Proof. We construct such a map $\hat F$ . Let $b\in B$ denote a recurrent point of f. Consider the horizontal distribution of our choice $H\subset TE$ . Take a small neighbourhood of $U\subset B$ that contains b such that for every $e\in p^{-1}b$ , $b\in U$ , there is a small disc $D_e\subset H_e$ so we have $\exp :D_e\to P(e)$ is a diffeomorphism where $P(e)$ is a disc of the dimension of the base but in the total space E. There exists an $N\in \mathbb {N}_{>0}$ such that $f^N(b)\in U$ . Then take a smaller neighbourhood $V\subset U$ containing b such that $f^N(V)\subset U$ . Now, by Theorem 2.6, there is a map $f'$ that is $C^1$ close to f on V, $f'=f$ outside V and has b as a periodic point. Without loss of generality, assume $(f')^Nb=b$ and $(f')^NV\subset U$ . Denote $\hat f=(f')^N$ .

Recall that we have defined $P(e)$ for $e\in p^{-1}b$ . Now let $P(e')=P(e)$ if $P(e)\cap p^{-1}(pe')=e'$ (there is a unique $P(e)$ because the neighbourhood is foliated by these discs). Now $P(F^N(e))$ is also well defined for all $e\in p^{-1}V$ . Then we define

$$ \begin{align*}\hat F(e)=P(F^N(e))\cap p^{-1}(\hat f(b'))\end{align*} $$

for each $e\in p^{-1}b'$ and $b'\in V$ . We can check that $\hat F$ covers $\hat f$ and is $C^1$ close to $F^N$ because V is small enough.

Proof. We use charts. Suppose for a fixed b, $f(b)\in U_j$ . Define $r_{j}=(\psi _jF)^{-1}.\psi _jG:E\to H$ . We show $r_j$ is independent of charts.

This is because if, at the same time, $f(b)\in U_k$ for some $k\neq j$ , then $\psi _k=h_{jk}.\psi _j$ . We have $r_j=r_k$ . These $r_j$ terms are continuous as compositions of continuous functions. Thus, we get a globally defined continuous function $\bar {r}:E\to H$ that agrees with the $r_j$ terms everywhere.

Proof. We have the following diagram.

(4.3)

The upper row comes from the long exact sequence. The trivial map on the left comes from the fact that the restriction of r to each fibre induces the zero map on $H_1(\mathbb {T}^d;\mathbb {R})$ .

Our goal is to define $v_{\#}$ . Once it is defined, from the $K(\pi ,1)$ -spaces fact (see, for example, Hatcher [Reference Hatcher7, 1B.9]), we get a unique v, up to homotopy, from any homomorphism $\pi _1(B)\to \pi _1(\mathbb {T}^d)$ .

For $\alpha \in \pi _1(B)$ , because $p_{\#}$ is surjective, there is a $\beta \in \pi _1(E)$ such that $p_{\#}\beta =\alpha $ . We define $v_{\#}\alpha =r_{\#}\beta $ .

To show this is well defined, suppose there is another $\beta '$ such that $p_{\#}\beta '=p_{\#}\beta =\alpha $ , we need to show that $r_{\#}\beta '=r_{\#}\beta $ .

We know $p_{\#}(\beta '\beta ^{-1})=1$ so $\beta '\beta ^{-1}\in \ker p_{\#}=\text {im}\, i_{\#}$ . There is a $\gamma \in \pi _1(\mathbb {T}^d)$ such that $i_{\#}\gamma =\beta '\beta ^{-1}$ . However, $r_{\#}i_{\#}\gamma =r_{\#}\beta '\beta ^{-1}=1$ .

Proof. We prove the statement by induction on k for principal torus bundles over $\mathbb {S}^k$ . After showing for spheres, a general statement is true because manifolds have CW approximations and we can just do inductions on the dimension of skeletons.

For $k=1$ , the bundle must be trivial, so there is trivially an $\mathrm {id}_{\mathbb {T}^d}$ -map.

Now suppose the statement for the bundle over $\mathbb {S}^k$ is true. Then for a bundle over $\mathbb {S}^{k+1}$ , we consider the clutching construction. The bundle splits into two trivial bundles over discs $\mathbb {D}^+$ and $\mathbb {D}^-$ of dimension $k+1$ . The restrictions of the torus bundle to the boundary spheres $\mathbb {S}^k$ of $\mathbb {D}^{+/-}$ and each admits an $\mathrm {id}_{\mathbb {T}^d}$ -map.

Let us denote the $\mathrm {id}_{\mathbb {T}^d}$ -map $g:\mathbb {S}^{k}\times \mathbb {T}^d\to \mathbb {S}^{k}\times \mathbb {T}^d$ . Parametrize the disc $\mathbb {D}^{k+1}$ by $(\rho ,\theta )$ , where $\rho $ gives the radius and $\theta $ is a k-dimensional vector of angles. In this coordinate, $\rho =1$ is the boundary sphere, and $g(\theta ,y)=(\theta ,\mathrm {id}+v(\theta ))$ on the boundary. The problem actually reduces to the extension of $v(\theta )$ to the entire disc.

However, because the pair $(\mathbb {D}^{k+1}\times \mathbb {T}^d,\mathbb {S}^{k}\times \mathbb {T}^d)$ is a good pair, which has the homotopy extension property, we are able to find a homotopy $h:\mathbb {D}\times \mathbb {T}^d\times I\to \mathbb {D}\times \mathbb {T}^d$ such that $h(1,\theta ,y,0)=g(\theta ,y)$ . Thus, $V(\rho ,\theta )=h(\rho ,\theta ,0,0)$ is a map that restricts to $v(\theta )$ on the boundary, where $y=0$ is just the zero section of the trivial bundle. Now $\mathrm {id}_{\mathbb {D}\times \mathbb {T}^d}+V$ is an $\mathrm {id}_{\mathbb {T}^d}$ -map on half of the bundle. We can glue via the clutching map.

