Question about the definition of a fibration In "Topology for Physicists" by Schwarz, the author writes:

Given a map $p$ from a space $E$ onto a space $B$, we can partition
  the domain into disjoint sets $F_b=p^{-1}(b)$, for $b\in B$. We say
  that $p$ defines a fibration if all the sets $F_b$ are homeomorphic
  to one another.

This is of course not the standard definition found in the literature, so my question is under what conditions the above definition can guarantee that $p$ has the homotopy lifting property so that it becomes a fibration (say in the sense of Serre).
 A: This definition of a "fibration" is pretty much useless except for pedagogical purposes, in order to ease you into the correct definition. In fairness, two pages later Schwarz gives the standard definition of a "locally trivial fibration", this is the one that he (and you, most other people) will use. 
Here are few exampless to consider which show how bad his initial notion of a "fibration" is:


*

*Consider the $U(1)$ action on $S^3$ given by $z\times (u,v)\mapsto (z^p u, z^q v)$, where $(u,v)$ is a unit vector in ${\mathbb C}^2$, $z$ is a unit complex number, $p, q$ are fixed coprime integers. The orbits of this action partition $S^3$ into circles. The quotient of this action is $S^2$. This construction is called a Seifert fibration. More complicated examples of Seifert fibrations have base which is, say, the 2-sphere, fibers which are circles and the total space which is aspherical (has contractible universal cover). 

*Consider the space $X$ which is the union $\{(x,y)\in R^2: x^2-y^2=0\}\cup \{(0,1)\}$. Define the map $p: X\to {\mathbb R}^2$ which is the projection to the $x$-axis. Then the fibers of this map are two-point subsets of $X$.   

*Take the surjective local diffeomorphism $p: {\mathbb C}\to {\mathbb C}$ given by the formula
$$
p(z)= \int_{0}^z \exp(z^2)dz.
$$
This map fails the path-lifting property. Fibers of this map will be discrete countably-infinite sets. 
