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.