most recent 30 from http://mathoverflow.net 2013-05-19T00:10:54Z http://mathoverflow.net/feeds/question/53729 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/53729/the-fiber-of-a-serre-fibration Jeff Strom 2011-01-29T17:41:51Z 2011-03-06T15:03:19Z

<p>If $p:E\to B$ is a Serre fibration (assume it is surjective), then for each $b\in B$ we get a comparison map $p^{-1}(b) \to F_b$, where $F_b$ is the homotopy fiber of $p$ over $b$. </p> <p>It is easy to see that these maps induce isomorphisms on $\pi_n$ for $n\geq 1$, but I wonder about $\pi_0$. </p> <p>Question: Is it true that $p^{-1}(b) \to F_b$ is a weak homotopy equivalence?</p>

http://mathoverflow.net/questions/53729/the-fiber-of-a-serre-fibration/53748#53748 Tom Goodwillie 2011-01-29T23:37:53Z 2011-01-30T16:07:19Z

<p>Yes, directly from the definition of fibration. And I see no advantage in assuming that the map is surjective. The fiber is empty if and if the homotopy fiber is empty.</p> <p>I am guessing that your proof for $\pi_n$ with $n$ positive is a five lemma argument?</p> <p>EDIT Now that I think about it, maybe my preferred proof goes by extending the 5 lemma argument a little. Like this: The homotopy fiber of $E\to B$ over $b\in B$ is the fiber of an associated fibration, call it $E'\to B$. There is a map $E\to E'$ over $B$ inducing a map of fibers $F\to F'$. You know that $E\to E'$ is a homotopy equivalence and so gives a bijection of $\pi_0$ and (for every basepoint in $E$) of $\pi_n$. To conclude that the associated map $F\to F'$ also induces such bijections, use a sort of modified 5 lemma argument. The key is that you have an action of $\pi_1(B,b)$ on $\pi_0(F)$ such that (a) for every point $f\in F$ the stabilizer of its class is the image of $\pi_1(E,f)$ and (b) two elements of $\pi_0(F)$ go to the same element of $\pi_0(E)$ if and only if they are in the same orbit.</p>

http://mathoverflow.net/questions/53729/the-fiber-of-a-serre-fibration/53749#53749 Harry Gindi 2011-01-29T23:59:07Z 2011-01-30T00:19:49Z

<p>It's a weak equivalence by the fact that it's a right proper model category (since all objects are fibrant), and therefore the fiber of a fibration is a homotopy pullback (and a hofiber). In particular, the induced map between two representatives of a homotopy pullback is necessarily a weak equivalence.</p>

http://mathoverflow.net/questions/53729/the-fiber-of-a-serre-fibration/57577#57577 Mark Grant 2011-03-06T15:03:19Z 2011-03-06T15:03:19Z

<p>Dear Jeff, The map you describe is a homotopy equivalence. This is proved as Proposition 1.1 in the paper</p> <p>Varadarajan, K. On fibrations and category. Math. Z. 88 1965 267–273.</p>