Is the mapping cylinder of a Serre fibration also a Serre fibration? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-25T00:44:11Z http://mathoverflow.net/feeds/question/40857 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40857/is-the-mapping-cylinder-of-a-serre-fibration-also-a-serre-fibration Is the mapping cylinder of a Serre fibration also a Serre fibration? Cary 2010-10-02T17:49:58Z 2010-10-02T20:52:05Z <p>If we have a Serre fibration $p: E \rightarrow B$ with fiber of homotopy type $S^{k-1}$, then we can create a fibration with contractible fiber by first taking the mapping cylinder $M_p$ of $p$ to get a map $M_p \rightarrow B$ (not necessarily a fibration) with contractible "fiber" and then applying the "space of paths" construction to get a fibration $M_p \times_B B^I \rightarrow B$. My question is, is this last part of the construction necessary, or is the mapping cylinder $M_p$ already a Serre fibration?</p> <p>I tried lifting a homotopy $f_t: X \times I \rightarrow B$ with starting point $\tilde f_0: X \rightarrow M_p$ by cutting $X$ into the closed preimage $C$ of $B \subset M_p$ and the open preimage $U$ of $E \times [0,1) \subset M_p$. On $C \times I$ we set $\tilde f_t(x) = f_t(x) \in B \subset M_p$. On $U \times I$ we lift $f_t|U: U \times I \rightarrow B$ to $g_t: U \times I \rightarrow E$ and then set $\tilde f_t(x) = (g_t(x),(1-t)(t$ coordinate of $\tilde f_0(x)) + t(1))$. This defines a continuous lift on $C$ and on $U$ separately. If the continuous lift on $U$ extends to the closure of $U$ then we're done. The map $U \rightarrow E$ could be nasty though near the boundary of $U$. Perhaps a better approach is to first construct a map from $X \times I$ that is only "close to" a lift, then use obstruction theory (I'm not an expert on this) to show that it is homotopic to some lift.</p> http://mathoverflow.net/questions/40857/is-the-mapping-cylinder-of-a-serre-fibration-also-a-serre-fibration/40862#40862 Answer by André Henriques for Is the mapping cylinder of a Serre fibration also a Serre fibration? André Henriques 2010-10-02T18:40:29Z 2010-10-02T18:40:29Z <p>I think that your idea works.</p> <p>You propose "$\tilde f_t(x) = (g_t(x),(1-t)(t$ coordinate of $\tilde f_0(x)) + t(1))$", which I don't quite understand. I would simply do: $\tilde f_t(x) = (g_t(x),t$ coordinate of $\tilde f_0(x))$, where $g_t(x)$ is the solution of the corresponding lifting problem for $p$. Note that you've implicitely used that $U$ is a CW-complex, but there's nothing wrong with that.</p> <p>As you say, this defines a continuous lift on $C$ and on $U$ separately, and they glue to a continuous lift on the whole.</p> http://mathoverflow.net/questions/40857/is-the-mapping-cylinder-of-a-serre-fibration-also-a-serre-fibration/40873#40873 Answer by John Rognes for Is the mapping cylinder of a Serre fibration also a Serre fibration? John Rognes 2010-10-02T20:52:05Z 2010-10-02T20:52:05Z <p>Waldhausen, Jahren and myself proved a fiber gluing lemma for Serre fibrations, in the context of simplicial sets, that may be useful. In Propositions 2.7.10 and 2.7.12 of "Spaces of PL manifolds and categories of simple maps"</p> <p><a href="http://folk.uio.no/rognes/papers/plmf.pdf" rel="nofollow">http://folk.uio.no/rognes/papers/plmf.pdf</a></p> <p>we prove that given:</p> <ul> <li><p>a diagram of simplicial sets $Z_1 \leftarrowtail Z_0 \to Z_2$, where one map is a cofibration,</p></li> <li><p>a sufficiently nice base simplicial set $B$ (a simplicial complex will do), and</p></li> <li><p>compatible maps $Z_i \to B$ that become Serre fibrations upon geometric realization,</p></li> </ul> <p>then the pushout map $Z_1 \cup_{Z_0} Z_2 \to B$ becomes a Serre fibration upon geometric realization.</p> <p>Mapping cylinders are a special case of pushouts. If $p \colon E \to B$ becomes a Serre fibration upon realization, then so do the obvious map $E \times \Delta^1 \to B$ and the identity map $B \to B$. The pushout map is your map $M_p \to B$, and our conclusion is that its realization is a Serre fibration.</p> <p>Our proof depends on working with simplicial sets. The technical condition on $B$ is that each nondegenerate simplex is embedded.</p> <ul> <li>John</li> </ul>