Suppose that $\pi:E \to B$ has the homotopy lifting property, so that for any space Y with a map $f:Y \to E$ and a homotopy $G$ of $g = \pi \circ f$, we have a homotopy $F: Y \times I \to E$ that lifts $G$. Every book I read claims that if $B$ is path connected, then $\pi^{-1}(b_0)$ is homotopy equivalent to $\pi^{-1}(b_1)$ for any $b_0,b_1 \in B$. I am aware that, in proving this claim, we are supposed to consider $Y = \pi^{-1}(b_0)$, with $f: \pi^{-1}(b_0) \hookrightarrow E$ being inclusion, and with $G: Y \times I \to B$ such that $G(y,t) = \gamma(t)$ for some path $\gamma$ from $b_0$ to $b_1$. It is then supposed to be easy to show that $\pi^{-1}(b_0) \times \{1\} \to \pi^{-1}(b_1)$ is a homotopy equivalence, but I do not see why this is true. Why is this true?
In particular, why does $\pi:D^2 \to I$ with $\pi(re^{i\theta}) = r$ not have the homotopy lifting property? I cannot seem to find any homotopy that cannot be lifted.
As commented by Eric Wofsey above, a proof of this fact may be found on page 405 of Hatcher's Algebraic Topology as the proof of Proposition 4.61. I would like to add that Hatcher's comment concerning the homeomorphism between $(I \times I, I \times \{0\} \cup \partial I \times I)$ and $(I \times I, I \times \{0\})$ is relevant with regard to his definition of the homotopy lifting property for a pair in terms of the lift extension property for a pair, which may be found on page 376.
Now, given paths $\gamma(t) = t$ and $\gamma'(t) = 1-t$, we have a homotopy $\gamma(s,t)$ from $\gamma * \gamma'$ to the constant loop, inducing a family $g_{s,t}: \pi^{-1}(\gamma(0)) \to I$ with $g_{s,t} = \gamma(s,t)$ (using the notation from Hatcher). We cannot lift $g_{s,t}$, for this would induce a homotopy equivalence between $\pi^{-1}(\gamma(0)) = \{0\}$ and $\pi^{-1}(\gamma'(0)) = S^1$, and we know that $S^1$ is not contractible.