## § Splitting of semidirect products in terms of projections

Say we have an exact sequence that splits:
$0 \rightarrow N \xrightarrow{i} G \xrightarrow{\pi} K \rightarrow 0$
with the section given by $s: K \rightarrow G$ such that $\forall k \in K, \pi(s(k)) = k$. Then we can consider the map $\pi_k \equiv s \circ pi: G \rightarrow G$. See that this firsts projects down to $K$, and then re-embeds the value in $G$. The cool thing is that this is in fact idempotent (so it's a projection!) Compute:
\begin{aligned} &\pi_k \circ \pi_k \\ &= (s \circ \pi) \circ (s \circ \pi ) \\ &= s \circ (\pi \circ s) \circ \pi ) \\ &= s \circ id \circ \pi \\ &= s \circ \pi = \pi_k \\ \end{aligned}
So this "projects onto the $k$ value". We can then extract out the $N$ component as $\pi_n: G \rightarrow G; \pi_n(g) \equiv g \cdot \pi(k)^{-1}$.