[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.[/step]

[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: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 $M$.[/step]

[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: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 $P$.[/step]

[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: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]