[proofplan] We first associate to a flat principal bundle with a chosen point $p$ in the fibre over $x$ its monodromy homomorphism by parallel transport around loops based at $x$. Flatness makes this assignment depend only on the homotopy class of the loop, and the principal right action makes changing the reference point in $P_x$ act by conjugation. Conversely, from a homomorphism $\rho:\pi_1(M,x)\to G$ we construct the quotient bundle $(\widetilde M\times G)/\pi_1(M,x)$, descend the product flat connection, and compute that its monodromy is exactly $\rho$. Finally, we check that these two constructions are inverse up to connection-preserving principal bundle isomorphism and that conjugate homomorphisms give isomorphic flat bundles. [/proofplan] [step:Define the monodromy representation from parallel transport] Let \begin{align*} \pi: P \to M \end{align*} be a smooth principal right $G$-bundle with a flat smooth principal connection. Choose a point $p \in P_x$. For a smooth loop \begin{align*} \gamma: [0,1] \to M \end{align*} with $\gamma(0)=\gamma(1)=x$, let \begin{align*} \operatorname{PT}_{\gamma}: P_x \to P_x \end{align*} denote parallel transport along $\gamma$. Because the connection is principal, parallel transport is $G$-equivariant: \begin{align*} \operatorname{PT}_{\gamma}(u g)=\operatorname{PT}_{\gamma}(u)g \end{align*} for all $u \in P_x$ and all $g \in G$. Since the right action of $G$ on the fibre $P_x$ is free and transitive, there is a unique element $\rho_p([\gamma]) \in G$ satisfying \begin{align*} \operatorname{PT}_{\gamma}(p)=p\,\rho_p([\gamma]). \end{align*} By the homotopy invariance theorem for parallel transport of smooth flat principal connections, parallel transport along smooth paths with fixed endpoints is unchanged under smooth endpoint-fixed homotopy. The hypotheses apply because the connection on $P$ is smooth and flat. Hence this element depends only on the class $[\gamma]\in \pi_1(M,x)$. We use the multiplication on $\pi_1(M,x)$ induced by the chosen identification with the deck group in the following inverse-endpoint convention: if $\gamma$ represents $\alpha$, then the lift of $\gamma$ starting at $\tilde x$ ends at $\alpha^{-1}\tilde x$. With this convention, if $\alpha$ is represented by $\gamma$ and $\beta$ is represented by $\eta$, then $\alpha\beta$ is represented by the based loop $\eta * \gamma$, where $\eta * \gamma$ means first traverse $\eta$ and then traverse $\gamma$. This is the convention compatible with the right quotient action used below. Parallel transport along a concatenated path composes in the same chronological order: for $\eta * \gamma$, \begin{align*} \operatorname{PT}_{\eta * \gamma}=\operatorname{PT}_{\gamma}\circ \operatorname{PT}_{\eta}. \end{align*} Therefore \begin{align*} \operatorname{PT}_{\eta * \gamma}(p)=\operatorname{PT}_{\gamma}\bigl(p\,\rho_p(\beta)\bigr)=\operatorname{PT}_{\gamma}(p)\,\rho_p(\beta)=p\,\rho_p(\alpha)\rho_p(\beta). \end{align*} Since $\eta * \gamma$ represents $\alpha\beta$, the defining uniqueness in the right $G$-torsor $P_x$ gives \begin{align*} \rho_p(\alpha\beta)=\rho_p(\alpha)\rho_p(\beta). \end{align*} Hence $\rho_p:\pi_1(M,x)\to G$ is a [group homomorphism](/page/Group%20Homomorphism). [guided] The fibre $P_x$ is a right $G$-torsor: for any two points $u,v\in P_x$, there is a unique $g\in G$ such that $v=ug$. This is why parallel transport around a loop produces a group