I have a question about a remark which provides an equivalent characterisation for (weak) Serre fibration:
By "classical" definition a map $p:X \to Y$ is a (weak) Serre Fibration if it has following homotopy lifting property:
For every diagramm
$$ \require{AMScd} \begin{CD} \{0\} \times I^{n-1} @>{a} >> X \\ @VViV @VVpV \\ I^n @>{b}>> Y \end{CD} $$
there exist $l:I^n \to X$ such that $a = l \circ i$ and $b = p \circ l$ holds.
But in a remark (source: "Einführung Topology" by Laures, Szymik; page 197, German edition) is mentioned that there exist an equivalent characterisation:
A map $p: X \to Y$ is fibration if it has homotopy lifting property for all inclusions $incl: \{0\} \times I^{n-1} \cup I \times \partial I^{n-1} \to \{0\} \times I^{n-1}$.
My question is why is this condition equivalent to the "classical" one?
The author says that it's because there exist an automorphism of $I^n$ which embedds $\{0\} \times I^{n-1} \cup I \times \partial I^{n-1} $ to$ \{0\} \times I^{n-1}$. The construction is already clear to me, but not why this provides the equivalence of this two characterisation of fibration property.
Here the original excerpt (I don't found an English version):
Take $\phi$ the automorphism of $I^n$ described above, and $i,i'$ the obvious inclusions such that the left square commutes:
$$ \require{AMScd} \begin{CD} \{0\}\times I^{n-1}\cup I\times\partial I^{n-1} @>{\text{incl}} >\sim> \{0\} \times I^{n-1} @>{a} >> X \\ @VVi'V @VViV @VVpV \\ I^n @>{\phi}>\sim> I^n @>b>> Y \end{CD} $$
Then if $i$ can be lifted to some $\ell:I^n\to X$ such that the diagram commutes, then we also have $\ell\circ \phi$ lifting $i'$. Conversely, if $i'$ can be lifted with some $\ell':I^n\to X$, then $\ell'\circ \phi^{-1}$ is a lifting of $i$.