[proofplan]
We pass between reductions and sections using the quotient map $\rho:P\to P/H$. An $H$-reduction has fibres that are right $H$-torsors, so taking the $H$-orbit of any point in the reduced fibre gives a well-defined point of $P/H$. Conversely, a section of $P/H$ specifies in each fibre $P_x$ exactly one $H$-orbit; the union of these orbits is the desired reduced subbundle. Local trivialisations reduce all smoothness and principal-bundle checks to the model statement that $G\to G/H$ is a principal right $H$-bundle.
[/proofplan]
[step:Construct a section from an $H$-reduction]
Let $Q\subset P$ be a smooth $H$-reduction. Thus $Q$ is a smooth principal right $H$-bundle over $M$ with projection $\pi_Q:=\pi|_Q:Q\to M$, and the inclusion map $i:Q\hookrightarrow P$ is $H$-equivariant for the restricted right action. Let $\rho:P\to P/H$ denote the quotient map $p\mapsto [p]_H$, and let $\bar{\pi}:P/H\to M$ denote the quotient bundle projection defined by $\bar{\pi}([p]_H)=\pi(p)$.
Define the map
\begin{align*}
s_Q:M &\to P/H
\end{align*}
as follows. For $x\in M$, choose any $q\in Q_x:=\pi_Q^{-1}(\{x\})$ and set
\begin{align*}
s_Q(x):=[q]_H.
\end{align*}
This definition is independent of the choice of $q$. Indeed, if $q_1,q_2\in Q_x$, then the principal right $H$-action on the fibre $Q_x$ is transitive, so there exists $h\in H$ such that $q_2=q_1h$. Therefore $[q_2]_H=[q_1h]_H=[q_1]_H$.
The map $s_Q$ is a section because
\begin{align*}
\bar{\pi}(s_Q(x))=\bar{\pi}([q]_H)=\pi(q)=x.
\end{align*}
It remains to check smoothness. Let $U\subset M$ be an [open set](/page/Open%20Set) over which $Q$ admits a smooth local section
\begin{align*}
\tau:U &\to Q.
\end{align*}
Then on $U$,
\begin{align*}
s_Q|_U=\rho\circ i\circ \tau.
\end{align*}
The maps $\tau$, $i$, and $\rho$ are smooth, so $s_Q|_U$ is smooth. Since such open sets cover $M$, the map $s_Q:M\to P/H$ is smooth.
[guided]
Let us unpack why the formula is well-defined. The fibre $Q_x$ of a principal right $H$-bundle is not just a set with an $H$-action; it is a right $H$-torsor. This means that for any two points $q_1,q_2\in Q_x$, there is a unique element $h\in H$ with $q_2=q_1h$. Since the quotient point $[q]_H$ records exactly the right $H$-orbit of $q$, all points of $Q_x$ determine the same element of $P/H$.
Formally, define
\begin{align*}
s_Q:M &\to P/H
\end{align*}
by choosing, for each $x\in M$, a point $q\in Q_x$ and putting
\begin{align*}
s_Q(x):=[q]_H.
\end{align*}
If $q_1,q_2\in Q_x$, transitivity of the right $H$-action on $Q_x$ gives $q_2=q_1h$ for some $h\in H$. Hence
\begin{align*}
[q_2]_H=[q_1h]_H=[q_1]_H,
\end{align*}
so the value $s_Q(x)$ does not depend on the chosen point of the fibre.
The map is a section because the quotient bundle projection $\bar{\pi}:P/H\to M$ is defined by $\bar{\pi}([p]_H)=\pi(p)$. Therefore, for $q\in Q_x$,
\begin{align*}
\bar{\pi}(s_Q(x))=\bar{\pi}([q]_H)=\pi(q)=x.
\end{align*}
Thus $\bar{\pi}\circ s_Q=\operatorname{id}_M$.
Finally, smoothness is local on $M$. On an open set $U\subset M$ where the principal $H$-bundle $Q$ has a smooth local section
\begin{align*}
\tau:U &\to Q,
\end{align*}
the section $s_Q$ is represented by the composite
\begin{align*}
s_Q|_U=\rho\circ i\circ \tau,
\end{align*}
where $i:Q\hookrightarrow P$ is the smooth inclusion and $\rho:P\to P/H$ is the smooth quotient map. A composition of smooth maps is smooth, so $s_Q|_U$ is smooth. Since such open sets cover $M$, $s_Q$ is smooth on all of $M$.
[/guided]
[/step]
[step:Construct an $H$-reduction from a section]
Let
\begin{align*}
s:M &\to P/H
\end{align*}
be a smooth section of $\bar{\pi}:P/H\to M$. Define
\begin{align*}
Q_s:=\{p\in P:\rho(p)=s(\pi(p))\}.