Proof. From the previous lemmas, we have found a homotopy $\bar {H}:E\times I\to \mathbb {T}^d$ such that $\bar {H}(e,0)=r(e)$ , where $\bar {F}(e)=\mathrm {id}_E(e)+r(e)$ , and $\bar {H}(e,1)=v\circ p(e)$ . Then, if we define $\tilde {H}:E\times I\to E$ as $\tilde H(e,t)=\mathrm {id}_E(e)+\bar H(e,t)$ , it gives a homotopy such that $\tilde {H}(e,0)=\bar {F}(e)$ and $\tilde {H}(e,1)=\mathrm {id}_E(e)+v\circ p(e)$ .

If we name the maps in Diagram (4.1), from the left to right, the first $\mathrm {id}_{\mathbb {T}^d}$ -map $g_1$ , the second $\mathrm {id}_{\mathbb {T}^d}$ -map $g_2$ and the $A^{-1}$ -map $g_3$ , then $H:=g_1^{-1}g_2^{-1}g_3^{-1}\tilde {H}:E\times I\to E$ gives a homotopy such that $H(e,0)=F(e)$ and $H(e,1)=G(e)$ for some A-map G.

Proof of Proposition 4.1

By the remarks we made right after the proof of Lemma 4.2, if $F:E\to E$ is fibrewise Anosov, then F induces the same hyperbolic automorphism A on the homology of the fibre $\mathbb {T}^d$ . Therefore, F is homotopic to G for some fibrewise affine Anosov diffeomorphism G from Proposition 4.7.

Proof. We define $\tilde w^s, \tilde w^u:E\to E$ ,

$$ \begin{align*} \tilde{w}^s(e)=-\sum_{n=0}^{\infty} A^n\tilde{r}^s(F^{-(n+1)}(e))\quad \text{and}\quad \tilde{w}^u(e)=\sum_{n=0}^{\infty} A^{-(n+1)}\tilde{r}^u(F^{n}(e)). \end{align*} $$

Then we have

$$ \begin{align*} A\tilde{w}^s(e)-\tilde{w}^s(F(e)) =-\sum_{n=0}^{\infty} A^{n+1}\tilde{r}^s(F^{-(n+1)}(e))+\sum_{n=0}^{\infty} A^{n}\tilde{r}^s(F^{-n}(e)) =\tilde{r}^s(e). \end{align*} $$

Similarly, we can check that $A\tilde {w}^u(e)-\tilde {w}^u(F(e))=\tilde {r}^u(e)$ . Let $\tilde {w}=\tilde {w}^s\oplus \tilde {w}^u$ . Then we have $A\tilde {w}(e)-\tilde {w}(F(e))=\tilde {r}(e)$ .

We show that $\tilde {w}:E\to E$ is continuous. Let $\mathbb {R}^d$ be endowed with the Riemannian metric lifted from $\mathbb {T}^d$ . Let $d_E$ denote the metric on E and $d_{\mathbb {R}^d}$ the metric on $\mathbb {R}^d$ . Since E is assumed to be compact, there is a uniform bound M of the distance from $\tilde {r}(e)$ to $0\in \mathbb {R}^d$ for all $e\in E$ . For a given $\varepsilon>0$ , take $N\in \mathbb {N}$ so we have $\sum _{n>N}\unicode{x3bb} ^nM<\varepsilon /4$ . Then, since both $-\sum _{n=0}^NA^n\tilde {r}^s(F^{-(n+1)}(e))$ and $\sum _{n=0}^N A^{-(n+1)}\tilde {r}^u(F^{n}(e))$ are uniformly continuous, there is a $\delta>0$ such that if $d_E(e_1,e_2)<\delta $ for any $e_1,e_2\in E$ , we have both

$$ \begin{align*} d_{\mathbb{R}^d}\bigg(\sum_{n=0}^N A^n\tilde{r}^s(F^{-(n+1)}(e_1)),\sum_{n=0}^N A^n\tilde{r}^s(F^{-(n+1)}(e_2))\bigg)<\varepsilon/4 \end{align*} $$

and

$$ \begin{align*} d_{\mathbb{R}^d}\bigg(\sum_{n=0}^N A^{-(n+1)}\tilde{r}^u(F^{n}(e_1)),\sum_{n=0}^N A^{-(n+1)}\tilde{r}^u(F^{n}(e_2))\bigg)<\varepsilon/4. \end{align*} $$

Then $d_{\mathbb {R}^d}(\tilde {w}(e_1),\tilde {w}(e_2))<\varepsilon $ .

We can then project $\tilde {w}:E\to \mathbb {R}^d$ and get $p_1\tilde {w}=w:E\to \mathbb {T}^d$ , where $p_1:\mathbb {R}^d\to \mathbb {T}^d$ denotes the covering projection. Note $p_1:\mathbb {R}^n\to \mathbb {T}^n$ is linear. Then,

$$ \begin{align*} Aw(e)-w(F(e))=Ap_1\tilde{w}(e)-p_1\tilde{w}(F(e))=p_1A\tilde{w}(e)-p_1\tilde{w}(F(e))=p_1\tilde{r}(e)=r(e). \end{align*} $$

Let $h: E\to E$ be such that $h(e)=e+w(e)$ , which obviously covers the identity. We have

$$ \begin{align*} G\circ h(e)&=G(e+w(e))=G(e)+Aw(e) =G(e)+r(e)-r(e)+Aw(e)\\ &=F(e)+w(F(e))=h\circ F(e). \end{align*} $$

Since $\tilde {w}$ is bounded, it is homotopic to a constant map $E\to \mathbb {R}^d$ , and thus h is homotopic to the identity via a homotopy that preserves each fibre. Then h induces a map of non-zero degree on the top homology, and thus it is surjective.

