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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 $b$.

It is easy to see that these maps induce isomorphisms on $\pi_n$ for $n\geq 1$, but I wonder about $\pi_0$.

Question: Is it true that $p^{-1}(b) \to F_b$ is a weak homotopy equivalence?

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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.

I am guessing that your proof for $\pi_n$ with $n$ positive is a five lemma argument?

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.

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Yeah, this is what I did, but the action only got me surjectivity on $\pi_0$. I'll try again. – Jeff Strom Jan 30 '11 at 22:52

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.

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Harry, You're using an anti-tank missile to prove a result that requires no more an a pea-shooter. – John Klein Jan 30 '11 at 0:48
To be fair, that does sound like something I would do. However, I don't really see it as a flaw of my answer, but rather as evidence of the power of Quillen's framework! – Harry Gindi Jan 30 '11 at 15:53
Before you can use the general framework, you have to know that Top is a model category, and this requires some basic topological results about weak homotopy equivalences and fibrations, in the same ballpark as what is being asked about here. – Tom Goodwillie Jan 30 '11 at 16:27
I like the proof that Top is a model category that goes through sSet (and doesn't rely on too much knowledge about $CGWH$!). That is, use Cisinski's framework or Quillen's proof using Kan's $Ex^\infty$ to construct the model structure on sSet and lift it to $CGWH$. – Harry Gindi Jan 30 '11 at 17:10
@Scott: That's obvious though. I've actually gone through this proof before, and nothing like that is necessary. Because boundary inclusions for simplices correspond under realization to boundary inclusions for disks, and horn inclusions correspond to the other correct maps. The nontrivial part of the proof is verifying the conditions for lifting. I like this method of constructing the model structure on Top because the model structure on sSet is extremely canonical (it is the unique model structure for which the cofibrations are exactly the monics, and such that all maps... – Harry Gindi Jan 31 '11 at 5:42

Dear Jeff, The map you describe is a homotopy equivalence. This is proved as Proposition 1.1 in the paper

Varadarajan, K. On fibrations and category. Math. Z. 88 1965 267–273.

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Thank you, I'll look into this. – Jeff Strom Mar 6 '11 at 17:11
There's also a proof of this for Hurewicz fibrations in Hatcher, Proposition 4.65. – Mark Grant Mar 9 '11 at 12:23
Varadarajan only talks about Hurewicz fibrations in his paper. – David Roberts Dec 13 '12 at 0:51
@David: Hurewicz fibrations are more general than Serre fibrations. – Mark Grant Dec 13 '12 at 7:02
Isn't it the other way around? A Hurewicz fibration is, in particular, a Serre fibration. Hence Serre fibrations are more general. – Chris Schommer-Pries Jun 24 '13 at 7:35

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