\end{align*}
Define the fibre product over $M$ by
\begin{align*}
M\times_M(P/H):=\{(x,y)\in M\times(P/H):x=\bar{\pi}(y)\}.
\end{align*}
Equivalently, $Q_s$ is the inverse image of the graph of $s$ under the smooth map
\begin{align*}
(\pi,\rho):P &\to M\times_M(P/H),\quad p\mapsto (\pi(p),\rho(p)).
\end{align*}
We verify locally that $Q_s$ is a smooth principal right $H$-subbundle of $P$. Let $U\subset M$ be an open set over which $P$ has a smooth principal $G$-bundle trivialisation
\begin{align*}
\Phi:P|_U &\to U\times G
\end{align*}
satisfying $\Phi(pg)=(x,ag)$ whenever $\Phi(p)=(x,a)$. This trivialisation induces a smooth quotient trivialisation
\begin{align*}
\bar{\Phi}:(P/H)|_U &\to U\times G/H
\end{align*}
given by
\begin{align*}
\bar{\Phi}([p]_H)=(x,aH)\quad \text{when } \Phi(p)=(x,a).
\end{align*}
Since $s$ is a section, there is a smooth map
\begin{align*}
\sigma:U &\to G/H
\end{align*}
such that
\begin{align*}
\bar{\Phi}(s(x))=(x,\sigma(x)).
\end{align*}
Under $\Phi$, the subset $Q_s|_U$ corresponds to
\begin{align*}
R_\sigma:=\{(x,a)\in U\times G:aH=\sigma(x)\}.
\end{align*}
By the Quotient Manifold Theorem for the closed Lie subgroup $H\subset G$, the quotient $G/H$ has a unique smooth manifold structure for which the projection
\begin{align*}
\rho_G:G &\to G/H,\quad a\mapsto aH
\end{align*}
is a smooth submersion. The right $H$-action on $G$ is free because $ah=a$ implies $h=e_H$. It is proper because the action graph map
\begin{align*}
A:G\times H &\to G\times G,\quad (a,h)\mapsto (a,ah)
\end{align*}
is a closed embedding onto the subset
\begin{align*}
\{(a,b)\in G\times G:a^{-1}b\in H\},
\end{align*}
which is closed since the map $(a,b)\mapsto a^{-1}b$ is continuous and $H\subset G$ is closed. Hence $\rho_G:G\to G/H$ is a smooth principal right $H$-bundle. Hence its pullback by $\sigma:U\to G/H$ is a smooth principal right $H$-bundle over $U$. The set $R_\sigma$ is precisely this pullback, with projection $(x,a)\mapsto x$ and right action $(x,a)\cdot h=(x,ah)$. Therefore $Q_s|_U$ is a smooth principal right $H$-bundle over $U$.
These local principal $H$-bundle structures agree on overlaps because they are all obtained by restricting the global subset $Q_s\subset P$ and the global right action of $H$ on $P$. Thus $Q_s\subset P$ is a smooth principal right $H$-subbundle over $M$.
[guided]
The geometric point is that a section of $P/H$ chooses one right $H$-orbit inside each principal $G$-fibre of $P$. To turn this fibrewise choice into a smooth subbundle, we check the statement in a local trivialisation.
Let $U\subset M$ be an open set over which $P$ has a smooth principal $G$-bundle trivialisation
\begin{align*}
\Phi:P|_U &\to U\times G.
\end{align*}
This trivialisation carries the quotient bundle to $U\times G/H$ by the induced map
\begin{align*}
\bar{\Phi}:(P/H)|_U &\to U\times G/H,
\end{align*}
where $\bar{\Phi}([p]_H)=(x,aH)$ whenever $\Phi(p)=(x,a)$. Since $s:M\to P/H$ is a smooth section, on $U$ it is represented by a smooth map
\begin{align*}
\sigma:U &\to G/H
\end{align*}
with
\begin{align*}
\bar{\Phi}(s(x))=(x,\sigma(x)).
\end{align*}
The defining condition $\rho(p)=s(\pi(p))$ then becomes, in the coordinates $(x,a)\in U\times G$, the condition
\begin{align*}
aH=\sigma(x).
\end{align*}
Thus $Q_s|_U$ corresponds exactly to
\begin{align*}
R_\sigma:=\{(x,a)\in U\times G:aH=\sigma(x)\}.
