We have a $2-$dimensional disk with a segment attached in its center (like a plane umbrella). Let's call this space $T$. And consider the map $\pi: T\to\mathbb{D}^2$ which consists on projecting onto a copy of the $2-$disk. In other words, $\pi$ is the identity everywhere except on the segment (in green), where it maps all points to the center of the disk.
Does this map have the homotopy lifting property?
My thoughts: it should, since homotopically $T\simeq \mathbb{D}^2$ and $\pi$ is at the level of homotopy the identity map. Nonetheless, according to the definition of the homotopy lifting property I don't see it that clearly. In other words, given a point that lifts the center of the disk; let's say $q$ in the picture, and the path consisting in going radially along the disk downstairs, it is clear that there doesn't exist a path upstairs projecting down to the given path (since the point $q$ is ''far'' from any point from the disk in $T$ upstairs). There exists, though, a homotopically equivalent path to the original one that does have a lift but this is an a priori ``weaker'' condition that the one in the definition. Thanks in advance.
$T$ together with the projection $\pi: T\to\mathbb{D}^2$" />
It does not have the homotopy lifting property. You have shown that it does not even have the path lifting property (which may be regarded as a special case of the homotopy lifting property for maps from a one-point space to $D^2$). You can generalize your counterexample as follows: Consider any non-empty space $X$ and define $h : X \times I \to D^2, h(x,t) = t\xi$, where $\xi \in D^2 \setminus \{0\}$. The map $h_0 = h(-,0) : X \to D^2$ has a lift $f : X \to T$ such that $f(x) = q \notin D^2$. But the homotopy $h$ does not have a lift $H : X \times I \to T$ such that $h_0 = f$. In fact, given any lift $H$ of $h$, we have $H(x,t) \in p^{-1}(h(x,t)) = p^{-1}(t\xi)$, in particular $H(x,t) \in D^2$ for $t > 0$, i.e. $H(X \times (0,1]) \subset D^2$. This implies $H(X \times I) = H(\overline{X \times (0,1]}) \subset \overline{H(X \times (0,1]}^T \subset D^2$.