# $\pi_i(E,F)\to \pi_{i-1}(F)$ injective if $F\hookrightarrow E$ is nullhomotpic

This question pertains to Grumpy Parsnip's answer here: Fiber bundle with null-homotopic fiber inclusion

The claim is: If $F\to E\to B$ is a fiber bundle with $F\to E$ nullhomotopic, then the LES of homotopy groups breaks into split SES's:

$$0\to \pi_i(E)\to \pi_i(B)\to \pi_{i-1}(F)\to 0$$

In the answer linked above, it is claimed that a splitting is given as follows:

\begin{align} & \pi_{i-1}(F)\to \pi_i(E,F)\stackrel{\cong}{\to}\pi_i(B) \\ &[f]\mapsto[(\widetilde{f},f)] \end{align}

where $\widetilde f:D^i\to E$ is an extension of the composite $S^{i-1}\stackrel{f}{\to} F\to E$ which is nullhomotopic.

What's not clear to me is why this is well-defined. That is, why is the boundary map $\pi_i(E,F)\to \pi_{i-1}(F)$ injective? Is this true for arbitrary pairs of spaces $(E,F)$ with $F\hookrightarrow E$ nullhomotpic?

First of all, the LES of the pair $(F,E)$ shows you that the boundary map you talk about is surjective, so it's rarely injective.
For the splitting, don't pick arbitrary extensions. Choose the ones coming from a fixed nullhomotopy $H\colon[0,1]\times F\rightarrow E$. In other words, if $[f]\in\pi_{i-1}(F)$ then compose $\operatorname{id}_{[0,1]}\times f$ with your fixed homotopy. This descends to an extension $\tilde{f}$ which will give you the splitting.