2
$\begingroup$

I'm currently reading Kirk's & Davis's Lecture notes in Algebraic Topology. On the page 190 they discuss obstruction for lifting $f:X \rightarrow B $ to $f':X \rightarrow E$ where $E \rightarrow B$ is a fibration. Suppose $f'$ is defined on $X_n$, then for any $n+1$ cell - $e^{n+1}$ - its characteristic map gives us a map $S^n \rightarrow F \hookrightarrow E$ which defines an element of $\pi_{n}(F)$ since we assume that $F$ is $n$ simple. Now the authors note that if$\pi_{1}(B) \neq 0$ then this assignement need not define a $cochain$ since $f$ does not preserve base points. Then they conclude that we can instead define a cocycle in cohomology with local coeaficient system induced by $p$. So we get an element of $Hom_{Z[\pi_1(X)]}(C_{n+1}(\tilde{X}),\pi_{n}(F))$. My question is - how do we define this cochain?

$\endgroup$
2
$\begingroup$

First consider the case that $X=B$, construct the class and then pull it back to $X$ via the map $f$. See section 5.2.1, pg 100 for remarks regarding the interaction between the cell structures of $B$ and its universal cover $\widetilde{B}$. If the cells $e^{n+1}_i$ are a $\mathbb{Z}$-basis for the cellular chain group $C^{n+1}(B)$ then $\tilde{e}^{n+1}_i$ are $\mathbb{Z\pi_1}$ basis for $C^{n+1}(\widetilde{B})$ (see 5.2.1 for notation). Given the lift partial lift $g:B_n\rightarrow E$ with $p\circ g=f|{B_n}$ define $\theta^{n+1}(g)\in C^{n+1}(\widetilde{B};\pi_nF)=Hom_{\mathbb{Z}\pi}(C_{n+1}(\widetilde{B}),\pi_nF)$ on the $\mathbb{Z}\pi$-basis elements by

$\theta^{n+1}(g)[\epsilon\cdot \tilde{e}^{n+1}_i]=\epsilon\cdot [g\circ\varphi_i]$

where $\epsilon\in\mathbb{Z}\pi$ with $\pi=\pi_1B$ acting on $\pi_nF$ through the representation $\rho$ (pg 189), and $\varphi_i:S^n\rightarrow B_n$ is the attaching map for the cell $e^{n+1}_i$ of $B$. Extend $\theta^{n+1}(g)$ to all of $C_{n+1}(\widetilde{B})$ by linearity and you have your cochain.

$\endgroup$
  • $\begingroup$ $[g \circ \varphi_i ] $ will land in $n$-th homotopy groups of fiber over different base points. How do you identify all those groups with one $\pi_n(F)$? $\endgroup$ – leg14able Jul 9 '17 at 15:44
  • $\begingroup$ Davis and Kirk use the assumption that $F$ is $n$-simple. In particular $F$ must be connected so $\pi_nF$ does not depend on any particular choice of basepoint. $\endgroup$ – Tyrone Jul 9 '17 at 16:07
  • $\begingroup$ Yes, of course. What I meant is the following. $[g \circ \varphi_i] $ defines an elemnt in $p^{-1}(b) \cong F$ but for different base element the point $b$ may be different. How do we identify $\pi_{n(}p^{-1}(b)$ for different b? $\endgroup$ – leg14able Jul 9 '17 at 16:19
  • $\begingroup$ This is Theorem 6.57, pg 158. $\endgroup$ – Tyrone Jul 9 '17 at 17:04
  • $\begingroup$ Could you please elaborate a little bit? i'll try to be more specific. We take a cell $e^{n-1}$. We have a homotopy $H:\partial e^{n-1} \times I \rightarrow B$ which takes the boundary into a point $b \in B$. This point does not have to be the same for different cells. then a lift of homotopy collapsing a cell to a point gives us an element in $\pi_{n}(p^{-1}(b)$. But since the point may be different for different cells my question is how do we bypass this obstacle? (because from what I understand this is exactly why we want to use local coeffients (to bypass this obstacle)) $\endgroup$ – leg14able Jul 9 '17 at 17:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.