[step:Identify the extension of structure group with the original bundle]
Let $Q \times_H G$ denote the associated bundle formed using the left action of $H$ on $G$ by multiplication. Thus $(qh,g)$ is identified with $(q,hg)$ for $q \in Q$, $h \in H$, and $g \in G$.
Define $F: Q \times_H G \to P$ by $F([q,g]) = qg$. This map is well-defined because
\begin{align*}
F([qh,g]) = qhg = F([q,hg]).
\end{align*}
It is smooth and right $G$-equivariant, where the right action on $Q \times_H G$ is $[q,g]k := [q,gk]$ and the right action on $P$ is the given principal action:
\begin{align*}
F([q,g]k) = F([q,gk]) = qgk = F([q,g])k.
\end{align*}
The map $F$ is surjective. If $p \in P$ and $x = \pi(p)$, choose $a \in A$ with $x \in V_a$. Since $s_a(x)$ lies in the same principal $G$-fiber as $p$, there is a unique $g \in G$ such that
\begin{align*}
p = s_a(x)g.
\end{align*}
Since $s_a(x) \in Q$, we have $p = F([s_a(x),g])$.
The map $F$ is injective. Suppose $F([q_1,g_1]) = F([q_2,g_2])$. Put $x := \pi(q_1g_1) = \pi(q_2g_2)$, and choose $a \in A$ with $x \in V_a$. Since $q_1,q_2 \in Q_x$, there exist unique $h_1,h_2 \in H$ such that
\begin{align*}
q_1 = s_a(x)h_1
\end{align*}
and
\begin{align*}
q_2 = s_a(x)h_2.
\end{align*}
The equality $q_1g_1 = q_2g_2$ becomes
\begin{align*}
s_a(x)h_1g_1 = s_a(x)h_2g_2.
\end{align*}
Freeness of the right $G$-action gives
\begin{align*}
h_1g_1 = h_2g_2.
\end{align*}
Using the defining [equivalence relation](/page/Equivalence%20Relation) in $Q \times_H G$, we get
\begin{align*}
[q_1,g_1] = [s_a(x)h_1,g_1] = [s_a(x),h_1g_1] = [s_a(x),h_2g_2] = [s_a(x)h_2,g_2] = [q_2,g_2].
\end{align*}
Thus $F$ is injective.
Finally, in the local trivialisation over $V_a$, every element of $Q \times_H G$ is represented uniquely as $[s_a(x),g]$, and $F$ sends it to $s_a(x)g$. Therefore locally $F$ is the identity map on $V_a \times G$. Hence $F$ is a diffeomorphism. We have constructed an $H$-reduction of $P$.
[/step]