element. Starting from the chosen reference point $p\in P_x$, transport it around a loop $\gamma:[0,1]\to M$ based at $x$. The endpoint again lies in $P_x$, so there is a unique element $\rho_p([\gamma])\in G$ such that \begin{align*} \operatorname{PT}_{\gamma}(p)=p\,\rho_p([\gamma]). \end{align*} We must check that this depends only on the homotopy class of $\gamma$. The connection is flat, meaning its curvature vanishes. The homotopy invariance theorem for parallel transport of smooth flat principal connections says that parallel transport along smooth paths with the same endpoints is invariant under smooth homotopy relative to endpoints. The hypotheses apply here because the connection is smooth and flat on a smooth principal bundle. Since loops representing the same element of $\pi_1(M,x)$ are homotopic with basepoint fixed, the element $\rho_p([\gamma])$ is well-defined on $\pi_1(M,x)$. Now we verify the multiplicative law and make the convention explicit. The chosen lift $\tilde x$ identifies $\pi_1(M,x)$ with the deck group by the inverse-endpoint convention: a loop representing $\alpha$ lifts from $\tilde x$ to $\alpha^{-1}\tilde x$. Under this convention, if $\alpha$ is represented by $\gamma$ and $\beta$ is represented by $\eta$, then $\alpha\beta$ is represented by $\eta * \gamma$, where $\eta * \gamma$ means first traverse $\eta$ and then traverse $\gamma$. Parallel transport along this concatenation is composition in chronological order: \begin{align*} \operatorname{PT}_{\eta * \gamma}=\operatorname{PT}_{\gamma}\circ \operatorname{PT}_{\eta}. \end{align*} Using $G$-equivariance of parallel transport, we compute \begin{align*} \operatorname{PT}_{\eta * \gamma}(p)=\operatorname{PT}_{\gamma}\bigl(\operatorname{PT}_{\eta}(p)\bigr). \end{align*} Since $\operatorname{PT}_{\eta}(p)=p\,\rho_p(\beta)$, this becomes \begin{align*} \operatorname{PT}_{\eta * \gamma}(p)=\operatorname{PT}_{\gamma}\bigl(p\,\rho_p(\beta)\bigr). \end{align*} The connection is principal, so parallel transport commutes with the right $G$-action: \begin{align*} \operatorname{PT}_{\gamma}\bigl(p\,\rho_p(\beta)\bigr)=\operatorname{PT}_{\gamma}(p)\,\rho_p(\beta). \end{align*} Finally, $\operatorname{PT}_{\gamma}(p)=p\,\rho_p(\alpha)$, and hence \begin{align*} \operatorname{PT}_{\eta * \gamma}(p)=p\,\rho_p(\alpha)\rho_p(\beta). \end{align*} Because $\eta * \gamma$ represents $\alpha\beta$, uniqueness of the group element carrying $p$ to the transported endpoint gives \begin{align*} \rho_p(\alpha\beta)=\rho_p(\alpha)\rho_p(\beta). \end{align*} Thus the monodromy associated to the chosen point $p$ is a genuine homomorphism $\rho_p:\pi_1(M,x)\to G$. [/guided] [/step] [step:Changing the reference point conjugates the monodromy] Let $h\in G$ and replace $p$ by $p h\in P_x$. For every based loop $\gamma$ representing $\alpha\in\pi_1(M,x)$, $G$-equivariance of parallel transport gives \begin{align*} \operatorname{PT}_{\gamma}(p h) = \operatorname{PT}_{\gamma}(p)h. \end{align*} Using the definition of $\rho_p$, this becomes \begin{align*} \operatorname{PT}_{\gamma}(p h) = p\,\rho_p(\alpha)h. \end{align*} Since $p\,\rho_p(\alpha)h=(p h)(h^{-1}\rho_p(\alpha)h)$, the monodromy based at $p h$ is \begin{align*} \rho_{p h}(\alpha)=h^{-1}\rho_p(\alpha)h. \end{align*} Thus forgetting the chosen point in the fibre sends a based monodromy representation to its [conjugacy class](/page/Conjugacy%20Class). [/step] [step:Construct