\end{align*}
We now identify $R_\sigma$ as a pullback principal bundle. The Quotient Manifold Theorem applies because $H\subset G$ is a closed Lie subgroup. It gives a smooth manifold structure on $G/H$ for which
\begin{align*}
\rho_G:G &\to G/H,\quad a\mapsto aH
\end{align*}
is a smooth submersion. The right action of $H$ on $G$ is free: if $ah=a$, left multiplication by $a^{-1}$ gives $h=e_H$. The action is proper because the map
\begin{align*}
A:G\times H &\to G\times G,\quad (a,h)\mapsto (a,ah)
\end{align*}
is a closed embedding onto $\{(a,b)\in G\times G:a^{-1}b\in H\}$, and this subset is closed since $(a,b)\mapsto a^{-1}b$ is continuous and $H$ is closed. Therefore $\rho_G:G\to G/H$ is a smooth principal right $H$-bundle.
Pulling this principal right $H$-bundle back along $\sigma:U\to G/H$ gives the smooth principal right $H$-bundle
\begin{align*}
\{(x,a)\in U\times G:\rho_G(a)=\sigma(x)\}\to U.
\end{align*}
This is precisely $R_\sigma\to U$, with projection $(x,a)\mapsto x$ and right action $(x,a)\cdot h=(x,ah)$. Hence $Q_s|_U$ is a smooth principal right $H$-bundle over $U$. Since these local structures are all restrictions of the same subset $Q_s\subset P$ and the same right $H$-action on $P$, they agree on overlaps. Consequently $Q_s\subset P$ is a smooth principal right $H$-subbundle over $M$.
[/guided]
[/step]
[step:Show that the induced extension recovers the original principal $G$-bundle]
We now verify that $Q_s$ is an $H$-reduction of $P$, not merely a principal $H$-subbundle. Define $Q_s\times_H G$ to be the quotient of $Q_s\times G$ by the [equivalence relation](/page/Equivalence%20Relation)
\begin{align*}
(q,g)\sim(qh,h^{-1}g)\quad \text{for } h\in H.
\end{align*}
Define
\begin{align*}
\Psi_s:Q_s\times_H G &\to P
\end{align*}
by
\begin{align*}
\Psi_s([q,g])=qg.
\end{align*}
This map is well-defined: if $(q,g)$ is replaced by $(qh,h^{-1}g)$ for $h\in H$, then
\begin{align*}
(qh)(h^{-1}g)=qg.
\end{align*}
The map is smooth because it is induced by the smooth right action map
\begin{align*}
Q_s\times G &\to P,\quad (q,g)\mapsto qg.
\end{align*}
For surjectivity, let $p\in P$ and set $x:=\pi(p)$. Since $s(x)\in (P/H)_x$, there exists $q\in P_x$ such that $s(x)=[q]_H$. Then $q\in Q_s$. Since $P_x$ is a principal right $G$-torsor, there exists $g\in G$ such that $p=qg$. Hence $p=\Psi_s([q,g])$.
For injectivity, suppose
\begin{align*}
\Psi_s([q_1,g_1])=\Psi_s([q_2,g_2]).
\end{align*}
Then $q_1g_1=q_2g_2$, so $q_2=q_1g_1g_2^{-1}$. Since $q_1,q_2\in Q_s$ lie over the same base point, their $H$-orbits in $P/H$ agree:
\begin{align*}
[q_1]_H=s(\pi(q_1))=s(\pi(q_2))=[q_2]_H.
\end{align*}
Thus there exists $h\in H$ such that $q_2=q_1h$. Comparing with $q_2=q_1g_1g_2^{-1}$ and using freeness of the right $G$-action on $P$, we get
\begin{align*}
h=g_1g_2^{-1}.
\end{align*}
Therefore $g_1=hg_2$, and
\begin{align*}
(q_2,g_2)=(q_1h,g_2)
\end{align*}
represents the same class in $Q_s\times_H G$ as $(q_1,hg_2)=(q_1,g_1)$. Hence $[q_1,g_1]=[q_2,g_2]$.
The map $\Psi_s$ is a smooth bijective principal $G$-bundle morphism over $M$. In the local trivialisation above, it is identified with the standard map from the associated bundle $R_\sigma\times_H G$ to $U\times G$, so its inverse is smooth locally. Hence $\Psi_s$ is a smooth principal $G$-bundle isomorphism. Therefore $Q_s$ is an $H$-reduction of $P$.
[guided]
The definition of an $H$-reduction requires more than the existence of a principal $H$-bundle inside $P$: extending the structure group from $H$ back to $G$ must recover $P$. We verify this directly.
Define
\begin{align*}
\Psi_s:Q_s\times_H G &\to P
\end{align*}
by
\begin{align*}
\Psi_s([q,g])=qg.
