Does $\Omega$ functor preserve fibration? If $p:E \rightarrow B$ is a fibration, is it necessarily that $\Omega p:\Omega E \rightarrow \Omega B$ a fibration? (where $\Omega E$ is the loopspace of $E$ and so is $\Omega B$)
 A: Write $f \perp g$ if $f$ has the left lifting property with respect to $g$. Then recall by definition $p$ is a Serre fibration if and only if $(D^n \to D^{n+1}) \perp p$ for every $n \ge 0.$
Remember that $\Omega$ is the right adjoint functor to $S^1 \wedge -$, and $S^1 \wedge D^n \cong D^{n+1}$. 
Then $(D^n \to D^{n+1}) \perp \Omega p \iff (S^1 \wedge D^n \to S^1 \wedge D^{n+1}) \perp p \iff (D^{n+1} \to D^{n+2}) \perp p$. Therefore, since $p$ is a Serre fibration, so is $\Omega p$.
A: This is a small elaboration on Thomas Belulovich's answer. 
In general, Serre fibrations (among other things; cf. the definition of a model category) are characterized via a diagrammatical condition: they have the right lifting property with respect to inclusions $D^n \to D^n \times I$ (cylindrical inclusions). Morphisms in some category defined by such a condition are generally closed under limits and under internal hom functors $\hom(X, \cdot)$ (which are an enriched version of limits anyway), for formal reasons. (Analogously, morphisms defined by a left lifting property tend to be closed under colimits and colimit-like processes.)
In this case, if $p: E \to B$ is a Serre fibration, it follows that the map on path spaces $PE \to PB$ is a fibration too (since the path space is the internal hom $\mathrm{Map}([0, 1], \cdot)$), and the loop space can be obtained by taking the fiber of the path space (i.e., $\Omega E$ is the fiber product $PE \times_{E \times E} \ast $).