the flat bundle associated to a representation] Let \begin{align*} \rho:\pi_1(M,x)\to G \end{align*} be a Lie group homomorphism. Define a right action of $\pi_1(M,x)$ on $\widetilde M\times G$ by \begin{align*} (\tilde m,g)\cdot\alpha=(\alpha^{-1}\tilde m,\rho(\alpha)^{-1}g). \end{align*} This is a right action: for $\alpha,\beta\in\pi_1(M,x)$, \begin{align*} \bigl((\tilde m,g)\cdot\alpha\bigr)\cdot\beta=(\beta^{-1}\alpha^{-1}\tilde m,\rho(\beta)^{-1}\rho(\alpha)^{-1}g). \end{align*} Since $\rho$ is a homomorphism, $\rho(\alpha\beta)^{-1}=\rho(\beta)^{-1}\rho(\alpha)^{-1}$, so this is exactly \begin{align*} (\tilde m,g)\cdot(\alpha\beta)=((\alpha\beta)^{-1}\tilde m,\rho(\alpha\beta)^{-1}g). \end{align*} The identity element acts as the identity because $\rho(e)=e$. The action is smooth, free, and proper because the deck action on $\widetilde M$ is smooth, free, and proper; hence the standard quotient theorem for smooth free proper actions gives a smooth quotient manifold. Let \begin{align*} P_{\rho}:=(\widetilde M\times G)/\pi_1(M,x) \end{align*} be the quotient by this right action, and write $[\tilde m,g]$ for the equivalence class of $(\tilde m,g)$. The projection \begin{align*} \pi_{\rho}:P_{\rho}\to M,\qquad [\tilde m,g]\mapsto q(\tilde m) \end{align*} is well-defined because $q(\alpha^{-1}\tilde m)=q(\tilde m)$ for every deck transformation $\alpha$. The right action \begin{align*} P_{\rho}\times G \to P_{\rho},\qquad ([\tilde m,g],h)\mapsto [\tilde m,gh] \end{align*} is well-defined because \begin{align*} (\tilde m,gh)\cdot\alpha = (\alpha^{-1}\tilde m,\rho(\alpha)^{-1}gh) \end{align*} is obtained from $(\tilde m,g)\cdot\alpha$ by right multiplication by $h$. Therefore $P_{\rho}$ is a smooth principal right $G$-bundle over $M$. On the product bundle $\widetilde M\times G\to\widetilde M$, take the product connection whose horizontal subspace at $(\tilde m,g)$ is \begin{align*} H_{(\tilde m,g)}:=T_{\tilde m}\widetilde M\oplus \{0\}\subset T_{\tilde m}\widetilde M\oplus T_gG. \end{align*} For each $\alpha\in\pi_1(M,x)$, denote by \begin{align*} R_{\alpha}:\widetilde M\times G\to \widetilde M\times G \end{align*} the smooth map defining the right action, \begin{align*} R_{\alpha}(\tilde m,g)=(\alpha^{-1}\tilde m,\rho(\alpha)^{-1}g). \end{align*} The differential of $R_{\alpha}$ sends \begin{align*} T_{\tilde m}\widetilde M\oplus\{0\} \end{align*} onto \begin{align*} T_{\alpha^{-1}\tilde m}\widetilde M\oplus\{0\}. \end{align*} Hence the horizontal distribution is invariant under $\pi_1(M,x)$ and descends by the standard descent criterion for invariant Ehresmann connections to a smooth principal connection on $P_{\rho}$. Since the product horizontal distribution is integrable and has zero curvature, the descended connection is flat. [guided] The construction uses the deck group as a left action on $\widetilde M$, and the displayed formula makes it into a right action on $\widetilde M\times G$ by using $\alpha^{-1}$ on the first coordinate and $\rho(\alpha)^{-1}$ on the second. The homomorphism identity gives \begin{align*} \rho(\alpha\beta)^{-1}=\rho(\beta)^{-1}\rho(\alpha)^{-1}, \end{align*} so applying $\alpha$ and then $\beta$ gives the same result as applying $\alpha\beta$. Thus the quotient $P_\rho$ is well-defined. The projection to $M$ is independent of representatives because deck transformations preserve $q:\widetilde M\to M$. The principal right $G$-action is