\end{align*}
Here $Q_s\times_H G$ is the quotient of $Q_s\times G$ by the relation
\begin{align*}
(q,g)\sim(qh,h^{-1}g)\quad \text{for } h\in H.
\end{align*}
The formula is well-defined because
\begin{align*}
(qh)(h^{-1}g)=qg.
\end{align*}
Thus changing representatives does not change the image in $P$.
To prove surjectivity, take any point $p\in P$ and define $x:=\pi(p)$. The section value $s(x)$ lies in the quotient fibre $(P/H)_x$, so it is an $H$-orbit in the fibre $P_x$. Hence there is some $q\in P_x$ with
\begin{align*}
s(x)=[q]_H.
\end{align*}
This condition is exactly the defining condition for membership in $Q_s$, because
\begin{align*}
\rho(q)=[q]_H=s(x)=s(\pi(q)).
\end{align*}
So $q\in Q_s$. Since $P_x$ is a principal right $G$-torsor, there exists $g\in G$ such that
\begin{align*}
p=qg.
\end{align*}
Therefore $p=\Psi_s([q,g])$, proving surjectivity.
For injectivity, suppose two classes have the same image:
\begin{align*}
\Psi_s([q_1,g_1])=\Psi_s([q_2,g_2]).
\end{align*}
This means
\begin{align*}
q_1g_1=q_2g_2.
\end{align*}
Rearranging inside the right $G$-torsor $P_{\pi(q_1)}$ gives
\begin{align*}
q_2=q_1g_1g_2^{-1}.
\end{align*}
Because both $q_1$ and $q_2$ lie in $Q_s$, their quotient points are prescribed by the same section value:
\begin{align*}
[q_1]_H=s(\pi(q_1))=s(\pi(q_2))=[q_2]_H.
\end{align*}
Thus $q_1$ and $q_2$ differ by the right action of an element of $H$. There exists $h\in H$ such that
\begin{align*}
q_2=q_1h.
\end{align*}
Comparing this with $q_2=q_1g_1g_2^{-1}$ and using freeness of the right $G$-action on $P$, we obtain
\begin{align*}
h=g_1g_2^{-1}.
\end{align*}
Equivalently, $g_1=hg_2$. The defining equivalence relation in $Q_s\times_H G$ gives
\begin{align*}
(q_1,g_1)=(q_1,hg_2)\sim(q_1h,g_2)=(q_2,g_2),
\end{align*}
so
\begin{align*}
[q_1,g_1]=[q_2,g_2].
\end{align*}
Hence $\Psi_s$ is injective.
Smoothness of $\Psi_s$ follows from smoothness of the right $G$-action on $P$. Its inverse is smooth locally because, in a local trivialisation of $P$, the map becomes the standard associated-bundle identification
\begin{align*}
R_\sigma\times_H G &\to U\times G.
\end{align*}
Therefore $\Psi_s$ is a smooth principal $G$-bundle isomorphism, and $Q_s$ is an $H$-reduction of $P$.
[/guided]
[/step]
[step:Verify that the two constructions are inverse]
Let $Q\subset P$ be an $H$-reduction. For $p\in P$,
\begin{align*}
p\in Q_{s_Q}
\end{align*}
if and only if
\begin{align*}
[p]_H=s_Q(\pi(p)).
\end{align*}
Since $Q$ is an $H$-reduction, every point of $P_{\pi(p)}$ lies in the right $G$-orbit of $Q_{\pi(p)}$, and the condition $[p]_H=s_Q(\pi(p))$ says precisely that $p$ lies in the right $H$-orbit of a point of $Q_{\pi(p)}$. Because $Q$ is stable under the right $H$-action, this is equivalent to $p\in Q$. Thus
\begin{align*}
Q_{s_Q}=Q.
\end{align*}
Conversely, let $s:M\to P/H$ be a smooth section. For $x\in M$, choose any $q\in (Q_s)_x$. By definition of $Q_s$,
\begin{align*}
[q]_H=\rho(q)=s(\pi(q))=s(x).
\end{align*}
Therefore
\begin{align*}
s_{Q_s}(x)=[q]_H=s(x).
\end{align*}
Since this holds for every $x\in M$,
\begin{align*}
s_{Q_s}=s.
\end{align*}
The two assignments are inverse bijections.
[/step]
[step:Record naturality of the correspondence]
The construction from $Q$ to $s_Q$ uses only the inclusion $Q\hookrightarrow P$ and the quotient map $\rho:P\to P/H$. The construction from $s$ to $Q_s$ uses only the equality condition
\begin{align*}
\rho(p)=s(\pi(p)).
\end{align*}
Consequently, the bijection is canonical with respect to the displayed constructions. This completes the proof.
[/step]
