I wonder why in this proof: since the fibre $p^{-1}(b_0)$ is weakly contractible. Hence $h′_{\alpha}$ extends to a map $h_{\alpha}:D^n\to E$?
I want to show if the boundary (the sphere of radius 1/2) is mapped to a weakly contractible space $p^{-1}(b_0)$, then we have an extension on $D^n(1/2)$ to $p^{-1}(b_0)$. But I have no clue how to use the property that $p^{-1}(b_0)$ is weakly contractible.
The general claim is this: Each map $f : S^k \to Z$ into a weakly contractible $Z$ has an extension $F : D^{k+1} \to Z$.
$Z$ weakly contractible means that all homotopy groups $\pi_k(Z,z_0)$ are trivial for some basepoint $z_0$. For $k=0$ this shows that $Z$ is path connected, hence the $\pi_k(Z,z)$ are trivial for all basepoints $z$. But this implies that each map $f : S^k \to Z$ is inessential. This provides the desired extension $F$.