also independent of representatives because multiplying $(\tilde m,g)$ by $h$ on the right changes the second coordinate after the $\pi_1(M,x)$-action from $\rho(\alpha)^{-1}g$ to $\rho(\alpha)^{-1}gh$. Finally, the product horizontal distribution $T\widetilde M\oplus\{0\}$ is preserved by the action, so it descends to a flat principal connection on the quotient. [/guided] [/step] [step:Compute the monodromy of the constructed bundle] Let \begin{align*} p_{\rho}:=[\tilde x,e]\in (P_{\rho})_x, \end{align*} where $e\in G$ is the identity element. Let $\alpha\in\pi_1(M,x)$, and let \begin{align*} \gamma:[0,1]\to M \end{align*} be a loop representing $\alpha$. Let \begin{align*} \tilde\gamma:[0,1]\to\widetilde M \end{align*} be the lift of $\gamma$ with $\tilde\gamma(0)=\tilde x$. By the chosen inverse-endpoint identification of the deck group with $\pi_1(M,x)$, \begin{align*} \tilde\gamma(1)=\alpha^{-1}\tilde x. \end{align*} The horizontal lift in $P_{\rho}$ starting at $p_{\rho}$ is \begin{align*} t\mapsto [\tilde\gamma(t),e], \end{align*} because its lift to $\widetilde M\times G$ has tangent vector in $T_{\tilde\gamma(t)}\widetilde M\oplus\{0\}$. Its endpoint is \begin{align*} [\alpha^{-1}\tilde x,e]. \end{align*} In the quotient relation for the corrected right action, \begin{align*} (\tilde x,\rho(\alpha))\cdot\alpha=(\alpha^{-1}\tilde x,e), \end{align*} so \begin{align*} [\alpha^{-1}\tilde x,e]=[\tilde x,\rho(\alpha)]. \end{align*} Therefore \begin{align*} \operatorname{PT}_{\gamma}(p_{\rho}) = p_{\rho}\,\rho(\alpha), \end{align*} and the monodromy representation of $(P_{\rho},p_{\rho})$ is $\rho$. [guided] Start at the point $p_\rho=[\tilde x,e]$. A loop representing $\alpha$ lifts from $\tilde x$ to $\alpha^{-1}\tilde x$ by the convention in the statement. In the product model, horizontal paths keep the $G$-coordinate fixed, so the horizontal lift ends at \begin{align*} [\alpha^{-1}\tilde x,e]. \end{align*} The quotient relation says \begin{align*} (\tilde x,\rho(\alpha))\cdot\alpha=(\alpha^{-1}\tilde x,e), \end{align*} so this endpoint is the same point as $[\tilde x,\rho(\alpha)]$. Since $[\tilde x,\rho(\alpha)]=[\tilde x,e]\rho(\alpha)$, parallel transport around the loop multiplies the reference point by $\rho(\alpha)$. Therefore the constructed flat bundle has monodromy $\rho$. [/guided] [/step] [step:Recover a flat bundle from its monodromy representation] Let \begin{align*} \pi:P\to M \end{align*} be a smooth principal right $G$-bundle with flat connection, and choose $p\in P_x$. Pull it back to the universal cover: \begin{align*} \widetilde P:=q^*P=\{(\tilde m,u)\in \widetilde M\times P:q(\tilde m)=\pi(u)\}. \end{align*} The pulled-back connection on $\widetilde P\to\widetilde M$ is flat. Define a map \begin{align*} \Phi:\widetilde M\times G\to \widetilde P \end{align*} as follows. For $\tilde m\in\widetilde M$, choose any smooth path \begin{align*} \tilde c:[0,1]\to\widetilde M \end{align*} from $\tilde x$ to $\tilde m$. Let $c=q\circ\tilde c$, and let $\operatorname{PT}_c(p)\in P_{q(\tilde m)}$ be parallel transport of $p$ along $c$. Set \begin{align*} \Phi(\tilde m,g):=(\tilde m,\operatorname{PT}_c(p)g). \end{align*} Because $\widetilde M$ is simply connected, any two smooth paths in $\widetilde M$ with the same endpoints are homotopic relative to endpoints. The pulled-back connection is smooth and flat, so the homotopy invariance theorem for parallel transport