# Obstruction to lifting a map from the base space to the total space.

Suppose $\pi :E \to B$ is a fibration with fibre $F$ above a chosen base point. Then I am trying to solve when a map $f$ from a manifold $M$ to $B$ lift to a map $g:M \to E$. The answer given is they are cohomology classes of form $H^{n+1}(M;\pi_n(F))$. I am unable to prove this.

In the special case where the manifold $M$ is $S^n$. This is equivalent to asking when is an element of $\pi_n(B)$ in the image of an element of $\pi_n(E)$. Now using the long exact sequence of homotopy groups for a fibration we get that corresponding element of $\pi_{n-1}(F)$ has to be zero. But this is precisely $H^{n}(S^n;\pi_{n-1}(F))$. But I am unable to prove in the case of a general manifold. I am guessing the idea is to use the CW structure of manifold but I do not really understand how. if anyone can give any hints it would be great. Thanks.

• Do you mean $H^n(M;\pi_{n-1}(F))$ perhaps or not? Is $n$ the dimension of $M$? – Malkoun Jul 9 '16 at 11:18
• @Malkoun I am not entirely sure if my interpretation is right but I think it is all cohomology classes of that form where $n$ ranges over $0$ to dim $M-1$ – happymath Jul 9 '16 at 11:22
• Look at the following post for a related discussion. – Malkoun Jul 9 '16 at 12:21
• You have the right idea: induction cell-by-cell is the way to go. See section 7.10 in Davis & Kirk's AT book for a reference. – JHF Jul 11 '16 at 15:04