[proofplan] We construct two explicit local trivialisations over two angular charts on $S^1$. One chart cuts the circle at $1$, so every point has a unique representative with parameter in $(0,1)$; the other cuts at $-1$, so every point has a unique representative with parameter in $(-1/2,1/2)$. On each chart the quotient is represented by a genuine product with $F$, and on the two overlap components the change of trivialisation is either the identity map of $F$ or one of the maps $f$ and $f^{-1}$. These transition maps lie in the cyclic subgroup $\langle f\rangle$, hence also in any specified Lie subgroup of $\operatorname{Diff}(F)$ containing $f$. [/proofplan] [step:Define angular charts and representatives for the quotient] By definition, the mapping torus $T_f$ is the quotient of $F\times\mathbb{R}$ by the $\mathbb{Z}$-action \begin{align*} \rho:\mathbb{Z}\times(F\times\mathbb{R})&\to F\times\mathbb{R}, & \rho(n,(y,t))&=(f^n(y),t-n). \end{align*} Let $q:F\times\mathbb{R}\to T_f$ denote the quotient map, so $q(y,t)=[y,t]$. This action is smooth because each iterate $f^n:F\to F$ is a diffeomorphism and $t\mapsto t-n$ is a smooth translation of $\mathbb{R}$. It is free because $\rho(n,(y,t))=(y,t)$ implies $t-n=t$, hence $n=0$. It is proper because the translation action of the discrete group $\mathbb{Z}$ on $\mathbb{R}$ is proper, and the projection $F\times\mathbb{R}\to\mathbb{R}$ is equivariant for this translation action. By the quotient-manifold theorem for free proper smooth actions of discrete Lie groups, $T_f$ has a unique smooth structure for which $q$ is a smooth covering map. The quotient relation is generated by $(y,t)\sim(f^n(y),t-n)$ for $y\in F$, $t\in\mathbb{R}$, and $n\in\mathbb{Z}$. The map $\pi:T_f\to S^1$ is well-defined because \begin{align*} e^{2\pi i(t-n)}=e^{2\pi i t}e^{-2\pi i n}=e^{2\pi i t} \end{align*} for every $n\in\mathbb{Z}$. Let $U_0:=S^1\setminus\{-1\}$ and let $U_1:=S^1\setminus\{1\}$. Define the angular coordinate maps $\theta_0:U_0\to(-1/2,1/2)$ and $\theta_1:U_1\to(0,1)$ by declaring $\theta_0(e^{2\pi i s})=s$ for the unique $s\in(-1/2,1/2)$ and $\theta_1(e^{2\pi i s})=s$ for the unique $s\in(0,1)$. These are smooth coordinate maps on the two open arcs. For each $[y,t]\in \pi^{-1}(U_j)$, where $j\in\{0,1\}$, there is a unique real number $s_j\in I_j$ such that $e^{2\pi i s_j}=\pi([y,t])$, where $I_0:=(-1/2,1/2)$ and $I_1:=(0,1)$. There is also a unique integer $n_j\in\mathbb{Z}$ such that $s_j=t-n_j$. The [equivalence relation](/page/Equivalence%20Relation) gives \begin{align*} [y,t]=[f^{n_j}(y),s_j]. \end{align*} Thus every point of $\pi^{-1}(U_j)$ has a unique representative of the form $(y_j,s_j)\in F\times I_j$. [guided] We first spell out the quotient structure because every later formula depends on using the same convention. The mapping torus $T_f$ is the quotient of $F\times\mathbb{R}$ by the action \begin{align*} \rho:\mathbb{Z}\times(F\times\mathbb{R})&\to F\times\mathbb{R}, & \rho(n,(y,t))&=(f^n(y),t-n). \end{align*} Let $q:F\times\mathbb{R}\to T_f$ be the quotient map defined by \begin{align*} q(y,t)=[y,t]. \end{align*} The action is smooth: for each $n\in\mathbb{Z}$, the map \begin{align*} F\times\mathbb{R}&\to F\times\mathbb{R}, & (y,t)&\mapsto (f^n(y),t-n) \end{align*} is a product of the diffeomorphism $f^n:F\to F$ and a smooth translation of $\mathbb{R}$. The action is free because if \begin{align*} \rho(n,(y,t))=(y,t), \end{align*} then $t-n=t$, so $n=0$. The action is proper because the translation action of the discrete group $\mathbb{Z}$ on $\mathbb{R}$ is proper and the projection \begin{align*} F\times\mathbb{R}&\to\mathbb{R}, & (y,t)&\mapsto t \end{align*} is equivariant for the two translation actions. Therefore the quotient-manifold theorem for free proper smooth actions of discrete Lie groups