Proof. It is sufficient to show that $h_b$ takes a local stable disk to a local stable disk, that is for any $e\in \mathbb {T}^d_b$ , if $e' \in W^s_{F_b,\varepsilon }(e)$ , we will have $h_b(e')\in W^s_{G_b,\varepsilon }(h_b(e))$ for some $\varepsilon>0$ .

Suppose the statement is not true, say $h_b(e')\notin W^s_{G_b,\varepsilon }(h_b(e))$ . Then,

$$ \begin{align*} d_{\mathbb{T}_{f^n(b)}^d}(G^n_b(h_b(e)),G^n_b(h_b(e')))=d_{\mathbb{T}_{f^n(b)}^d}(h_{f^n(b)}F^n_b(e), h_{f^n(b)}F^n_b(e'))\to 0, \end{align*} $$

as $n\to \infty $ . However, the $G_b$ is the fibrewise affine Anosov diffeomorphism which has straight leaves in the eigen-direction of A, so the left-hand side of the equation goes to infinity. We have reached a contradiction.

Sketch of proof.

Fix a stable leaf S of $\tilde F_b$ , and we want to show that the set

$$ \begin{align*} Q:=\{\tilde e\ |\ \widetilde W^u_{\tilde F_b}(\tilde e)\cap S\neq\emptyset\} \end{align*} $$

is $\mathbb {R}^d_b$ . Because of the local product structure, there is an open set O that contains S such that $Q=\{\tilde e\ |\ \widetilde W^u_{\tilde F_b}(\tilde e)\cap O\neq \emptyset \}$ . If $\tilde e\in Q$ , there is a foliated neighbourhood U containing $\tilde e$ which we could extend to a path of foliated neighbourhoods whose union contains $\widetilde W^u_{\tilde F_b}(\tilde e)$ , so $U\subset Q$ . Thus, Q is open. We want to show that Q is also closed.

Suppose x is a point in the closure of Q but not in Q. We show that $\widetilde W^u_{\tilde F_b}(x)\cap S\neq~\emptyset $ . Because of the local product structure, we are able to find a sequence of points in Q approaching x which is also contained in the same stable leaf $\widetilde W^s_{\tilde F_b}(x)$ . We pick one point from the sequence and denote it by y. Pick another point z in $\widetilde W^u_{\tilde F_b}(y)\cap S$ . We connect y and z by a path $\gamma :[0,1]\to \widetilde W^u_{\tilde F_b}(y)$ and connect x and y by a path $\rho :[0,1]\to \widetilde W^s_{\tilde F_b}(x)=\widetilde W^s_{\tilde F_b}(y)$ .

We aim to define a function $\theta :[0,1]\times [0,1]\to \mathbb {R}_b^d$ such that $\theta (r,0)=\gamma (r)$ and $\theta (0,t)=\rho (t)$ , and $\theta (r,t)=\widetilde W^s_{\tilde F_b}(\gamma (r))\cap \widetilde W^u_{\tilde F_b}(\rho (t))$ . If we are able to define $\theta $ at $(1,1)$ , then $\widetilde W^s_{\tilde F_b}(z)\cap \widetilde W^u_{\tilde F_b}(x)=\widetilde W_{\tilde F_b}^u(x)\cap S\neq \emptyset $ (see Figure 1).

Figure 1 Definition of $\theta $ .

Because of the local product structure, the set where $\theta $ can be defined is open. Now let

$$ \begin{align*} t_0=\sup\{\hat t\ |\ \theta(r,t) \text{ is defined for } 0 \leq r\leq 1\text{ and } 0\leq t\leq \hat t\},\quad \text{and} \end{align*} $$

$$ \begin{align*} r_0=\sup\{\hat r\ |\ \theta(r,t_0)\text{ is defined for } 0\leq r\leq\hat r\}. \end{align*} $$

If we let $J:=\{(r,t)\ |\ t<t_0\}$ , $\tilde h_b(\theta (J))$ is bounded because the projections of it to the linear leaves $\widetilde W^u_{\tilde G_b}(\tilde h_b(y))$ and $\widetilde W^s_{\tilde G_b}(\tilde h_b(y))$ are bounded. Then, $\theta (J)$ is bounded because $\tilde h_b$ is proper. Next pick a sequence $\{(r_n,t_n)\ |\ r_n\leq r_{n+1}\}$ in J converging to $(r_0,t_0)$ such that $\theta (r_n,t_n)$ converges to a point $w\in \mathbb {R}^d_b$ . We define $\theta (r_0,t_0):=w$ by continuity. What is left is to check that $\theta $ is well defined.

Let N be a product neighbourhood of w. We can assume $\theta (r_n,t_n)\in N$ for all $n>0$ (see Figure 2).

Figure 2 Extending $\theta $ to $(r_0,t_0)$ .

Since $\theta $ is defined at $(r_n,t_n)$ for $n>0$ and is, by definition, $\widetilde W^s_{\tilde F_b}(\gamma (r_n))\cap \widetilde W^u_{\tilde F_b}(\rho (t_n))$ . Thus, $\theta (r_1,t_n)$ lies on $\widetilde W^s_{\tilde F_b}(\gamma (r_1))$ for all $n>0$ . Then, $\theta (r_1,t_n)$ converges to $\widetilde W^s_{\tilde F_b}(\theta (r_1,t_1))\cap \widetilde W^u_{\tilde F_b}(w)$ . (Suppose not. If $\theta (r_1,t_n)$ converges to $\bar w:=\widetilde W^s_{\tilde F_b}(\theta (r_1,t_1))\cap \widetilde W^u(w')$ for a point $w'\neq w$ , then $\theta (r_n,t_n)$ would converge to a point inside $\widetilde W^u_{\tilde F_b}(\bar w)\neq \widetilde W^u_{\tilde F_b}(w)$ .) Similarly, $\theta (r_n,t_1)$ converges to $\widetilde W^s(w)_{\tilde F_b}\cap \widetilde W^u_{\tilde F_b}(\theta (r_1,t_1))$ . Thus, it is either defined already so we have, or we could define that,

$$ \begin{align*} \theta(t_1,r_0)=\widetilde W^s_{\tilde F_b}(\theta(r_1,t_1))\cap \widetilde W^u_{\tilde F_b}(w),\quad \theta(t_0,r_1)=\widetilde W^s(w)_{\tilde F_b}\cap\widetilde W^u_{\tilde F_b}(\theta(r_1,t_1)). \end{align*} $$

