Proof. Let $C_0=\max \{\sup _{x\in A}|Df_x|,\sup _{x\in A}|Dg_x|\}<\infty $ and put

$$ \begin{align*}m(x,g,E):=m(Dg_x|_{E(x,f)})=\inf_{v\in E(x,f)\setminus\{0\}}\dfrac{|Dg_x v|}{|v|},\end{align*} $$

then one has that $m(x,g,E)>0$ for every $x\in \Gamma $ , since $Dg_x$ is injective and $E(x,f)$ is finite dimensional.

We claim that for $\mu $ -a.e. $x\in \Gamma $ ,

(3.1) $$ \begin{align} \lim_{n\rightarrow \pm \infty} \frac{1}{n} \log m(f^{n}(x),g,E)=0. \end{align} $$

We proceed to prove the lemma by assuming that the claim is true. By definition, we have $m(x,g,E)|v|\leq |Dg_x v| \leq C_0|v|$ for every $x\in \Gamma $ and $v\in E(x,f)$ , which together with the community of $f,g$ yields that

$$ \begin{align*} m(f^{n}(x),g,E)|Df^n_x v| \leq |Dg_{f^{n}(x)} Df^n_x v|=|Df^n_{g(x)}Dg_x v|\leq C_0 |Df^n_x v|. \end{align*} $$

It follows from the claim immediately that $\unicode{x3bb} (g(x),Dg_xv,f)=\unicode{x3bb} (x,v,f)$ for $\mu $ -a.e. $x\in \Gamma $ .

To finish the proof of the lemma, it suffices to prove the above claim.

Proof of the claim

Let $\Gamma (m)=\{x\in \Gamma : m(x,f)=m\}$ , then $\Gamma =\bigcup _{m\in \mathbb {N}} \Gamma (m)$ and $\Gamma (m)$ is f-invariant for every $m\in \mathbb {N}$ . Take a positive integer $m\in \mathbb {N}$ with $\mu (\Gamma (m))>0$ and let $\mu _m$ be the normalized measure of $\mu $ restricted on $\Gamma (m)$ . Hence, $\mu _m$ is an f-invariant probability measure supported on $\Gamma (m)$ . Note that

$$ \begin{align*}Dg_{f(x)}|_{E(f(x),f)}=Df_{g(x)} \circ Dg_{x} \circ (Df_x |_{E(x,f)})^{-1},\end{align*} $$

it is easy to show that $m(f(x),g,E)\leq |Df_{g(x)}|\cdot m(x,g,E) \cdot |(Df_x |_{E(x,f)})^{-1}|$ . Consequently, one can show that

(3.2) $$ \begin{align} (\log m(f(x),g,E)-\log m(x,g,E))^{+}\leq \log C_0 + \log^{+} |(Df_x |_{E(x,f)})^{-1}|. \end{align} $$

To complete the proof of the lemma, it suffices to show that (3.1) holds for $\mu $ -a.e. $x\in \Gamma (m)$ . By Lemma 3.5 and equation (3.2), it suffices to show that

(3.3) $$ \begin{align} \int \log^{+} |(Df_x |_{E(x,f)})^{-1}|\,d\mu_m(x)<\infty. \end{align} $$

For every $x\in \Gamma (m) $ and every $v\in E(x,f)$ with $|v|=1$ , by [Reference Blumenthal and Young3, Lemma 3.11], we have that

(3.4) $$ \begin{align} \det(Df_x |_{E(x,f)}) \leq m^{m/2} |Df_x |_{E(x,f)} |^{m-1} |Df_x v|, \end{align} $$

where one can see [Reference Blumenthal and Young3, §3.2] for the definition of the determinant of a linear operator on Banach spaces. This, together with the fact $ |Df_x |_{E(x,f)} |\leq C_0$ , yields that

$$ \begin{align*}|(Df_x |_{E(x,f)})^{-1}|\leq C_m \cdot (\det(Df_x |_{E(x,f)}))^{-1},\end{align*} $$

where $C_m=m^{m/2} \cdot C_0^{m-1}$ .

