at.algebraic-topologyfibrationhomotopy-theory
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?
There are many "homotopy fiber bundles" which are not fibrations. For example take any homotopy equivalence $E \to B$ that is not a fibration, then it is a "homotopy fiber bundle" with a one-point fiber.
On the other hand the other implication is (almost) true. The following works for Hurewicz fibrations. I don't whether it is true that a fiber of a Serre fibration between CW-complexes has a homotopy type of a CW-complex. If this is the case, then the proof works also for Serre fibrations.
Let $\pi : E \to B$ be a Hurewicz fibration between CW-complexes and $x \in B$. CW-complexes are locally contractible, so there is a contractible neighborhood $U$ of $x$. Let $\pi_U : E_U \to U$ be the restriction of $\pi$ to $U$. If $E_x$ is the fiber of $\pi$ at $x$, then the inclusion $E_x \to E_U$ is a pullback of the inclusion $\{x\} \to U$ along a Hurewicz fibration, so it is a homotopy equivalence and admits a homotopy inverse $f : E_U \to E_x$. Thus $(\pi_U, f) : E_U \to U \times E_x$ is a homotopy equivalence over $U$.
$\newcommand{\real}[1]{\left\lvert #1 \right\rvert}$$\newcommand{\Sing}[1]{\operatorname{Sing}(#1)}$$\newcommand{\counit}{\epsilon}$$\newcommand{\To}{\longrightarrow}$$\newcommand{\proj}{\mathrm{proj}}$$\newcommand{\NN}{\mathbb{N}}$$\newcommand{\RR}{\mathbb{R}}$Yes, the map $\real{\Sing{X}} \to X$ is a Serre fibration.
[Disclaimer: This answer is very long. A lot of what I will write is contained in Oscar's answer and in the comments. I present it here for completeness and convenience.]
The present answer adapts the constructions given by Oscar Randall-Williams in his answer. The missing point is to prove that the maps generalizing the ones appearing in Oscar's answer are always inclusions of retracts. We will actually prove that they are trivial cofibrations, which will fundamentally require the fact that finite cell complexes are Euclidean neighbourhood retracts. Please upvote Oscar's answer.
Let $X$ be a topological space, and let $\counit_X:\real{\Sing{X}}\to X$ be the counit map of the adjunction between the singular complex functor and the geometric realization functor. We want to show that $\counit_X$ is a Serre fibration. Let then $h:\real{\Delta^n} \to \real{\Sing{X}}$ and $H:\real{\Delta^n} \times I = \real{\Delta^n \times \Delta^1}\to X$ be continuous maps. We need only prove that it is possible to provide a diagonal lift for the diagram $$ \begin{array}{ccc} \real{\Delta^n} & \overset{h}{\To} & \real{\Sing{X}} \\ \Big\downarrow\rlap{\scriptstyle \iota_0} & & \Big\downarrow\rlap{\scriptstyle \counit_X} \\ \real{\Delta^n}\times I & \underset{H}{\To} & X \end{array} $$
The space $C$. Since simplices are compact, the image of $h:\real{\Delta^n} \to \real{\Sing{X}}$ is contained in the realization $\real{K}$ of some finite sub-simplicial set $K$ of $\Sing{X}$. Then $h$ restricts to a map $h:\real{\Delta^n}\to\real{K}$, and we can take the mapping cylinder of $h$: $$ C = M_h = \real{K} \coprod_{\real{\Delta^n}} \bigl( \real{\Delta^n}\times I \bigr) $$ This mapping cylinder generalizes the space $C$ described by Oscar Randall-Williams in his answer.
The space $D$. The preceding space $C$ includes naturally into the mapping cylinder $$ D = M_{\proj_{\real{K}}} = \real{K} \coprod_{\real{K}\times\real{\Delta^n}} \bigl( \real{K}\times\real{\Delta^n}\times I \bigr) $$ of the projection map $\proj_{\real{K}} : \real{K}\times\real{\Delta^n} \to \real{K} $. Importantly, observe that since geometric realization preserves colimits and finite products, the space $D$ is also the geometric realization $$ D = \real{M_{\proj_K}} $$ of the simplicial mapping cylinder $M_{\proj_K} = K \coprod_{K\times\Delta^n} (K\times\Delta^n\times\Delta^1)$ of the projection $\proj_K: K\times\Delta^n \to K$.
The space $D$ plays here the role of the join appearing in Oscar Randall-Williams' answer: note that the join $\real{K}\ast\real{\Delta^n}$ is naturally a quotient of $D$.
The maps. The inclusion map $j:C \to D$ is given by:
There is a further map $G:C\to X$ determined by:
Now that we have adapted Oscar's construction, the main part of the argument consists of showing that $j: C\to D$ is a trivial cofibration.
First, the map $j$ is easily seen to be injective. Since $C$ is compact (because both $\real{K}$ and $\real{\Delta^n}\times I$ are compact) and $D = \real{M_{\proj_K}}$ is Hausdorff, it follows that $j: C\to D$ is a closed map, and in particular a homeomorphism onto its image.
Second, the map $j$ is a homotopy equivalence. Simply note that the composition of $\real{K} \hookrightarrow C \overset{j}{\to} D$ is the canonical inclusion $\real{K} \hookrightarrow M_{\proj_{\real{K}}} = D$ into the mapping cylinder, and thus a homotopy equivalence. Similarly, $\real{K}\hookrightarrow C$ is the inclusion into the mapping cylinder, and thus a homotopy equivalence. By the two-out-of-three property for homotopy equivalences, $j$ is itself a homotopy equivalence.