Now suppose somehow we are able to define for another sequence $\{(r_n',t_n')\}$ that converges to $(r_0,t_0)$ such that $\theta (r_0,t_0)=\lim _{n\to \infty }\theta (r_n',t_n')$ . Then, we have

$$ \begin{align*} \lim_{n\to\infty}\theta(r_n',t_n') &= \lim_{n\to\infty}\widetilde W^u_{\tilde F_b}(\theta(r_1,t_n'))\cap \widetilde W^s_{\tilde F_b}(\theta(r_n',t_1)) \\ &=\widetilde W^u_{\tilde F_b}(\theta(r_1,t_0))\cap\widetilde W^s_{\tilde F_b}(\theta(r_0,t_1))=w. \end{align*} $$

This is saying that we could always extend $\theta $ to a limit point, so Q is closed.

Proof. We use the fact that the leaves are ‘very close’ to the leaves of an Anosov diffeomorphism.

Assume $\widetilde {W}^u_{\tilde {F}_b}(\tilde {e}),\widetilde {W}^s_{\tilde {F}_b}(\tilde {e})$ intersect at two points, say $\tilde {e},\tilde {e}'$ . Then, $\tilde {F}^n_b(\tilde {e}')\in \widetilde {W}^u_{\tilde {F}_{b}}(\tilde {F}^n_b(\tilde {e}))\cap \widetilde {W}^s_{\tilde {F}_{b}}(\tilde {F}^n_b(\tilde {e}))$ , which means the leaves along the orbit also intersect twice.

We are able to find a subsequence $\{f^{n_k}(b)\}$ that converges to b. By Lemma 4.2, for each large enough $n_k$ , there is an $\hat F^{(k)}$ such that $F^{n_k}_b$ is $C^1$ close to $\hat F_b^{(k)}$ . Denote $\bar F_b^n=\varphi _b\circ \psi \circ F_b^n: p^{-1}b\to p^{-1}b$ , the projection of $F^n$ at b to the fibre $p^{-1}b$ . Since $F^{n_k}_b$ and $\hat F_b^{(k)}$ are $C^1$ close, the vectors in the distributions of $\bar F_b^{n_k}$ at a point in $p^{-1}b$ lie in the $\gamma _k$ -cones of the distributions of $\hat F_b^{(k)}$ for some small $\gamma _k$ , that is, the angles between the respective stable and unstable distributions are small. To suppress notation, let us still use $\tilde e,\tilde e'$ for the projection of $\tilde e,\tilde e'$ to the fibre $p^{-1}b$ .

For any $\varepsilon>0$ , we can take k large enough so we have small enough $\gamma _k$ such that

$$ \begin{align*} d(s;\bar{F}_b^{n_k}(\tilde{e}),\bar{F}_b^{n_k}(\tilde{e}'))\cdot \gamma_k < \varepsilon. \end{align*} $$

This means $\widetilde {W}^s_{\hat {F}_b^{(k)}}(\hat {F}_b^{(k)}(\tilde {e}))$ lies in the normal neighbourhood $N(\widetilde {W}^s_{\bar {F}_b}(\bar {F}_b^{n_{k}}(\tilde {e})),\varepsilon )$ .

Now the unstable leaf $\widetilde {W}^u_{\hat {F}_b^{(k)}}(\hat {F}_b^{(k)}(\tilde {e}))$ must pass through $N(\widetilde {W}^s_{\bar {F}_b}(\bar {F}_b^{n_k}(\tilde {e})),\varepsilon )$ near $\bar {F}_b^{n_k}(\bar {e}')$ because of the existence of the unstable cone. Then, $\widetilde {W}^s_{\hat {F}_b^{(k)}}(\hat {F}_b^{k}(\tilde {e}))$ and $\widetilde {W}^u_{\hat {F}_b^{(k)}}(\hat {F}_b^{(k)}(\tilde {e}))$ should intersect the second time because of the local product structure, which is not the case. Thus, $\widetilde {W}^u_{\tilde {F}_b}(\tilde {e})$ and $\widetilde {W}^s_{\tilde {F}_b}(\tilde {e})$ cannot intersect more than once.

Proof. We show it for $\tilde {\mathscr {F}}_2$ . From the global product structure, we get a homeomorphism $\alpha =(\alpha _1,\alpha _2):\mathbb {R}^d\to \mathbb {R}^l\times \mathbb {R}^{d-l}$ , where $\alpha _1,\alpha _2$ are the projections, $\tilde {\mathscr {F}}_2$ is of dimension $d-l$ and codimension l, and the latter $\mathbb {R}^l\times \mathbb {R}^{d-l}$ has the usual topology.

Suppose there is not such an upperbound $M_K$ for the distance along leaves. Then there exists a sequence of pairs $\{(\tilde {e}_n,\tilde {e}_n')\}$ such that $\tilde {e}_n,\tilde {e}_n'\in K\cap \widetilde {W}^2(\tilde {e}_n)$ and $d_1(\tilde {e}_n,\tilde {e}_n')>n$ .