gives a smooth structure on $T_f$ for which $q$ is a smooth covering map. In this quotient, the equivalence relation is generated by \begin{align*} (y,t)\sim(f^n(y),t-n) \end{align*} for $y\in F$, $t\in\mathbb{R}$, and $n\in\mathbb{Z}$. The map $\pi:T_f\to S^1$ is well-defined because equivalent representatives have the same image. Indeed, for every $n\in\mathbb{Z}$, \begin{align*} e^{2\pi i(t-n)}=e^{2\pi i t}e^{-2\pi i n}=e^{2\pi i t}. \end{align*} Thus replacing $(y,t)$ by $(f^n(y),t-n)$ does not change $e^{2\pi i t}$. Now we choose representatives in two angular coordinate intervals. Let \begin{align*} U_0&:=S^1\setminus\{-1\}, & U_1&:=S^1\setminus\{1\}, \end{align*} and define \begin{align*} I_0&:=\left(-\frac{1}{2},\frac{1}{2}\right), & I_1&:=(0,1). \end{align*} The coordinate map $\theta_0:U_0\to I_0$ sends $e^{2\pi i s}$ to the unique $s\in I_0$, and $\theta_1:U_1\to I_1$ sends $e^{2\pi i s}$ to the unique $s\in I_1$. These are smooth coordinate maps on the two open arcs. Fix $j\in\{0,1\}$ and $[y,t]\in\pi^{-1}(U_j)$. Since $\pi([y,t])\in U_j$, there is a unique $s_j\in I_j$ such that \begin{align*} e^{2\pi i s_j}=\pi([y,t])=e^{2\pi i t}. \end{align*} Hence $t-s_j\in\mathbb{Z}$. Define $n_j:=t-s_j\in\mathbb{Z}$, so $s_j=t-n_j$. Applying the quotient relation with this integer gives \begin{align*} [y,t]=[f^{n_j}(y),t-n_j]=[f^{n_j}(y),s_j]. \end{align*} The uniqueness of $s_j\in I_j$ gives uniqueness of $n_j=t-s_j$, and then the fibre representative is forced to be $f^{n_j}(y)$. Thus every point of $\pi^{-1}(U_j)$ has a unique representative of the form $(y_j,s_j)\in F\times I_j$. This uniqueness is the reason the local trivialisation will be independent of the initially chosen representative $(y,t)$. [/guided] [/step] [step:Build product trivialisations over the two arcs] For $j\in\{0,1\}$, define $\Phi_j:\pi^{-1}(U_j)\to U_j\times F$ by \begin{align*} \Phi_j([y,t])=\left(e^{2\pi i t}, f^{n_j}(y)\right), \end{align*} where $n_j\in\mathbb{Z}$ is the unique integer such that $t-n_j\in I_j$. Define also $\Psi_j:U_j\times F\to \pi^{-1}(U_j)$ by \begin{align*} \Psi_j(z,y)=[y,\theta_j(z)]. \end{align*} Then $\Phi_j$ and $\Psi_j$ are inverse maps. Indeed, if $z\in U_j$ and $y\in F$, then $\theta_j(z)\in I_j$, so \begin{align*} \Phi_j(\Psi_j(z,y))=(z,y). \end{align*} Conversely, if $[y,t]\in\pi^{-1}(U_j)$ and $s_j=t-n_j\in I_j$, then \begin{align*} \Psi_j(\Phi_j([y,t]))=[f^{n_j}(y),s_j]=[y,t]. \end{align*} To justify smoothness, define \begin{align*} q_j:F\times I_j&\to\pi^{-1}(U_j), & q_j(y,s)&=[y,s]. \end{align*} The uniqueness of representatives from the previous step shows that $q_j$ is bijective. Since $q:F\times\mathbb{R}\to T_f$ is a smooth covering map, its restriction $q_j$ is a diffeomorphism onto the [open set](/page/Open%20Set) $\pi^{-1}(U_j)$; equivalently, the quotient smooth structure over this slice is the product smooth structure on $F\times I_j$. Under this diffeomorphism, $\Psi_j$ is the map \begin{align*} U_j\times F&\to\pi^{-1}(U_j), & (z,y)&\mapsto q_j(y,\theta_j(z)), \end{align*} which is smooth because $\theta_j$ is smooth and $q_j$ is smooth. Under $q_j^{-1}$, $\Phi_j$ is the map \begin{align*} F\times I_j&\to U_j\times F, & (y,s)&\mapsto (e^{2\pi i s},y), \end{align*} which is smooth. Hence $\Phi_j$ and $\Psi_j$ are diffeomorphisms inverse to each other. Let \begin{align*} \operatorname{pr}_{U_j}:U_j\times F&\to U_j, & \operatorname{pr}_{U_j}(z,y)&=z \end{align*} denote the projection onto the first factor. Moreover, \begin{align*} \operatorname{pr}_{U_j}\circ \Phi_j=\pi \end{align*} on $\pi^{-1}(U_j)$, so $\Phi_j$ is a local trivialisation of $\pi$ over $U_j$ in the sense of the definition of a smooth fibre bundle. [/step] [step:Compute the transition functions on the overlap] The overlap $U_0\cap U_1$ has two connected components $V_+:=\{e^{2\pi i s}:0<s<1/2\}$ and $V_-:=\{e^{2\pi i s}:1/2<s<1\}$. On $V_+$ the angular coordinates agree: $\theta_0(z)=\theta_1(z)$ for $z\in V_+$. Therefore, for $(z,y)\in V_+\times F$, \begin{align*} (\Phi_1\circ \Psi_0)(z,y)=(z,y). \end{align*} Thus the transition function over $V_+$ is $\operatorname{id}_F$. On $V_-$, write $z=e^{2\pi i s}$ with $s\in(1/2,1)$. Then $\theta_1(z)=s$ and $\theta_0(z)=s-1$. Hence, for $(z,y)\in V_-\times F$, \begin{align*} (\Phi_1\circ \Psi_0)(z,y)=(z,f^{-1}(y)), \end{align*} because $s=(s-1)-(-1)$ and therefore the unique representative in $F\times(0,1)$ is $(f^{-1}(y),s)$. Thus the transition function over $V_-$ is $f^{-1}$. [guided] We now check the only place where the bundle is not globally a product: the overlap of the two angular charts. The intersection $U_0\cap U_1$ is the circle with both $1$ and $-1$ removed, so it has two connected components $V_+=\{e^{2\pi i s}:0<s<1/2\}$ and $V_-=\{e^{2\pi i s}:1/2<s<1\}$. These two components behave differently because one lies on the same side of the cut in both coordinate systems, while the other crosses the cut at $1$. First let $z\in V_+$. Then $z=e^{2\pi i s}$ for a unique $s\in(0,1/2)$, and both angular charts assign the same parameter: \begin{align*} \theta_0(z)=s=\theta_1(z). \end{align*} For $(z,y)\in V_+\times F$, the change of trivialisation from the $U_0$-chart to the $U_1$-chart is \begin{align*} (\Phi_1\circ \Psi_0)(z,y)=(z,y). \end{align*} Thus the transition function on $V_+$ is the identity diffeomorphism $\operatorname{id}_F:F\to F$. Now let $z\in V_-$. Then $z=e^{2\pi i s}$ for a unique $s\in(1/2,1)$. The $U_1$-coordinate is $s$, but the $U_0$-coordinate must lie in $(-1/2,1/2)$, so it is $s-1$. Thus $\theta_1(z)=s$ and $\theta_0(z)=s-1$. For $(z,y)\in V_-\times F$, the $U_0$-chart gives the quotient point $[y,s-1]$. To express this point in the $U_1$-chart, we need the representative with parameter in $(0,1)$. Since \begin{align*} -1\cdot(y,s-1)=(f^{-1}(y),s) \end{align*} under the convention $n\cdot(y,t)=(f^n(y),t-n)$ would give $f^{-1}$ if changing by $-1$. Equivalently, using the relation $(a,r)\sim(f(a),r-1)$ with $a=f^{-1}(y)$ gives \begin{align*} [y,s-1]=[f^{-1}(y),s]. \end{align*} Therefore the transition from the $U_0$-trivialisation to the $U_1$-trivialisation on $V_-$ is \begin{align*} (\Phi_1\circ \Psi_0)(z,y)=(z,f^{-1}(y)). \end{align*} If one instead writes the inverse transition $\Phi_0\circ\Psi_1$, the fibre map is $f$. In either orientation, the transition function belongs to the cyclic subgroup $\langle f\rangle$ of $\operatorname{Diff}(F)$. [/guided] [/step] [step:Conclude the structure group reduction] The maps $\Phi_0$ and $\Phi_1$ form a smooth bundle atlas for $\pi:T_f\to S^1$ because they are smooth local trivialisations over the open cover $\{U_0,U_1\}$. For indices $\alpha,\beta\in\{0,1\}$ and a connected component $W\subset U_\alpha\cap U_\beta$, define the transition function \begin{align*} g_{\alpha\beta,W}:W&\to\operatorname{Diff}(F) \end{align*} by the identity \begin{align*} (\Phi_\beta\circ\Psi_\alpha)(z,y)=(z,g_{\alpha\beta,W}(z)(y)) \end{align*} for every $(z,y)\in W\times F$. The preceding computation gives $g_{\alpha\beta,W}(z)\in\{\operatorname{id}_F,f,f^{-1}\}\subset \langle f\rangle$ for every such $z$, where $\langle f\rangle$ denotes the cyclic subgroup of $\operatorname{Diff}(F)$ generated by $f$. Therefore the transition functions may be chosen to take values in $\langle f\rangle$. If $G\subset\operatorname{Diff}(F)$ is a specified Lie subgroup containing $f$, then it also contains $f^{-1}$ and $\operatorname{id}_F$, so the same bundle atlas has transition functions taking values in $G$. This proves that $\pi$ is a smooth fibre bundle with fibre $F$ and with the asserted structure group reduction. [/step]