It remains to show that the inclusion of $j(C)$ into $D$ is a Hurewicz cofibration. Observe that both $C$ and $D$ are finite cell complexes, and in particular are Euclidean neighbourhood retracts (ENRs). This can be proved in the same way as corollary A.10 in the appendix to Allen Hatcher's book "Algebraic topology", which states that finite CW-complexes are ENRs. The desired result is now encoded in the following lemma.
Lemma: Assume $Y$ is a closed subspace of the topological space $Z$. Assume also that $Y$ and $Z$ are both ENRs. Then the inclusion of $Y$ into $Z$ is a Hurewicz cofibration.
This result actually holds in general for ANRs, and is stated as proposition A.6.7 in the appendix of the book "Cellular structures in topology" by Fritsch and Piccinini. Nevertheless, for completeness, I will provide a proof of the lemma at the end of this answer which uses only the results in the appendix of Hatcher's book when $Z$ is compact.
I will now conclude the proof that $\counit_X$ is a Serre fibration.
First, observe that $j:C\to D$ is the inclusion of a strong deformation retract because it is a homotopy equivalence and a Hurewicz cofibration. In particular, $j$ admits a right inverse $r:D\to C$. The composite $\overline{G} = G\circ r:D\to X$ is then an extension of $G:C\to X$ along $j$, i.e. $\overline{G}\circ j = G$.
Now we use the description of $D$ as the realization of the simplicial set $M_{\proj_K}$ to give a diagonal lift for the square diagram at the beginning of this answer. Note that to give a map $\overline{G}:D=\real{M_{proj_K}} \to X$ is by adjunction the same as giving a map $F:M_{proj_K}\to\Sing{X}$. I claim that the composite $$ \widetilde{H} : \real{\Delta^n}\times I \hookrightarrow C \overset{j}{\To} D \overset{\real{F}}{\To} \real{\Sing{X}} $$ is such a diagonal lift:
$\counit_X \circ\real{F}= \overline{G}$ by construction of $F$ via adjunction. Consequently, $\counit_X \circ \real{F}\circ j = G$.
In particular, $\counit_X \circ \real{(F|_K)} = \counit_X \circ (\real{F})|_{\real{K}} = \counit_X \circ \real{F} \circ j|_{\real{K}} = G|_{\real{K}} = \counit_X$. This implies by adjunction that $F|_K$ is the inclusion of $K$ into $\Sing{X}$. Therefore, the restriction $(\real{F})|_{\real{K}} = \real{(F|_K)}$ is the inclusion of $\real{K}$ into $\real{\Sing{X}}$.
It follows that $\widetilde{H}\circ \iota_0 = \real{F}\circ j|_{\real{K}} \circ h = (\real{F})|_{\real{K}} \circ h = h$, i.e. the map $\widetilde{H}$ makes the upper triangle commute.
Furthermore, it follows from item 1 that $\counit_X \circ\widetilde{H} = \counit_X \circ\real{F}\circ j|_{(\real{\Delta^n}\times I)} = G|_{(\real{\Delta^n}\times I)} = H$. So $\widetilde{H}$ makes the lower triangle commute.
We will use (a simplification of) the characterization of Hurewicz cofibrations given in theorem 2 of Arne Strøm's article "Note on cofibrations" (published in Mathematica Scandinavica 19 (1966), pages 11-14).
The inclusion of a closed subspace $Y$ of a metrizable topological space $Z$ is a cofibration if there exists a neighbourhood $U$ of $Y$ in $Z$ which deforms in $Z$ to $Y$ rel $Y$. More explicitly, there must exist a homotopy $F:U\times I\to Z$ such that $F(x,0)=x$ and $F(x,1)\in Y$ for $x\in U$, and $F(y,t)=y$ for $(y,t)\in Y\times I$.
This characterization of closed cofibrations is very similar to (and follows easily from) the usual characterization in terms of NDR-pairs, except that it does not demand the homotopy to be defined on the whole space.
Assuming that $Y$ and $Z$ are ENRs, we will now prove the existence of the neighbourhood $U$ and the homotopy $F$ as above.
A space $A$ is a ENR exactly when:
[For reference, the aforementioned appendix of Allen Hatcher's book "Algebraic topology" explains these two points when $A$ is compact.]
So we may assume without loss of generality that $Z$ is a closed subspace of $\RR^N$, and that some neighbourhood $V$ of $Z$ in $\RR^N$ retracts to $Z$ via a retraction $r_Z:V\to Z$. Since $Y\subset Z\subset\RR^N$ is closed in $\RR^N$ and $Y$ is a ENR, there also exists a neighbourhood $W$ of $Y$ in $\RR^N$ which admits a retraction $r_Y:W\to Y$. Using the convexity of $\RR^N$, we can produce a straight line homotopy $SLH:W\times I\to\RR^N$ between the identity of $W$ and $r_Y$. We may assume without loss of generality (by shrinking $W$ if necessary) that the image of the homotopy $SLH$ is contained in the open $V$. Define then the desired neighbourhood by $U=W\cap Z$ and the homotopy by $F = r_Z \circ SLH$.
Best Answer
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.