If I have two (Serre-)fibrations over the same base, and a weak equivalence of the total spaces that is also a map over the base, could I hope that the induced map on the spaces of sections would also be a weak equivalence? 
If I restrict to CW-spaces, this is fairly easy, but I don't really know how to handle the space of sections if only weak equivalences are available.



Space of sections of a fibration under weak homotopy equivalence
This is not true in general, unless you assume the base is sufficiently nice (eg a CW-complex). Here is a counter-example. 


Let $B = \mathbb{Q}$, the rationals with its topology as a subspace of the reals $\mathbb{R}$. Let $E\_2 = \mathbb{Q}$ as well and let $E\_1 = \mathbb{Q}\_\delta$, the rationals with the discrete topology. 


Then $E\_2 \to B$ is the identity map (a homeomorphism) and is a Serre fibration. The set-theoretic identity map $E\_1 \to B$ is also continuous. Since any map from a disk into $E\_1$ or $B$ factors through a constant map, you can check that it is also a Serre fibration. The map $E\_1 \to E\_2$ is also a weak homotopy equivalence and a map over $B$, as you require. 


However the space of sections of $E\_2$ is the one point space, while the space of sections of $E\_1$ is empty. In particular they are not weak homotopy equivalent. 