By the compactness of K, there is a subsequence $\{(\tilde {e}_{n_k},\tilde {e}^{\prime }_{n_k})\}$ that converges to a pair of points $(\tilde {e}_0,\tilde {e}_0')$ . Since $\alpha _1(\tilde {e}_n)=\alpha _1(\tilde {e}^{\prime }_n)$ (because they are on the same leaf of $\tilde {\mathscr {F}}_2$ ), we know $\alpha _1(\tilde {e}_0)=\alpha _1(\tilde {e}_0')$ , that is, $\tilde {e}^{\prime }_0\in \widetilde {W}^2(\tilde {e}_0)$ .

Now we can cover the path between $\tilde {e}_0$ and $\tilde {e}^{\prime }_0$ that realizes $d_2(\tilde {e}_0,\tilde {e}^{\prime }_0)$ in $\widetilde {W}^2(\tilde {e}_0)$ with a finite number of cubes with length of edges smaller than the constant of the local product structure, and denote the cubes by $\{P_1,\ldots , P_t\}$ . Then, $\alpha (\bigcup _{i=1}^t P_i)$ covers a neighbourhood $B_{\varepsilon } \times \alpha _2(\widetilde {W}^2(\tilde {e}_0)\cap K)$ , where $B_{\varepsilon }$ is a small l-dimensional disk. Thus, $\alpha ^{-1}(B_{\varepsilon } \times \alpha _2(\widetilde {W}^2(\tilde {e}_0)\cap K))$ also covers a piece between $\tilde {e}_{n_k}$ and $\tilde {e}^{\prime }_{n_k}$ of $\widetilde {W}^2(\tilde {e}_{n_k})$ for all $k>L$ for some large L, because of the local product structure. However, $d_2(\tilde {e}_{n_k},\tilde {e}^{\prime }_{n_k})>k$ . They cannot be all covered by a finite number of cubes of a fixed diameter. We have reached a contradiction.

Proof. We are able to run the above argument, starting with a homeomorphism $\alpha = id\times (\alpha _1,\alpha _2):C\times \mathbb {R}^d\to C\times (\mathbb {R}^l\times \mathbb {R}^{d-l})$ because of the global product structure.

If we want to check for the unstable foliation, then instead of $B_{\varepsilon } \times \alpha _2(\widetilde {W}^2(\tilde {e}_0)\cap K)$ , we consider a neighbourhood $A\times B_{\varepsilon } \times \alpha _2(\widetilde {W}^u_{b_0}(\tilde {e}_0)\cap K)$ covered by a finite number of cubes given by the local product structure, where $p^{-1}b_0$ contains $\tilde e_0,\tilde e_0'$ , A is a small neighbourhood around $b_0$ and $B_{\varepsilon }$ is an l-dimensional disk. The rest of the argument is the same.

Proof. Note that if we have the global product structure at one point b, then we also have it along the orbit of b, if considering the lifts fibrewise. We are then able to follow a similar argument to Franks [Reference Franks3] to deduce the injectivity of $\tilde h_b$ .

Recall that $\tilde h_b$ preserves the stable and unstable leaves (Property 5.2). Suppose after fixing the lifts, we can find $\tilde {e}\neq \tilde {e}'$ such that $\tilde {h}_b(\tilde {e})=\tilde {h}_b(\tilde {e}')$ . Let $\tilde {e}":=\widetilde {W}^u_{\tilde {F}_b}(\tilde {e})\cap \widetilde {W}^s_{\tilde {F}_b}(\tilde {e}')$ . We know $\widetilde {W}^u_{\tilde {G}_b}(\tilde {h}_b (\tilde {e})) \cap \widetilde {W}^s_{\tilde {G}_b}(\tilde {h}_b (\tilde {e}'))=\tilde {h}_b(\tilde {e}")$ , that is, they also intersect only once at $\tilde {h}_b(\tilde {e}")$ . Then, $\tilde {h}_b(\tilde {e})=\tilde {h}_b(\tilde {e}')=\tilde {h}_b(\tilde {e}")$ . Thus, we have found $\tilde {e},\tilde {e}"\in \widetilde {W}^u_{\tilde {F}_b}(\tilde {e})$ such that $\tilde {h}_b(\tilde {e})=\tilde {h}_b(\tilde {e}")$ .

Since B is assumed to be a compact manifold (which is normal), for each finite open cover, we can find a finite refinement that contains compact sets. Now we use compact trivialized neighbourhoods, and $h_b$ is lifted continuously over each of them, denoted as $\tilde {h}_C$ on a compact set C. Then, $\tilde {h}_C$ is homotopic to the identity along the fibres, and thus is proper (Property 5.4). Denote the set $(\mathrm {id},\tilde {h}_C)^{-1}(C\times [0,1]^d)$ as D, which is compact.

Then, by Corollary 5.9, we have an $M_{C}>0$ such that

$$ \begin{align*} \sup_{b\in C}\,\sup\{d(u;\tilde{e},\tilde{e}"): \tilde{e},\tilde{e}"\in (\tilde{h}_b^{-1} [0,1]^d\cap \widetilde{W}^u_{\tilde{F}_b}(\tilde{e}))\}<M_{C}. \end{align*} $$

Then again by the compactness of B, there is a uniform bound $M>0$ such that

$$ \begin{align*} \sup_{b\in B}\,\sup\{d(u;\tilde{e},\tilde{e}"): \tilde{e},\tilde{e}"\in (\tilde{h}_b^{-1} [0,1]^d\cap \widetilde{W}^u_{\tilde{F}_b}(\tilde{e}))\}<M. \end{align*} $$

Denote $\tilde {w}_n:=\tilde {h}_{f^n(b)}\tilde {F}_{b}^n(\tilde {e})=\tilde {h}_{f^n(b)}\tilde {F}_{b}^n(\tilde {e}")$ . We can always use deck transformations to translate the $\tilde {w}_n$ terms to $[0,1]^d$ . Thus, we should also have $d(u;\tilde {F}_{b}^n(\tilde {e}),\tilde {F}_{b}^n(\tilde {e}"))<M$ for all n. We have reached a contradiction.

