CW-complex, Hausdorff space, second-countable space, sober space
connected space, locally connected space, contractible space, locally contractible space
Serre fibration $\Leftarrow$ Hurewicz fibration $\Rightarrow$ Dold fibration $\Leftarrow$ shrinkable map
Write
for the set of inclusions of the topological n-disks, into their cylinder objects, along (for definiteness) the left endpoint inclusion.
A Serre fibration is a $J_{Top}$-injective morphism, def. 1, hence a continuous function $f \colon X \longrightarrow Y$ that has the right lifting property with respect to all inclusions of the form $(id,0) \colon D^n \hookrightarrow D^n \times I$ that include the standard topological n-disk into its standard cylinder object.
I.e. $f$ is a Serre fibration if for every commuting square of continuous functions of the form
then there exists a continuous function $h \colon D^n \to X$ such as to make a commuting diagram of the form
The condition in def. 2 is part of the condition on a Hurewicz fibration, hence every Hurewicz fibration is in particular a Serre fibration.
The class of Serre fibrations serves as the class of abstract fibrations in the classical model structure on topological spaces (whence “Serre-Quillen model structure”).
A Serre fibration has the right lifting property against all retracts of $J_{Top}$-relative cell complexes (def. 1).
By general closure properties of projective and injective morphisms, see there this proposition for details.
Since Serre fibrations are the abstract fibrations in the Serre-classical model structure on topological spaces, the following statement follows from general model category theory. But it may also be seen by direct inspection, as follows.
For $X$ a finite CW-complex, then its inclusion $X \overset{(id, \delta_0)}{\longrightarrow} X \times I$ into its standard cylinder is a $J_{Top}$-relative cell complex (def. 1).
First erect a cylinder over all 0-cells
Assume then that the cylinder over all $n$-cells of $X$ has been erected using attachment from $J_{Top}$. Then the union of any $(n+1)$-cell $\sigma$ of $X$ with the cylinder over its boundary is homeomorphic to $D^{n+1}$ and is like the cylinder over the cell “with end and interior removed”. Hence via attaching along $D^{n+1} \to D^{n+1}\times I$ the cylinder over $\sigma$ is erected.
Let $f\colon X \longrightarrow Y$ be a Serre fibration, def. 2, let $y \colon \ast \to Y$ be any point and write
for the fiber inclusion over that point. Then for every choice $x \colon \ast \to X$ of lift of the point $y$ through $f$, the induced sequence of homotopy groups
is exact, in that the kernel of $f_\ast$ is canonically identified with the image of $\iota_\ast$:
It is clear that the image of $\iota_\ast$ is in the kernel of $f_\ast$ (every sphere in $F_y\hookrightarrow X$ becomes constant on $y$, hence contractible, when sent forward to $Y$).
For the converse, let $[\alpha]\in \pi_{\bullet}(X,x)$ be represented by some $\alpha \colon S^{n-1} \to X$. Assume that $[\alpha]$ is in the kernel of $f_\ast$. This means equivalently that $\alpha$ fits into a commuting diagram of the form
where $\kappa$ is the contracting homotopy witnessing that $f_\ast[\alpha] = 0$.
Now since $x$ is a lift of $y$, there exists a left homotopy
as follows:
(for instance: regard $D^n$ as embedded in $\mathbb{R}^n$ such that $0 \in \mathbb{R}^n$ is identified with the basepoint on the boundary of $D^n$ and set $\eta(\vec v,t) \coloneqq \kappa(t \vec v)$).
The pasting of the top two squares that have appeared this way is equivalent to the following commuting square
Because $f$ is a Serre fibration and by lemma 1 and prop. 1, this has a lift
Notice that $\tilde \eta$ is a basepoint preserving left homotopy from $\alpha = \tilde \eta|_1$ to some $\alpha' \coloneqq \tilde \eta|_0$. Being homotopic, they represent the same element of $\pi_{n-1}(X,x)$:
But the new representative $\alpha'$ has the special property that its image in $Y$ is not just trivializable, but trivialized: combining $\tilde \eta$ with the previous diagram shows that it sits in the following commuting diagram
The commutativity of the outer square says that $f_\ast \alpha'$ is constant, hence that $\alpha'$ is entirely contained in the fiber $F_y$. Said more abstractly, the universal property of fibers gives that $\alpha'$ factors through $F_y\overset{\iota}{\hookrightarrow} X$, hence that $[\alpha'] = [\alpha]$ is in the image of $\iota_\ast$.
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long exact sequence of homotopy groups
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