# How to prove that a map is a Serre fibration?

I want to prove that the homotopy groups of some topological space $B$ of interest to me (not a CW complex) are trivial. I have a strategy of proof that consists in introducing another space $E$ that is contractible, and easily comes with a continuous surjection $\pi :E\to B$. If I can prove that any continuous map $f:I^k\to B$ lifts to a continuous map $\tilde f : I^k\to E$, then I'm done.

If I am not mistaken, this lifting property is true as soons as $\pi$ is a Serre fibration. Here is my question: are there classical way to prove such a thing, and were can I learn them (or simply learn about Serre fibrations)? Of course, any reference for the initial problem, which seems slightly weaker, is welcome too.

I guess that I should be able to manage my case by hand, but I think it may be an opportunity to learn more mathematics.

-
Benoit -- it is one of the equivalent versions of the definition of a Serre fibration that, given a map from the cube to the base and a lifting map of one of the faces to the total space, one can lift the rest to the total space. In particular, one can lift any map of a cube from the base of a Serre fibration to the total space. This however does not guarantee that given that the total space is contractible, so is the base: think of the classifying spaces. One needs the fiber to be contractible as well. –  algori Sep 3 '10 at 13:44
Or sometimes you can get away with less than being a fibration. But the first question is: are the fibers contractible or, at least, are their homotopy groups trivial? –  Tom Goodwillie Sep 3 '10 at 15:39
I believe what Tom is getting at is that even for maps that aren't fibrations, provided the fibers are all contractible you can frequently prove the map is a homotopy-equivalence. A standard reference for these types of arguments is Lacher's "Cell-like maps" paper. –  Ryan Budney Sep 3 '10 at 18:53
Ok, I clearly neglected a point: I in fact hoped that I could lift a map that sends the boundary of the cube to a point, into a map with the same property, which is usually foolish. But in fact, in my particular case there is a point $b\in B$ that has a unique antecedent $e\in E$. I can therefore consider maps $f:(I_k,\partial I_k)\to(B,b)$, lift them to $\tilde f:I_k\to E$ (the important point being that automatically, $\tilde f(\partial I_k)=\{e\})$. Then the composition of $\tilde f$ with the homotopy from $Id_E$ to $e$ then with $\pi$ is precisely a homotopy of $f$ to $b$ and I'm done. –  Benoît Kloeckner Sep 3 '10 at 20:10
Benoit -- if you have a Serre fibration such that the fiber over some point is contractible, then all fibers will automatically be pathwise connected and will have trivial homotopy groups, since they are all weakly homotopy equivalent. If in addition to that the total space is contractible, then all homotopy groups of the base vanish. –  algori Sep 4 '10 at 0:40