Put $\phi (x)=\log \det (Df_x |_{E(x,f)})$ . To show equation (3.3), it suffices to prove that $ \phi \in L^{1}(\mu _m)$ . Since $\det (Df_x |_{E(x,f)})\leq m^{m/2} C^{m}_0$ by equation (3.4), one has $\phi ^{+}\in L^1(\mu _m)$ . It remains to show that $\phi ^{-}\in L^1(\mu _m)$ . By Birkhoff’s ergodic theorem, there exists a measurable function $\hat {\phi }:\Gamma _m\rightarrow [-\infty ,\infty )$ with $\hat {\phi }^{+}\in L^1(\mu _m)$ such that

$$ \begin{align*} \hat{\phi}(x)=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=0}^{n-1} \phi(f^i(x))=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=0}^{n-1} \phi(f^{-i}(x))\end{align*} $$

for $\mu _m$ -a.e. $x\in \Gamma (m)$ . Note that

$$ \begin{align*} \hat{\phi}(x)&=\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=0}^{n-1} \phi(f^{-i}(x))\\ &=\lim_{n\rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}\log \det(Df_{f^{-i}(x)} |_{E(f^{-i}(x),f)})\\ &=\lim_{n\rightarrow \infty} \frac{1}{n} \log \det(Df^n_{f^{-n}(x)}|_{E(f^{-n}(x),f)})\\ &=\lim_{n\rightarrow \infty} -\frac{1}{n} \log \det(Df^{-n}_{x}|_{E(x,f)}) \end{align*} $$

and, by [Reference Blumenthal and Young3, Lemma 3.11], one can choose $\{v_1,\ldots ,v_m \}$ to be a unit basis for $E(x,f)$ such that

$$ \begin{align*}\det(Df^{-n}_{x}|_{E(x,f)})\leq m^{m/2} \prod_{i=1}^{m}|Df^{-n}_{x} v_i|.\end{align*} $$

By Theorem 3.2(a), we have that

Then, we have that $\int \hat {\phi }(x) dx>-\infty $ . Note that $\int \hat {\phi }(x)\,dx= \int \phi (x)\,dx$ . Hence, we have $\phi (x) \in L^1(\mu _m).$

This completes the proof of this lemma.

Proof. By Lemma 3.7, for $\mu $ -a.e. $x\in \Gamma $ and every $u\in E(x,f)\setminus \{0\}$ , we have that

This implies that $Dg_x u \notin E_{\alpha }(g(x),f)$ , that is, $E_{\alpha }(g(x),f)\cap Dg_x E(x,f)=\{0\}$ . Hence, one has that

$$ \begin{align*}m(x,f)\leq \text{co}\dim E_{\alpha}(g(x),f)=m(g(x),f)\end{align*} $$

for $\mu $ -a.e. $x\in \Gamma $ . Since $\mu $ is $(f,g)$ -ergodic and $m(x,f)=m(f(x),f)$ , by Proposition 2.3, we have that $m(\cdot ,f)$ is a constant $\mu $ -almost everywhere.

Take a point x with $m(x,f)=m(g(x),f)$ , one can show that $\unicode{x3bb} _{j}(x,f)\le \unicode{x3bb} _{j}(g(x),f)$ by Lemma 3.7 and $m_j(x,f)\le m_{j}(g(x),f)$ by the injectivity of $Dg_x$ for $j=1,\ldots ,r(x,f)$ . Consequently, one has that $r(x,f)\le r(g(x),f)$ since we have already obtained $r(x,f)$ number of exponents at point $g(x)$ . Note that $r(\cdot ,f),\ \unicode{x3bb} _{j}(\cdot ,f)$ and $m_j(\cdot ,f)$ are also f-invariant functions, so it follows from the $(f,g)$ -ergodity that they are constants for $\mu $ -a.e. x.

Since $E_{\alpha }(g(x),f)\cap Dg_x E(x,f)=\{0\}$ and $Dg_x$ is injective for $\mu $ -a.e. x, one may consider a splitting of the tangent space at point x as follows:

$$ \begin{align*} \mathcal{B}_x=\bigoplus_{i=1}^{r}Dg_{g^{-1}x}E_i(g^{-1}(x),f)\oplus E_{\alpha}(g(x),f). \end{align*} $$

It satisfies Theorem 3.2. By the unique existence of Oseledets splitting, we have that $Dg_x E_{j}(x,f)=E_{j}(g(x),f)$ for $j=1,\ldots ,r(x,f)$ and $Dg_x E_{\alpha }(x,f)\subset E_{\alpha }(g(x),f)$ for $\mu $ -a.e. x.

Proof. Note that $f,g$ are $C^2$ maps, so by the assumption and Lemma 3.8, we have that

$$ \begin{align*}\log^{+} |(Df|_{V})^{\pm}|\in L^{1}(\mu) \ \text{and} \ \log^{+} |(Dg|_{V})^{\pm}|\in L^{1}(\mu).\end{align*} $$

This implies that $\log ^{+} |Df^sg^t|_{V}|\in L^{1}(\mu )$ for every $s,t\in \mathbb {Z}$ . For every $s,t\in \mathbb {Z}$ , consider the finite dimensional $D(f^sg^t)$ -invariant sub-bundle V and apply the multiplicative ergodic theorem for the cocycle $\{D(f^sg^t)^{n}\}_{n\ge 0}$ , which is defined on A with respect to $f^sg^t$ , for $\mu $ -a.e. $x\in \Gamma $ , the following limit:

$$ \begin{align*}\unicode{x3bb}(x,v,f^sg^t)=\lim_{n\rightarrow \pm \infty}\frac{1}{n}\log |D(f^sg^t)^n_x v|\end{align*} $$

exists for every $v\in V(x)\setminus \{0\}$ .

