set, so the spaces $X_n$ are CW-complexes). Suppose that all the face and degeneracies maps are homotopy equivalences. Is it true that the geometric realization $|X_\bullet|$ is homotopy equivalent to the space $X_0$? (Possibly under some mild extra-hypotheses, e.g. connectedness of the $X_n$, etc.?)

Regard $X_0$ as the constant simplicial space at $X_0$. The natural simplicial map $X_*\to X_0$ induces a homotopy equivalence of realizations provided that $X$ is Reedy cofibrant, which means that the inclusion of the degeneracy subspace $sX_{n-1}$ in $X_n$ is a cofibration for each $n$. (e.g. Theorem A.4 in [13] on my web page). The Reedy condition is unnecessary if you use the fat realization (ignore degeneracy operations).
• Thank you very much for your answer Professor May! I am not really familiar with the Reedy condition, so may I ask if the Reedy condition is in this case ($X_\bullet$ coming from a bisimplicial set) always satisfied? Does it have to do with the fact that every inclusion of simplicial sets is a cofibration? Thanks again – AlexP Mar 16 '13 at 14:04
• I see. Maybe I was not precise enough in my last comment, so let me restate it in a more detailed way. My simplicial space $X_\bullet$ comes from a bisimplicial set $X_{\bullet\bullet}$, in such a way that each space $X_n$ is the geometric realization of the simplicial \it set \rm $X_{n,\bullet}$, i.e., for each $n$ we have $X_n=|X_{n,\bullet}|$. So the simplicial space to which I refer in my question arises in fact as the levelwise geometric realization of a bisimplicial set. My latter question was: if this is the case, is the Reedy condition satisfied? It seems so to me. Thank you again. – AlexP Mar 16 '13 at 17:03