Proof of Proposition 5.12

We first fix the notation. We will denote leaves at $\tilde e$ of $\tilde {\mathscr {F}}_i$ by $\widetilde W_1^i(\tilde e)$ and leaves of $\tilde {\mathscr {F}}_i'$ by $\widetilde W_2^i(\tilde e)$ , $i=1,2$ . Then, for any $\tilde e_0\in \mathbb {R}^d$ , we want to show that $\widetilde W^1_2(\tilde e_0)\cap \widetilde W^2_2(\tilde e_0)=\{\tilde e_0\}$ . We start from selecting the following constants.

• • $\varepsilon $ : Let $\varepsilon>0$ be smaller than the constant of the local product structure.

• • $M_1$ : Let $M_1$ be the constant such that $\|h-\mathrm {id}\|< M_1$ .

• • D: We fix a $D>10(M_1+\varepsilon )$ .

• • $M_2$ : Suppose K is a compact set of diameter $\leq D$ . There is an $M_2$ such that

  $$ \begin{align*}\sup\{d_1(\tilde{e},\tilde{e}'): \tilde{e},\tilde{e}'\in (K\cap \widetilde{W}^1_1(\tilde{e})),\ \widetilde{W}^1_1\text{ is a leaf of }\tilde{\mathscr{F}}_1 \}<M_2,\quad \text{and} \end{align*} $$
  $$ \begin{align*}\sup\{d_2(\tilde{e},\tilde{e}'): \tilde{e},\tilde{e}'\in (K\cap \widetilde{W}^2_1(\tilde{e})),\ \widetilde{W}^2_1\text{ is a leaf of }\tilde{\mathscr{F}}_2 \}<M_2,\end{align*} $$
  from Corollary 5.8.

• • M: Let $M=10\max \{M_2,D\}$ .

We could choose $\alpha <\varepsilon /10M$ .

We denote $\mathcal {O}:=\tilde {h}(\tilde {e}_0)$ and the linear leaves of $\tilde {\mathscr {F}}_i",\ i=1,2$ passing through $\mathcal {O}$ by $\widetilde {W}^1_0,\ \widetilde {W}^2_0$ . There is then a natural coordinate of every point if we consider the projection of it onto these straight leaves. For a point $\tilde e\in \mathbb {R}^d$ , we want to denote the distance between the projection of $\tilde e$ to $\widetilde W^1_0$ and $\mathcal {O}$ along $\widetilde W^1_0$ as $\|\tilde e-\mathcal {O}\|_1$ and the distance between the projection of $\tilde e$ to $\widetilde W^2_0$ and $\mathcal {O}$ along $\widetilde W^2_0$ as $\|\tilde e-\mathcal {O}\|_2$ .

We claim that for any point $\tilde {e}_1\in \widetilde {W}_2^1(\tilde {e}_0)$ and any point $\tilde {e}_2\in \widetilde {W}_2^2(\tilde {e}_0)$ of the foliations $\tilde {\mathscr {F}}_i'$ ,

$$ \begin{align*}\text{either}\ \|\tilde{e}_1-\tilde{e}_2\|_1>0 \ \text{or}\ \|\tilde{e}_1-\tilde{e}_2\|_2>0.\end{align*} $$

Case 1. We start with the special case when $\|\tilde {e}_1-\mathcal {O}\|_1$ , $\|\tilde {e}_1-\mathcal {O}\|_2$ , $\|\tilde {e}_2-\mathcal {O}\|_1$ and $\|\tilde {e}_2-\mathcal {O}\|_2$ are all $\leq M_1+\varepsilon $ .

There exist points $\tilde {e}_1'\in \widetilde {W}^1_1(\tilde {e}_0)$ and $\tilde {e}_2'\in \widetilde {W}^2_1(\tilde {e}_0)$ such that $\|\tilde {e}_1-\mathcal {O}\|_1=\|\tilde {e}_1'-\mathcal {O}\|_1$ , $\|\tilde {e}_2-\mathcal {O}\|_2=\|\tilde {e}_2'-\mathcal {O}\|_2$ . With the $\alpha $ we chose, $\widetilde {W}^2_2(\tilde {e}_1)$ stays in a cone of slope $\leq {\varepsilon }/{10M}$ around the leaf $\widetilde {W}^2_1(\tilde {e}_0)$ for at least a distance of $M_2$ , which is the upperbound of the distance along a leaf of $\tilde {\mathscr {F}}_i$ inside a set of diameter $D>10M_1$ . We then have $\|\tilde {e}_2-\tilde {e}_2'\|_1\leq 2M_2({\varepsilon }/{10M})<<\varepsilon /2$ . Also, $\|\tilde {e}_1-\tilde {e}_1'\|_1=0$ .

Note that either $\|\tilde {e}_1'-\tilde {e}_2'\|_1\geq \varepsilon $ or $\|\tilde {e}_1'-\tilde {e}_2'\|_2\geq \varepsilon $ because of the local product structure and the fact that $\tilde {W}^2_1(\tilde {e}_0)$ and $\tilde {W}^1_1(\tilde {e}_0)$ (with the GPS) do not meet a second time besides at $\tilde {e}_0$ . If both $\|\tilde e^{\prime }_1-\tilde e_2'\|_1<\varepsilon $ and $\|\tilde e_1'-\tilde e_2'\|_2<\varepsilon $ , this means $W_1^i(\tilde e_0)$ have entered a product neighbourhood where they must intersect for another time, contradicting the global product structure.

