A fibration need not be a locally trivial bundle

For any ordinary parameter space $Z$, let the initial map be $f(z)=(a(z),b(z))\in E$ and let $h:Z\times I\to [0,1]$ satisfy $h(z,0)=a(z)$. Define $L(z,t)=(h(z,t),\min (b(z),h(z,t)))$. Its coordinates are continuous by F3–F4; the same coordinate check into the Euclidean subspace gives continuity into $E$. Indeed $0\le \min (b(z),h(z,t))\le h(z,t)\le 1$, so its range is in $E$.

The fiber over zero is the singleton $\{(0,0)\}$, while for every $x>0$ the fiber contains the distinct points $(x,0)$ and $(x,x)$ and is homeomorphic to $[0,x]$. Any relatively open neighbourhood of zero in $[0,1]$ contains some $x>0$. A chart over such a neighbourhood would, by F2, identify both fibers with one fixed fiber, forcing a singleton to be in bijection with a set containing two distinct points. This is impossible, so no bundle chart exists at zero.

At time zero, $b(z)\le a(z)$ implies $\min (b(z),a(z))=b(z)$, hence $L(z,0)=f(z)$. Also $pL=h$ at every time. This establishes F1 for every space $Z$, proving $p$ Hurewicz without any universal path-space theorem or choice. It also proves the CGWH test conclusion because the triangle and interval are ordinary compact Hausdorff spaces and their interval cylinders have the same topology.

Empty test spaces in step 1.1 give the empty lift; one-point tests give the same explicit path lift. If $a(z)=0$ then $b(z)=0$ and the lift is $(h(z,t),0)$, so emergence from the collapsed fiber is continuous. If $h=0$, the lift is $(0,0)$; if $h=a$ is constant in time, the formula fixes the original point. At $x=1$ and at both homotopy endpoints the same inequalities remain valid. Thus the singleton/interval change does not obstruct HLP but does obstruct local triviality, as claimed.