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.?)

Thanks in advance!


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).

  • $\begingroup$ Beat me to it! :) $\endgroup$ – Dylan Wilson Mar 16 '13 at 13:38
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    $\begingroup$ (With a stronger result and more details too...) $\endgroup$ – Dylan Wilson Mar 16 '13 at 13:40
  • $\begingroup$ 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 $\endgroup$ – AlexP Mar 16 '13 at 14:04
  • $\begingroup$ No, not always, and your question is about spaces not simplicial sets. The most verifiable condition is usually that each degeneracy map s_i is a cofibration. $\endgroup$ – Peter May Mar 16 '13 at 16:30
  • $\begingroup$ 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. $\endgroup$ – AlexP Mar 16 '13 at 17:03

An easy spectral sequence argument tells us that the natural map induces an isomorphism for every generalized cohomology theory... so that deals with everything modulo fundamental groups. In general maybe you have to ask for the degeneracies to be cofibrations, then this might follow from a model structure argument (the geometric realization would be a hocolim.) (The spectral sequence mentioned above is in, for example, Segal's paper "Classifying Spaces and Spectral Sequences.")


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