If $\|\tilde {e}_1'-\tilde {e}_2'\|_1\geq \varepsilon $ , then we have

$$ \begin{align*}\|\tilde{e}_1-\tilde{e}_2\|_1\geq \|\tilde{e}_1'-\tilde{e}_2'\|_1- \|\tilde{e}_1-\tilde{e}_1'\|_1-\|\tilde{e}_2-\tilde{e}_2'\|_1 \geq \varepsilon-2M_2\frac{\varepsilon}{10M}>0.\end{align*} $$

Similarly, if we have $\|\tilde {e}_1'-\tilde {e}_2'\|_2\geq \varepsilon $ , then $\|\tilde {e}_1-\tilde {e}_2\|_2\geq \varepsilon $ .

Case 2. When any of $\|\tilde {e}_1-\mathcal {O}\|_1$ , $\|\tilde {e}_1-\mathcal {O}\|_2$ , $\|\tilde {e}_2-\mathcal {O}\|_1$ or $\|\tilde {e}_2-\mathcal {O}\|_2$ is greater than $M_1+\varepsilon $ , we can prove this claim by a double induction. See Figure 3.

Figure 3 Leaves.

A brief idea of this is that we show leaves of $\tilde {\mathscr {F}}_i'$ follow along the leaves of $\tilde {\mathscr {F}}_i$ for distance D with an error $<\varepsilon $ . Then we switch to other leaves of $\tilde {\mathscr {F}}_i$ to follow, to keep the error of our estimate small. At the same time, note that the pair of leaves of $\tilde {\mathscr {F}}_i$ is far apart, so they never meet again by GPS, and hence the leaves of $\tilde {\mathscr {F}}_i'$ would not meet neither.

Here, we see the necessity of the existence of the homeomorphism. If it is not injective, it may take a entire leaf to a bounded subset of the linear leaf. Then, we would not have the leaves of $\tilde {\mathscr {F}}^{\prime }_i$ terms separating from each other.

The detailed induction is as follows.

First, for the base case, we let $\tilde e_1,\ \tilde e_2$ be points such that

$$ \begin{align*} M_1+\varepsilon<\|\tilde{e}_1-\mathcal{O}\|_1\leq M_1+\varepsilon+D\quad \text{and}\quad M_1+\varepsilon<\|\tilde{e}_2-\mathcal{O}\|_2\leq M_1+\varepsilon+D. \end{align*} $$

Note that we would not need to consider cases where $M_1+\varepsilon <\|\tilde {e}_1-\mathcal {O}\|_1\leq M_1+\varepsilon +D$ and $\|\tilde {e}_2-\mathcal {O}\|_2\leq M_1+\varepsilon $ , or $\|\tilde {e}_1-\mathcal {O}\|_1\leq M_1+\varepsilon $ and $M_1+\varepsilon <\|\tilde {e}_2-\mathcal {O}\|_2\leq M_1+\varepsilon +D$ , since they have been dealt with in Case 1. For any such points $\tilde {e}_1\in \widetilde W_2^1(\tilde e_0)$ and $\tilde e_2\in \widetilde W_2^2(\tilde e_0)$ , we again have $\tilde {e}_1'\in \widetilde {W}^1_1(\tilde {e}_0)$ and $\tilde {e}_2'\in \widetilde {W}^2_1(\tilde {e}_0)$ , such that $\|\tilde {e}_1-\mathcal {O}\|_1=\|\tilde {e}_1'-\mathcal {O}\|_1$ and $\|\tilde {e}_2-\mathcal {O}\|_2=\|\tilde {e}_2'-\mathcal {O}\|_2$ .

We have the estimate $\|\tilde {e}_1-\tilde {e}_1'\|_2<4M_2\alpha <4M_2({\varepsilon }/{10M})<\varepsilon $ , and similarly $\|\tilde {e}_2-\tilde {e}_2'\|_1<\varepsilon $ , while $\|\tilde e_1-\tilde e_1'\|_1=\|\tilde e_2-\tilde e_2'\|_2=0$ . We have $\|\tilde {e}_1'-\mathcal {O}\|_2\leq M_1$ because again $\tilde {h}$ has a bounded distance $M_1$ from the identity and, as a result, any point on $\widetilde {W}^1_1(\tilde {e}_0)$ has a bounded distance $M_1$ from the linear leaf $\widetilde W_0^1$ . In this case, we always have $\|\tilde {e}_1-\mathcal {O}\|_2<M_1+\varepsilon $ and $\|\tilde {e}_2-\mathcal {O}\|_1<M_1+\varepsilon $ because, for instance,

$$ \begin{align*} \|\tilde{e}_1-\mathcal{O}\|_2\leq \|\tilde{e}_1-\tilde{e}_1'\|_2+\|\tilde{e}_1'-\mathcal{O}\|_2 <M_1+\varepsilon. \end{align*} $$

Then,

$$ \begin{align*} \|\tilde{e}_1-\tilde{e}_2\|_1 \geq\|\tilde{e}_1-\mathcal{O}\|_1-\|\tilde{e}_2-\mathcal{O}\|_1>(M_1+\varepsilon)-(M_1+\varepsilon)>0. \end{align*} $$

Now we denote by $\tilde {e}^{(k)}_1$ the point on $\widetilde {W}^1_2(\tilde {e}_0)$ such that $\|\tilde {e}_1^{(k)}-\mathcal {O}\|_1=M_1+kD<({1}/{10}+k)D$ ; $\tilde {e}^{(k)}_2$ the point on $\widetilde {W}^2_2(\tilde {e}_0)$ such that $\|\tilde {e}_2^{(k)}-\mathcal {O}\|_2=M_1+kD<({1}/{10}+k)D$ , $k\geq 1$ . We then take leaves $\widetilde {W}^1_1(\tilde {e}_1^{(k)})$ and $\widetilde {W}^2_1(\tilde {e}_2^{(k)})$ for each $k\geq 1$ . These are the ‘new leaves’ of $\tilde {\mathscr {F}}_i$ we switch to track.

