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Let $p:E \to X$ be a fibration with the fiber $F$ where $X$ is a CW-complex. Denote by $U$ the set $U:=D^n \times \{0\} \cup S^{n-1} \times I$ (part of a cylinder) and let $\tilde{f}:U \to E$ be a section in the sense that $p \circ \tilde{f}=\Phi$ where $\Phi$ is an attaching map for a cell $e$ in $X$. Suppose that $c \in [\partial(D^n \times I),F]$ be a given class.

How to show that it is possible to extend $\tilde{f}$ on $\partial(D^n \times I)$ such that the obstruction cocycle associated with this extended $\tilde{f}$ evaluated on $e$ is exactly $c$?

One can find a proof of the similar statement in the context of obstructions for extending maps, see page 176 of these notes (in this context the values of the obstruction cocycle are in $Y$, in our notation it would be $E$-but in the context of fibratons we want these values to be in $F$ thus my question, despite smiliarity, is not a special case of this theory).

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  • $\begingroup$ Why isn't it a special case of the theory? The obstruction cochain to extending a section from the base to a fibration is defined by associating a homotopy element over each cell -- where we can then identify the fibration restricted to the cell to one fiber, as all fibers are homotopy equivalent (to make this globally well defined, one has to generally use local coefficients or make some assumptions about the base/fiber). $\endgroup$
    – Tobias Shin
    Jul 26, 2018 at 19:01
  • $\begingroup$ In the case of fibrations the obstruction cocycle is defined in a different way: in both cases one proceed inductively, assuming that the map is already defined on $n-1$-th skeleton. In the general extension problem you take the boundary of your cell: on this boundary your map (call it $g$-and call $Y$ the codomain of $g$) is already defined. You compose your map with the attaching map and obtain a map from $S^{n-1}$ to $Y$ and thats all. In the case of fibrations you have your section defined over the $n-1$-th skeleton. You consider the attaching map restricted to the sphere and compose ... $\endgroup$
    – truebaran
    Jul 26, 2018 at 19:29
  • $\begingroup$ ... with your section: call this composition $\tilde{g}$. You can also consider a nullhomotopy $g_t$ obtained as follows: $g_t(x)=\Phi((1-t)x)$ where $\Phi$ is the attaching map. This homotopy has values in $X$ and from the lifting property you can lift this homotopy to $\tilde{g_t}$ such that $\tilde{g_0}=\tilde{g}$. As $g_1$ was constant and $\tilde{g_1}$ is a lift of $g_1$, this means that $\tilde{g_1}$ has values in some fiber. This the class $[\tilde{g_1}]$ gives an element in $\pi_n(F)$ (note that our original section has values in $E$ but ultimately we obtain a class in $\pi_n(F)$).. $\endgroup$
    – truebaran
    Jul 26, 2018 at 19:34
  • $\begingroup$ and not in $\pi_n(E)$. For simplicity I assume that the action of the fundamental group of $F$ is trivial on all $\pi_n(F)$ as well as the action of $\pi_1(B)$ on all $\pi_n(F)$. This allows me to not worry about the different choice of fibers (don't need to deal with local coefficients) and not worry about base points. $\endgroup$
    – truebaran
    Jul 26, 2018 at 19:37
  • $\begingroup$ You can apply the same argument presented in the notes except you can compose your attaching map with the section you're interested in extending, so that you have a map from U to the fiber (since by composing the section with the attaching map, you're really restricting yourself to a cell, which is contractible). $\endgroup$
    – Tobias Shin
    Jul 27, 2018 at 0:02

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