of smooth flat principal connections implies that parallel transport between two fixed points of $\widetilde M$ is independent of the path. Hence $\Phi$ is well-defined. To see smoothness, choose a simply connected coordinate ball $B\subset\widetilde M$ and a smooth family of paths from a fixed point of $B$ to points of $B$, for instance radial coordinate paths. The ordinary differential equation for horizontal lifting has smooth coefficients, so its endpoint depends smoothly on the endpoint in $B$. Path-independence then patches these local smooth descriptions to a smooth global map. The same horizontal-lift construction shows that $\Phi$ sends $T\widetilde M\oplus\{0\}$ to the pulled-back horizontal distribution and is $G$-equivariant for the right action. Therefore $\Phi$ is a connection-preserving principal bundle trivialization of $\widetilde P$. Let $\rho_p:\pi_1(M,x)\to G$ be the monodromy representation of $(P,p)$. We compute the deck action under the trivialization $\Phi$. Fix $\alpha\in\pi_1(M,x)$ and $\tilde m\in\widetilde M$, and choose a smooth path $\tilde c:[0,1]\to\widetilde M$ from $\tilde x$ to $\tilde m$. Let $\tilde\ell_{\alpha}:[0,1]\to\widetilde M$ be the lift of a based loop representing $\alpha$ with $\tilde\ell_{\alpha}(0)=\tilde x$, so $\tilde\ell_{\alpha}(1)=\alpha^{-1}\tilde x$. Define the concatenated path $\tilde d:[0,1]\to\widetilde M$ from $\tilde x$ to $\alpha^{-1}\tilde m$ by first traversing $\tilde\ell_{\alpha}$ and then traversing $\alpha^{-1}\tilde c$. Its projection $d=q\circ\tilde d$ is the concatenation of the based loop representing $\alpha$ with $c=q\circ\tilde c$. Parallel transport first sends $p$ to $p\rho_p(\alpha)$, and then along $c$ sends this point to $\operatorname{PT}_{c}(p)\rho_p(\alpha)$ by right $G$-equivariance of parallel transport. Therefore, for an arbitrary coordinate $g'\in G$, \begin{align*} \Phi(\alpha^{-1}\tilde m,g')=(\alpha^{-1}\tilde m,\operatorname{PT}_{c}(p)\rho_p(\alpha)g'). \end{align*} The deck-transformed point corresponding to $\Phi(\tilde m,g)=(\tilde m,\operatorname{PT}_{c}(p)g)$ is $(\alpha^{-1}\tilde m,\operatorname{PT}_{c}(p)g)$ in $\widetilde P$. To find its $\Phi$-coordinate, solve \begin{align*} \rho_p(\alpha)g'=g, \end{align*} which gives $g'=\rho_p(\alpha)^{-1}g$. Thus the deck transformation induced on $\widetilde M\times G$ is the right action \begin{align*} (\tilde m,g)\mapsto(\alpha^{-1}\tilde m,\rho_p(\alpha)^{-1}g). \end{align*} The formula at general $g\in G$ follows from right $G$-equivariance of $\Phi$, and the formula at general $\tilde m$ follows from the path-independence of flat parallel transport on $\widetilde M$ just proved. Consequently \begin{align*} P\cong \widetilde P/\pi_1(M,x)\cong (\widetilde M\times G)/\pi_1(M,x)=P_{\rho_p}, \end{align*} and the isomorphism preserves both the principal right $G$-action and the connection. [guided] Pull the given flat bundle \begin{align*} \pi:P\to M \end{align*} back to the universal cover, obtaining \begin{align*} \widetilde P=q^*P=\{(\tilde m,u)\in\widetilde M\times P:q(\tilde m)=\pi(u)\}. \end{align*} Choose $p\in P_x$. For $\tilde m\in\widetilde M$ and $g\in G$, choose a smooth path \begin{align*} \tilde c:[0,1]\to\widetilde M \end{align*} from $\tilde x$ to $\tilde m$, put $c=q\circ\tilde c$, and define \begin{align*} \Phi(\tilde m,g)=(\tilde m,\operatorname{PT}_c(p)g). \end{align*} Since $\widetilde M$ is simply connected and