From the induction hypothesis: for $i\geq 1,\ j\geq 1$ , and such selected $\tilde e^{(i)}_1,\ \tilde e^{(j)}_2$ , we have

$$ \begin{align*} \|\tilde e_1^{(i)}-\mathcal{O}\|_2< (2i-1)(M_1+\varepsilon)\quad \text{and}\quad \|\tilde e_2^{(j)}-\mathcal{O}\|_1< (2j-1)(M_1+\varepsilon). \end{align*} $$

Note that this is implicit in the base case.

Now consider the case where $i\geq 1$ and $j\geq 1$ , and the points $\tilde e_1,\ \tilde e_2$ such that

$$ \begin{align*} M_1+\varepsilon+iD<\|\tilde e_1-\mathcal{O}\|_1\leq M_1+\varepsilon +(i+1)D\quad \text{and} \end{align*} $$

$$ \begin{align*} M_1+\varepsilon+jD<\|\tilde e_2-\mathcal{O}\|_2\leq M_1+\varepsilon +(j+1)D. \end{align*} $$

For any $\tilde {e}_1\in \widetilde W^1_2(\tilde e_0)$ and $\tilde {e}_2\in \widetilde W^2_2(\tilde e_0)$ , we could also find $\tilde {e}_1'\in \widetilde {W}^1_1(\tilde {e}_1^{(i)})$ and $\tilde {e}_2'\in \widetilde {W}^2_1(\tilde {e}_2^{(j)})$ such that $\|\tilde {e}_1-\mathcal {O}\|_1=\|\tilde {e}_1'-\mathcal {O}\|_1$ and $\|\tilde {e}_2-\mathcal {O}\|_2=\|\tilde {e}_2'-\mathcal {O}\|_2$ .

Again, $\|\tilde {e}_1-\tilde {e}_1'\|_2\leq 2M_2\alpha <2M_2({\varepsilon }/{10M})<\varepsilon $ and $\|\tilde {e}_2-\tilde {e}_2'\|_1 <\varepsilon $ . Here, we have

$$ \begin{align*} \|\tilde{e}_1-\mathcal{O}\|_2 &\leq\|\tilde{e}_1-\tilde{e}_1'\|_2+\|\tilde{e}_1'-\tilde{e}_1^{(i)}\|_2+\|\tilde{e}_1^{(i)}-\mathcal{O}\|_2\\ &\leq \varepsilon+2M_1+(2i-1)(M_1+\varepsilon)\\ &< (2i+1)(M_1+\varepsilon), \end{align*} $$

where $\|\tilde e_1'-\tilde e_1^{(i)}\|_2\leq 2M_1$ because every point on the leaf $\widetilde W_1^1(\tilde e_1^{(i)})$ is of bounded distance $<M_1$ from a linear leaf passing through $h(\tilde e_1^{(1)})$ and then $\tilde e^{\prime }_1$ can only be in a cylinder of radius $M_1$ that contains this linear leaf. Similarly, we have $\|\tilde {e}_2-\mathcal {O}\|_1< (2j+1)(M_1+\varepsilon)$ .

If $i\geq j$ , then

$$ \begin{align*} \|\tilde{e}_1-\tilde{e}_2\|_1 &\geq\|\tilde{e}_1-\mathcal{O}\|_1-\|\tilde{e}_2-\mathcal{O}\|_1 \\ &>\|\tilde{e}_1^{(i)}-\mathcal{O}\|_1-\|\tilde{e}_2-\mathcal{O}\|_1\\ & \geq iD-(2j+1)(M_1+\varepsilon)\\ &\geq (10i-(2j+1))(M_1+\varepsilon)>0. \end{align*} $$

If $i\leq j$ , then $\|\tilde {e}_1-\tilde {e}_2\|_2>0$ .

Proof. We apply the above proposition, where $\mathscr {F}_i,\ i=1,2$ is the pair of the stable and unstable foliations of $F_{b_0}$ and $\mathscr {F}_i',\ i=1,2$ is the stable and unstable foliations of $F_{b}$ . We get an $\alpha $ .

Because of the smoothness of F and thus of its induced foliations and the compactness of the bundle, we are able to pick a $\delta>0$ such that if $d(b,b_0)<\delta $ with $b,b_0$ in the same trivialized neighbourhood, then at any point $e\in p^{-1}b_0$ and $e'\in p^{-1}b$ such that $\psi (e)=\psi (e')$ , we have $\max \angle _e(p_i(e),d\psi (p_i'(e)))<\alpha $ , where $\{p_i\}_{i=1}^d$ and $\{p_i'\}_{i=1}^d$ are bases of the tangent space of $p^{-1}b_0$ and $p^{-1}b$ , respectively. Also, $p_i,\ p_i'$ are bases elements of the stable bundle if $i=1,\ldots , l$ and are bases elements of the unstable bundle if $i=l+1,\ldots ,d$ .

Proof. This also follows from Corollary 5.14, because for that fixed $\delta $ , we could find an N such that $d(b_0,b_N)<\delta $ .

Proof of Theorem 1.1

The set

$$ \begin{align*}\{b\in B\ |\ \text{the foliations induced by } F_b \text{ has the global product structure}\}\end{align*} $$

is non-empty by Lemma 5.7, is open by Corollary 5.14 and is closed by Corollary 5.15. Hence, this set is the full set of B. By Lemma 5.10, $\tilde {h}_b$ is injective for all $b\in B$ .

We know h is surjective from Proposition 5.1. If we take $e,e'$ such that $h(e)=h(e')$ , then e and $e'$ must be in the same fibre. Let $b:=p(e)=p(e')$ . Consider the lifts to $\mathbb {R}^d_b$ . Then there is a deck transformation $j\in \mathbb {Z}^d$ , such that $\tilde {h}_b(\tilde {e})=\tilde {h}_b(\tilde {e}')+j=\tilde {h}_b(\tilde {e}'+j)$ , so $\tilde {e}=\tilde {e}'+j$ . Thus, $e=e'$ . Therefore, h is also injective.