For every $\varepsilon>0$ , let

$$ \begin{align*}A_{\varepsilon}=\{x\in \Gamma: \unicode{x3bb}(x,v,f^sg^t)-s\unicode{x3bb}(f)-t\unicode{x3bb}(g)>4\varepsilon \ \text{for some} \ v\in V(x)\}.\end{align*} $$

Assume that $\mu (A_{\varepsilon })>0$ . For every $\ell \in \mathbb {N}$ , let

$$ \begin{align*} {A}^{\prime}_{\varepsilon,\ell}=\{&x\in A_{\varepsilon}: \text{ there exists } \ u_x \in V(x) \ \text{such that}\ |D(f^sg^t)^n_x u_x|\\&\geq \ell^{-1} e^{n(\unicode{x3bb}(x,u_x,f^sg^t)-\varepsilon)}|u_x| \text{ for all } \ n\geq 1 \} \end{align*} $$

and

$$ \begin{align*} {A}^{\prime\prime}_{\varepsilon,\ell}=\{x\in A_{\varepsilon}: \ |Dg^{tn}_x u|\leq \ell e^{n(t\unicode{x3bb}(g)+\varepsilon)}|u| \text{ for all } u \in V(x), \text{ for all } n\geq 1 \}. \end{align*} $$

Since $\mu (A_{\varepsilon })>0$ , one may choose $\ell $ large enough such that $\mu ({A}^{\prime }_{\varepsilon ,\ell }\cap {A}^{\prime \prime }_{\varepsilon ,\ell })>0$ . For every $x\in {A}^{\prime }_{\varepsilon ,\ell }\cap {A}^{\prime \prime }_{\varepsilon ,\ell }$ , applying Poincaré’s recurrence theorem with respect to $f^s$ , there exists $n>2({\log \ell }/{\varepsilon })$ such that $f^{sn} x\in {A}^{\prime }_{\varepsilon ,\ell }\cap {A}^{\prime \prime }_{\varepsilon ,\ell }$ and

$$ \begin{align*}|Df^{sn}_x u|\leq |u|\cdot e^{n(s\unicode{x3bb}(f)+\varepsilon)}\end{align*} $$

for every $u\in V(x)$ . Hence, for every $0\neq u\in V(x)$ , we have that

$$ \begin{align*} |D(f^sg^t)^n_x u|&=|Dg^{tn}_{f^{sn}(x)}Df^{sn}_x u|\\ &\leq \ell e^{n(t\unicode{x3bb}(g)+s\unicode{x3bb}(f)+2\varepsilon)}|u|\\ &\leq \ell e^{-n\varepsilon} e^{n(\unicode{x3bb}(x,u,f^sg^t)-\varepsilon)}|u|\\ &< \ell^{-1} e^{n(\unicode{x3bb}(x,u,f^sg^t)-\varepsilon)}|u|, \end{align*} $$

which contradicts with the fact $x\in {A}^{\prime }_{\varepsilon ,\ell }$ . Hence, we have $\mu (A_{\varepsilon })=0$ for every $\varepsilon> 0$ . In a similar fashion, one can show that the following set:

$$ \begin{align*}B_{\varepsilon}=\{x\in \Gamma: \unicode{x3bb}(x,v,f^sg^t)-s\unicode{x3bb}(f)-t\unicode{x3bb}(g)<4\varepsilon \ \text{for some} \ v\in V(x)\}\end{align*} $$

satisfies that $\mu (B_{\varepsilon })=0$ for every $\varepsilon> 0$ . By the arbitrariness of $\varepsilon> 0$ , we have that

(3.5) $$ \begin{align} \lim_{n\rightarrow \pm \infty}\frac{1}{n}\log |D(f^sg^t)^n_x v|=s\unicode{x3bb}(f)+t\unicode{x3bb}(g)\quad \mu\text{-a.e.}\, x \end{align} $$

for every $0\neq v\in V(x)$ .

