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I'm interested in understanding the importance of the local coefficients in the definition of the obstruction cocycle for a lift of a map $f\colon X \to B$ along a fibration $p \colon E \to B$. I'm following the explanation given at page $189$ in Davis & Kirk's Lecture Notes in Algebraic Topology.

Aim of this question: understand the role of the local coefficients system i.e. why we need them and how we use them.

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Sketch of the construction: Let $X$ be a connected CW-complex. Suppose $g$ is constructed over the $n$-skeleton of $X$. Given an $(n + 1)$- cell of $X$, $e^{n+1}$, the composite of the attaching map and $g$ gives a map $\partial e^{n+1}\cong S^n \to X \to E$. Notice that the composite $S^n \to X \to E \to B$ is null homotopic since we can (clearly) extend $f$ to the interior of such cell. Let $H\colon S^n\times I \to B$ such null-homotopy.By the homotopy lifting property, there exists a map homotopic to $g \colon S^n \to E$ whose image is contained entirely in the fibre over $H(1,S^n)=b_0$, call it $F_{b_0}$. So in a highly non-canonical way we associated to each $(n+1)$-cell, an element of $[S^n,F_{b_0}]$. Then if we assume that the fibre is $n$-simple, we have $[S^n,F_{b_0}]\cong \pi_n(F_{b_0}, \ast)$ for any base point.

Then they conclude:

However, if $\pi_1B \neq 0$, then some ambiguity remains, namely it was not necessary that $f$ preserved base points, and hence, even if $F$ is $n$-simple, we do not obtain a cochain in $C^n_{cell}(X;\pi_nF)$. However, one does get a cochain with local coefficients. Thus obstruction theory for fibrations requires the use of cohomology with local coefficients, as we will now see.

and they build the local coefficient system: if F is n-simple, then the fibration $F \to E \to B$ defines a local coefficient system over $B$ with fiber $\pi_nF$. We can associate to each loop $a \in \pi_1B$ a homotopy class $h_a$ of self-homotopy equivalences of $F$. Then $h_a$ induces an automorphism of $[S^n,F]$ by post composing. Since we are assuming F is n-simple, the fibration determines a representation $\rho \colon \pi_1B \to Aut(\pi_n F )$. Precomposing with $f_*$ we have a representation $\rho' \colon \pi_1X \to Aut(\pi_n F )$.

So the questions:

1) Why don't we obtain a cochain in $C^n_{cell}(X;\pi_nF)$?

2) Why instead we obtain a cochain in $C^n_{cell}(X;(\pi_nF)_{\rho'})$? (in the local coefficient system defined by the representation $\rho'$.)

3) Is this juggling with local coefficients really necessary? I can't find any other references for this approach, and in fact I've found some notes like this one at page $136$ with the same statements and hypothesis but not mentioning local coefficients in any way. Clearly there is some mistake somewhere, here on on Davis & Kirk's book, since the latter claims one has to go through local coefficients

My Answers:

1) since the $f$'s are not based, at the end of the procedure we might land in different fibres with the liftings, and there is no canonical isomorphism between the fundamental groups of fibre over different points. If $\pi_1B=0$ we can overcome this problem since every isomorphism between fundamental groups of different fibres is the same, and therefore we have a canonical identification.

2) I think that the most natural thing to do is to define a local coefficient system as the bundle $\pi \colon\coprod_{b\in B} \pi_1(F_{b}) \to B$ and under the additional assumption that $B$ is locally simply connected (pretty natural of we work with CW-complex or manifolds) one can prove that $\pi$ is indeed locally trivial. I don't see why this solution should lead at the same result as the one proposed by the authors thought. Secondly, I don't get why their solution work.

3) I think the authors of the note missed something or I missed some additional hypothesis he mande. Can't find them thought

Thanks in advance for any hint/solutions

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    $\begingroup$ Every map (between path-connected spaces, say) is always homotopic to a based one. The point is that if $\pi_1 B$ is simply connected, then the isomorphism $\pi_n F \to \pi_n F'$ is canonical (where as otherwise, picking a different path down below might conjugate the above isomorphism). This is also why $F$ being $n$-simple is sufficient: There's no action of $\pi_1 B$, even though it's nonzero. $\endgroup$ – user98602 Apr 24 '16 at 18:41
  • $\begingroup$ @MikeMiller thanks for your comment, I agree with you for the first part, but then I don't follow you when you speak about $n$-simple. I mean, the highlighted part says that $n$-simplicity is not enough. So is the book wrong/what am I missing? $\endgroup$ – Riccardo Apr 24 '16 at 19:52
  • $\begingroup$ Hm, you've got me there. It's been a while since I thought about this. To make up for my mistake, I'll answer this when I get home tonight if nobody else has. $\endgroup$ – user98602 Apr 24 '16 at 20:17
  • $\begingroup$ @MikeMiller thank you very much! $\endgroup$ – Riccardo Apr 24 '16 at 20:31
  • $\begingroup$ Sorry for the long delay, BTW. I'm not sure what the deal with #3 is, but admittedly, I don't want to dig through that document and look for hidden assumptions. $\endgroup$ – user98602 Apr 26 '16 at 17:33
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The reason that being $n$-simple is not enough is that, not only do we need to have a canonical identification of $\pi_n(F,*)$ for every choise of basepoint (which is where the $n$-simplicity condition shows up), we also need a canonical identification of the homotopy groups of different fibers $\pi_n(F,*) \cong \pi_n(F',*)$. Put another way, there's an action of $\pi_1(B,b)$ on the fiber $F=\pi^{-1}(b)$. We need that not only is $F$ $n$-simple, it's also "$\pi_1(B)$-simple" - this action needs to be trivial. Once we assume $F$ is $n$-simple there's no reason to worry about whether or not the maps are based, like you said, because we can just homotope them so they are. (Unless the fibers are disconnected, in which case we run into extra trouble from that, in case $\pi_1(B)$ acts nontrivially on $\pi_0(F)$!)

The point of local coefficients is that, because we've got different (but isomorphic) coefficients with each fiber, and each path gives us an isomorphism between the fibers, there's no need to bother with the above and get canonical isomorphisms. I'm not sure I understand your question (2) - your solution (provided that the fundamental groupoid has appropriate parallel transports) is exactly the same as theirs, since all you're saying is "We identify $\pi_n(F)$ and $\pi_n(F')$ by lifting the map $f: F \times I \to B$, $f(x,t) = \gamma(t)$, upstairs, with initial lift $\tilde f(x,0) = x$". But when $\gamma$ is a loop, this is precisely the same thing as the induced map of $\pi_1(B)$ on $F$, and hence the same thing as the induced map of $\pi_1(B)$ on $\pi_n(F)$!

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  • $\begingroup$ Thank you! as a side note, I realised I build a local coefficients system on $B$ and not on $X$, but I think I managed to convince myself that what I need is the pullback via $f$ of my bundle over $B$. I think I got the point you are making, thanks again! $\endgroup$ – Riccardo Apr 26 '16 at 21:09
  • $\begingroup$ @Riccardo: Ah, right, I hadn't caught that. Normally we should have a local coefficient system on X, because the choice of basepoint (both in B, and in the fiber above b) will matter, but with the n-simple assumption only the choice of basepoint in B matters, so the local coefficient system on X is pulled back from one on B. $\endgroup$ – user98602 Apr 26 '16 at 21:17

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