Hi Tilman, it seems as we are teaching the same course at the same time! I will do this in about three weeks.
There are proofs of that theorem in the book of May and in tom Dieck's new volume. May explains the conceptual idea a bit, and tom Dieck puts it into a context, relating it to other proofs (for example the bundle homotopy theorem).
I think the underlying idea is very natural and simple. The glueing is definitly required because the theorem is a local-to-global statement and paracompactness of the base is needed since paracompact spaces are the right class of spaces for local-to-global arguments.
Here is my version of the idea. Filling all the details is technical and probably leads to the proofs that you find in the literature, but it might be helpful to know the idea (and it could be an interesting exercise for you or me to work out the details, only consulting the first section of chapter 13 of tom Diecks book). Let $p:E \to B$ be a map, $\gamma$ be a path in $B$ and $p(y)=\gamma(0)$. Consider the space $X_{y,\gamma}$ of paths $\delta:[0,1] \to E$ beginning at $y$ and covering. $p$ has the path-lifting property with respect to a point iff $X_{x,\gamma}$ is $-1$-connected (i.e. nonempty) for all $x$ and $\gamma$. If $p$ is a local fibration, then $X_{y,\gamma}$ is nonempty: If the path $\gamma$ is short (stays in one of the sets where $p$ is known to be a fibration), this is clear. Otherwise, the path goes through a finite number of these sets and you have to subdivide the intervals to see that $X_{y,\gamma}$ is nonempty. But you see why you have to subdivide the interval and this subdivision is needed for the same reason as in the proof of the excision theorem in homology (but the direction is reverse: for excision, you need to localize, here we globalize).
In order to prove the Hurewicz-Huebsch theorem, you have to produce such lifts of paths in families. Let $X \times [0,1] \to B$ be a homotopy; I like to think of it as a family of paths parametrized by $X$. For any point on $X$, you find a nonempty space of local lifts. Pick such lifts, for a locally finite cover $(U_i)$, $i\in I$, of $X$; what you want is to glue them together with a partition of unity. For that, you need homotopies on the overlaps $U_{ij}$, so you better know in advance that the space of lifts on any $U_i$ is connected. But there might be triple overlaps $U_{ijk}$ and you want to glue the three lifts on $U_{ij}$, $U_{ik}$ and $U_{jk}$. So the space of local lifts is better $1$-connected. In an obvious continuation of this pattern, you see that you need to know that the space of local lifts is $\infty$-connected (contractible). In fact, if $p$ is a fibration, then the space of lifts is contratible; you need some formal nonsense tricks and the definition of a fibration for that.
Let us go back to a local fibration $E \to B$ and a path $\gamma$ in $X$. As long as your path is short, the above argument for fibrations tells you that the space of lifts is not merely $-1$- but $\infty$-connected. This is also true if the path is not short, again by subdividing the interval. Do the same argument in small families and globalize by the following construction: for any finite nonemtpy $S \subset I$, pick a map from the simplex on $S$ to the space of lifts over $U_S =\cap_{i \in S} U_i$, in a compatible way for inclusions of index sets (induction on $|S|$ and contractibility of the spaces of local lifts). Finally, use a partition of unity.
Once your students understand this idea (to construct something with a local flavour on a space, construct it locally, show the the space of local constructions is contractible, use this contractibility and partitions of unity to glue these local things together), then the bundle homotopy theorem is easy-going. In fact, if you think about the content of the bundle homotopy theorem and the theorem "fibre bundles are Hurewicz fibrations", you'll find out that they follow from each other by rather formal arguments. In my course, I will teach the bundle homotopy theorem and then "fibre bundles are Hurewicz fibrations" as a consequence, omitting the general case of the Hurewicz-Huebsch theorem.
I hope this helps and you see why all these annyoing technicalities enter. As for the axiom of choice: Im afraid I cannot help you since I never care about this :-))
$\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.
Setup
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
$$ \begin{array}{ccc}
\real{\Delta^n} & \overset{h}{\To} & \real{\Sing{X}} \\
\Big\downarrow\rlap{\scriptstyle \iota_0} & & \Big\downarrow\rlap{\scriptstyle \counit_X} \\
\real{\Delta^n}\times I & \underset{H}{\To} & X
\end{array} $$
Constructions (adapted from Oscar Randall-Williams' answer)
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:
- $j$ restricted to $\real{K}$ is the canonical inclusion $\real{K}\hookrightarrow D$ of the end of the mapping cylinder;
- $j([x,t]) = [h(x),x,t]$ for $(x,t)\in \real{\Delta^n} \times I$ (where we see $D$ as a quotient of $\real{K}\times\real{\Delta^n}\times I$).
There is a further map $G:C\to X$ determined by:
- $G$ restricted to $\real{K}$ coincides with $\counit_X$;
- $G$ restricted to $\real{\Delta^n}\times I$ coincides with $H$.
Main argument: $j$ is a trivial Hurewicz cofibration
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.
Conclusion
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.
Proof of the lemma
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:
- $A$ is homeomorphic to a closed subspace of $\RR^N$ for some $N\in\NN$;
- for any $N\in\NN$, if $B$ is a closed subspace of $\RR^N$ which is homeomorphic to $A$, then some neighbourhood of $B$ in $\RR^N$ retracts onto $B$.
[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
You've already answered your own question, but here is another example.
Let $f: \mathbb{Q}^\delta \to \mathbb{Q}$ be the obvious map from the rational numbers with the discrete topology to the rational numbers with the usual topology. Let $M_f$ be the mapping cylinder. Then the projection $$p:M_f \to [0,1]$$ is a Serre fibration. This follows because any map of a disc into $\mathbb{Q}$ factors through f. Hence as far as discs are concerned, $M_f$ might as well be $\mathbb{Q}^\delta \times [0,1]$.
However this projection is not a Dold fibration. It is easy to construct a diagram using $Y = \mathbb{Q}$ which will have no weak homotopy lift. Indeed consider the projection map $\mathbb{Q} \times [0,1] \to [0,1]$ with the obvious initial lift. Any other initial lift vertically homotopic to this one in fact coincides with this one, hence it is easy to see that there is no weak lift of this map.
By replacing $\mathbb{Q}^\delta$ and $\mathbb{Q}$ with their cones, we get a similar example where now the base, total space, and fibers are contractible. Hence they are path-connected and also have the homotopy type of CW complexes. So this also answers Ronnie Browns question (but surely the answer to that has been known for some time).