the pulled-back connection is flat, parallel transport between two fixed points of $\widetilde M$ is independent of the chosen path. Hence the displayed formula for $\Phi$ is well-defined. Smoothness is local: on a simply connected coordinate ball $B\subset\widetilde M$, choose a smooth family of paths to points of $B$; the horizontal-lift ordinary differential equation has smooth coefficients, so $\operatorname{PT}_c(p)$ depends smoothly on the endpoint. These local descriptions agree by path-independence, so $\Phi$ is smooth. It is right $G$-equivariant because \begin{align*} \Phi(\tilde m,gh)=(\tilde m,\operatorname{PT}_c(p)gh)=\Phi(\tilde m,g)h. \end{align*} It preserves the product horizontal distribution because varying $\tilde m$ horizontally is exactly the horizontal lift used to define parallel transport. Now compute how a deck transformation acts in the $\Phi$-coordinates. Fix $\alpha\in\pi_1(M,x)$ and $\tilde m\in\widetilde M$. Let $\tilde c$ run from $\tilde x$ to $\tilde m$, and let $\tilde\ell_\alpha$ be the lift of a based loop representing $\alpha$, so \begin{align*} \tilde\ell_\alpha(0)=\tilde x,\qquad \tilde\ell_\alpha(1)=\alpha^{-1}\tilde x. \end{align*} The path to $\alpha^{-1}\tilde m$ first follows $\tilde\ell_\alpha$ and then follows $\alpha^{-1}\tilde c$. Its projection first transports $p$ to \begin{align*} p\rho_p(\alpha) \end{align*} and then transports along $c=q\circ\tilde c$, giving \begin{align*} \operatorname{PT}_c(p)\rho_p(\alpha). \end{align*} Therefore, for an arbitrary coordinate $g'\in G$, \begin{align*} \Phi(\alpha^{-1}\tilde m,g')=(\alpha^{-1}\tilde m,\operatorname{PT}_c(p)\rho_p(\alpha)g'). \end{align*} The deck transform of the point with coordinate $g$ has second component \begin{align*} \operatorname{PT}_c(p)g. \end{align*} To express this transformed point in $\Phi$-coordinates, solve \begin{align*} \rho_p(\alpha)g'=g. \end{align*} Thus \begin{align*} g'=\rho_p(\alpha)^{-1}g. \end{align*} So the induced action on $\widetilde M\times G$ is \begin{align*} (\tilde m,g)\mapsto(\alpha^{-1}\tilde m,\rho_p(\alpha)^{-1}g), \end{align*} which is exactly the quotient action defining $P_{\rho_p}$. Hence \begin{align*} P\cong \widetilde P/\pi_1(M,x)\cong(\widetilde M\times G)/\pi_1(M,x)=P_{\rho_p}, \end{align*} and the isomorphism preserves the right $G$-action and the connection. [/guided] [/step] [step:Identify isomorphisms with conjugacy of representations] Let \begin{align*} F:P\to P' \end{align*} be a connection-preserving principal bundle isomorphism between two flat principal right $G$-bundles over $M$. Choose $p\in P_x$. Since $F$ maps $P_x$ to $P'_x$ and is $G$-equivariant, $F(p)\in P'_x$ is a reference point. Because $F$ preserves horizontal subspaces, it carries horizontal lifts of paths in $P$ to horizontal lifts of the same paths in $P'$. Hence for every $\alpha\in\pi_1(M,x)$, \begin{align*} \rho_{F(p)}'(\alpha)=\rho_p(\alpha). \end{align*} If instead $p'\in P'_x$ is another reference point, then there is a unique $h\in G$ such that \begin{align*} F(p)=p'h. \end{align*} By the conjugation computation above, \begin{align*} \rho_{p'}'(\alpha)=h\,\rho_p(\alpha)h^{-1} \end{align*} for every $\alpha\in\pi_1(M,x)$. Thus isomorphic flat bundles determine the same conjugacy class of representations. Conversely, suppose \begin{align*} \rho'(\alpha)=h\rho(\alpha)h^{-1} \end{align*} for a fixed $h\in G$ and every $\alpha\in\pi_1(M,x)$. Define \begin{align*} F_h:P_{\rho}\to P_{\rho'} \end{align*} by \begin{align*} F_h([\tilde m,g])=[\tilde m,hg]. \end{align*} This is well-defined: if $(\tilde m,g)$ is replaced by $(\tilde m,g)\cdot\alpha=(\alpha^{-1}\tilde m,\rho(\alpha)^{-1}g)$, then \begin{align*} F_h([\alpha^{-1}\tilde m,\rho(\alpha)^{-1}g]) = [\alpha^{-1}\tilde m,h\rho(\alpha)^{-1}g]. \end{align*} Using $\rho'(\alpha)=h\rho(\alpha)h^{-1}$, we have $\rho'(\alpha)^{-1}h=h\rho(\alpha)^{-1}$. Thus the displayed representative is exactly the result of applying the $\alpha$-action in $P_{\rho'}$ to $(\tilde m,hg)$, and hence it is equivalent in $P_{\rho'}$ to \begin{align*} [\tilde m,hg]. \end{align*} The map $F_h$ is smooth, covers $\operatorname{id}_M$, commutes with the right $G$-action, and sends the descended horizontal distribution from $T\widetilde M\oplus\{0\}$ to itself. Hence $F_h$ is a connection-preserving principal bundle isomorphism. Combining the preceding steps, the constructions \begin{align*} (P,p)\mapsto \rho_p \end{align*} and \begin{align*} \rho\mapsto P_{\rho} \end{align*} are inverse after passing from chosen fibre points to conjugacy classes. This gives the claimed natural bijection between isomorphism classes of smooth flat principal right $G$-bundles over $M$ and conjugacy classes of homomorphisms $\pi_1(M,x)\to G$. [guided] Let \begin{align*} F:P\to P' \end{align*} be a connection-preserving principal bundle isomorphism over $\operatorname{id}_M$, and choose $p\in P_x$. Since $F$ is $G$-equivariant and maps fibres over $x$ to fibres over $x$, the point $F(p)\in P'_x$ can be used as the reference point for $P'$. If $\gamma$ represents $\alpha\in\pi_1(M,x)$, then $F$ carries the horizontal lift of $\gamma$ through $p$ to the horizontal lift of $\gamma$ through $F(p)$. Therefore \begin{align*} \rho'_{F(p)}(\alpha)=\rho_p(\alpha). \end{align*} If another reference point $p'\in P'_x$ is chosen, there is a unique $h\in G$ with \begin{align*} F(p)=p'h. \end{align*} The reference-point conjugation formula gives \begin{align*} \rho'_{p'}(\alpha)=h\rho_p(\alpha)h^{-1}. \end{align*} Thus isomorphic flat bundles determine the same conjugacy class of homomorphisms. Conversely, suppose two homomorphisms $\rho,\rho':\pi_1(M,x)\to G$ satisfy \begin{align*} \rho'(\alpha)=h\rho(\alpha)h^{-1} \end{align*} for every $\alpha\in\pi_1(M,x)$. Define \begin{align*} F_h:P_\rho\to P_{\rho'} \end{align*} by \begin{align*} F_h([\tilde m,g])=[\tilde m,hg]. \end{align*} To check well-definedness, replace $(\tilde m,g)$ by \begin{align*} (\tilde m,g)\cdot\alpha=(\alpha^{-1}\tilde m,\rho(\alpha)^{-1}g). \end{align*} Then \begin{align*} F_h([\alpha^{-1}\tilde m,\rho(\alpha)^{-1}g]) = [\alpha^{-1}\tilde m,h\rho(\alpha)^{-1}g]. \end{align*} The conjugacy identity implies \begin{align*} \rho'(\alpha)^{-1}h=h\rho(\alpha)^{-1}, \end{align*} so the last displayed representative is obtained from $(\tilde m,hg)$ by the $\alpha$-action defining $P_{\rho'}$. Hence it represents the same class as \begin{align*} [\tilde m,hg]. \end{align*} The map $F_h$ is smooth, covers $\operatorname{id}_M$, commutes with the right $G$-action, and preserves the descended horizontal distribution $T\widetilde M\oplus\{0\}$. Therefore $F_h$ is a connection-preserving principal bundle isomorphism, and conjugate representations give the same flat bundle class. Combining this with the inverse construction proves the desired bijection. [/guided] [/step]