Let $F\hookrightarrow Y\stackrel{f}{\longrightarrow} B$ be a fibration. If $F$ is contractible in $Y$ via some homotopy $H: F\times I\rightarrow Y$, we get split short exact sequences:

$0\rightarrow \pi_n(Y, y)\rightarrow \pi_n(B,b)\rightarrow \pi_{n-1}(F,y)\rightarrow 0$ $(y\in F, b=f(y))$

The splitting map $\sigma: \pi_n(B)\leftarrow \pi_{n-1}(F)$ is given as follows:

Take $[\phi]\in \pi_{n-1}(F)$ and consider $S^{n-1}\times I\stackrel{\phi\times id}{\longrightarrow} F\times I\stackrel{H}{\longrightarrow} Y \stackrel{f}{\longrightarrow} B$. This descends to a map $\psi: S^{n}\rightarrow B$ via the quotient map collapsing the bottom and top of $S^{n-1}\times I$. Then $\sigma([\phi]):=[\psi]$.

$\textbf{what I don't understand}:$ I have great difficulty verifying the fact that $\textbf{this splitting map is a group homomorphism}$ though I am told the proof is trivial.

The problem I encounter is just that when I think of $\sigma([\phi_1]\ast[\phi_2])$ in the naive way as the map $I^{n-1}\times I \stackrel{\phi_1 \ast \phi_2 \times id}{\longrightarrow} F\times I\stackrel{H}{\longrightarrow} Y\stackrel{f}{\longrightarrow} B$, I have the image of the path (under the homotopy composed with $f$) of the basepoint, y, running up the vertical sides of the hypercube, $I^n$, rather than a constant map to the basepoint. Any suggestions would be very appreciated. Thank you for reading this far.

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    $\begingroup$ Are you assuming $H$ is a homotopy through based maps? At the very least, $H(F\times \{1\})$ needs to land in $F\subset Y$ for your description of $\sigma$ to make sense. Assuming this is true, it might be possible to demand that $H$ consist of based maps, probably under the condition that $F$ is "well-pointed" (i.e. the inclusion of the basepoint is a cofibration -- this is a standard assumption). Where did you come across this? $\endgroup$ – Aaron Mazel-Gee Feb 29 '12 at 11:12
  • $\begingroup$ @AaronMazel-Gee: wow, if $H$ respects the base point then it really is trivial. Maybe thats what the author means. Here's the statement from Bredon's Topology and Geometry (p. 453): If $p: Y\rightarrow B$ is a fibration with fiber $F$ and if there is a homotopy $H:F\times I \rightarrow Y$ between the inclusion and a constant map, then the sequence $0\rightarrow \pi_n (Y) \rightarrow \pi_n (B) \rightarrow \pi_{n-1} (F) \rightarrow 0$ is split exact for $n\geq 2$. $\endgroup$ – schopenhauer Feb 29 '12 at 12:08
  • $\begingroup$ @AaronMazel-Gee: Here's what seems like the same statement from Whitehead's Elements of Homotopy Theory (p. 188): Let $f: Y\rightarrow B$ be a fibration whose fibre $F$ is contractible in $Y$. Then the homotopy sequence breaks up into a family of splittable short exact sequences $0\rightarrow\pi_n(Y)\stackrel{f_{\ast}}{\longrightarrow} \pi_n (B) \stackrel{\Delta_{\ast}}{\longrightarrow}\pi_{n-1} (F)\rightarrow 0$. Proof: Let $h: (T\wedge F, F)\rightarrow (Y,F)$ be a nullhomotopy of the inclusion map. Then $T\wedge F$ is contractible, so that $\partial_\ast :\pi_n(T\wedge F, F)=\pi_{n-1}(F)$ $\endgroup$ – schopenhauer Feb 29 '12 at 12:32
  • $\begingroup$ ... The composite $\pi_{n-1} (F) \stackrel{\partial_\ast ^{-1}}{\longrightarrow} \pi_n(T\wedge F, F) \stackrel{h_\ast}{\longrightarrow} \pi_n(Y,F) \stackrel{f_\ast}{\longrightarrow}$ is the desired left inverse for $\Delta_\ast$." His use of $T\wedge F$ would suggest the homotopy respects the basepoint right? (I think $T\wedge F$ is the reduced cone on $F$, btw) $\endgroup$ – schopenhauer Feb 29 '12 at 12:36
  • $\begingroup$ Yes, definitely smash products are all about basepoints. If $T\wedge F$ is indeed the reduced cone, then the idea is that all the basepoints of $F$ are stuck together, so they have no choice but to stay fixed! (Note, by the way, that smash product isn't the coproduct in $Top_*$. I didn't realize this for a while. Rather, its primal use lies in the exponential adjunction $Map_*(X,Map_*(Y,Z))\cong Map_*(X\wedge Y,Z)$. Think of it as the "tensor product".) $\endgroup$ – Aaron Mazel-Gee Feb 29 '12 at 15:33

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