Proof. Let $h=f$ or g. Since $Dh_x V(x)=V(h(x))$ and $Dh_x$ is injective for every x, we have that

$$ \begin{align*}Dh_x \circ (\pi_{V}(x)) v= (\pi_{V}(hx)) \circ Dh_x v\end{align*} $$

for every $v\in \mathcal {B}_x$ . Hence, we have that $|\pi _{V}(x)|\leq |(Dh_x|_{V(x)})^{-1}| \cdot |\pi _{V}(h(x))| \cdot |Dh_x|$ . Consequently, one has that

$$ \begin{align*}\log |\pi_{V}(x)|-\log|\pi_{V}(h(x))|\leq \log |(Dh_x|_{V(x)})^{-1}|+ \log C_0,\end{align*} $$

where $C_0=\max \{\sup _{x\in A}|Df_x|,\sup _{x\in A}|Dg_x|\}\hspace{-1pt}<\hspace{-1pt}\infty $ . Since $\log ^{+}|(Dh_{x}|_{V(x)})^{-1}|\hspace{-1pt}\in\hspace{-1pt} L^1(\mu )$ , we have that $(\log |\pi _{V}(x)|-\log |\pi _{V}(h(x))|)^{+}\in L^1(\mu )$ . It follows from Lemma 3.5 that

$$ \begin{align*} \lim_{n\rightarrow \pm\infty} \frac{1}{n}\log |\pi_{V}(h^{n}(x))|=0 \quad \mu\text{-a.e.}\,x.\\[-40pt] \end{align*} $$

Proof. For every $i\in \{1,\ldots ,r(f)\}$ , $j\in \{1,\ldots ,j_i\}$ and every $v\in E_{i,j}(x)\setminus \{0\}$ , we have that

and $\unicode{x3bb} _{i}(f)>\unicode{x3bb} _{\alpha }$ , $\unicode{x3bb} _{i,j}(g)>-\infty $ by the assumptions. By Lemma 3.8, one has that

$$ \begin{align*}\log|(Df|E_{i,j})^{-1}|\in L^1(\mu) \quad \text{and} \quad \log|(Dg|E_{i,j})^{-1}|\in L^1(\mu).\end{align*} $$

It follows form Lemma 3.12 that the norm of projection $\pi _{i,j}^{E}$ is ( $\mu $ -)temperate with respect to f and g, respectively. In a similar fashion, one can show that the norm of projection $\pi _{i,j}^{G}$ is ( $\mu $ -)temperate with respect to f and g, respectively.

Proof. Note that the Lyapunov exponent of the cocycle $\{Df_x^n\}_{n\ge 0}$ on the sub-bundle $E_i(x,f)$ is $\unicode{x3bb} _i(f)$ , and the Lyapunov exponent of the cocycle $\{Dg_x^n\}_{n\ge 0}$ on the sub-bundle $E_{\alpha ,j}(x)$ is $\unicode{x3bb} _j(g)$ , so both $\unicode{x3bb} _i(f)$ and $\unicode{x3bb} _j(g)$ are finite. By Lemma 3.8, we have that

$$ \begin{align*} \int \log^{+}|(Df_x|E_i(x,f))^{-1}|\,d\mu<\infty \quad \text{and} \quad \int \log^{+}|(Dg_x|E_{\alpha,j}(x))^{-1}| \,d\mu<\infty\end{align*} $$

for every $1\leq i \leq r(f),1\leq j \leq r(g)$ . Consequently, by condition (H4), we have that

$$ \begin{align*} \int \log^{+}|(Dg_x|E_i(x,f))^{-1}|\,d\mu<\infty \quad \text{and} \quad \int \log^{+}|(Df_x|E_{\alpha,j}(x))^{-1}|\,d\mu<\infty\end{align*} $$

for every $1\leq i \leq r(f),1\leq j \leq r(g)$ . The integrability of $\log ^{+}|(Dg_x|E_i(x,f))^{-1}|$ implies that $\unicode{x3bb} _{i,j}(g)>-\infty $ for every $i=1,\ldots ,r(f)$ and $j=1,\ldots , j_i$ . Similarly, the integrability of $ \log ^{+}|(Df_x|E_{\alpha ,j}(x))^{-1}|$ yields that $\unicode{x3bb} _{i,j}(f)>-\infty $ for every $j=1,\ldots ,r(g)$ and $i=1,\ldots , i_j$ .

Hence, by Corollary 3.13, the norms of the projections $\pi _{i,j}^{E}(x)$ and $\pi _{i,j}^{G}(x)$ are ( $\mu $ -)temperate with respect to f and g, respectively. Finally, the fact

$$ \begin{align*}\mathrm{Id}_x=\sum_{i=1}^{r(f)}\sum_{j=1}^{j_i}\pi_{i,j}^{E}(x)+\sum_{j=1}^{r(g)}\sum_{i=1}^{i_j}\pi_{i,j}^{G}(x) +\pi_{\alpha}(x)\end{align*} $$

yields that the norm of the projection $\pi _{\alpha }(x)$ is ( $\mu $ -)temperate with respect to f and g, respectively.

Proof. For simplicity, we write $\pi ^{\tau }_{h^{k}(x),h}$ , $|\cdot |_{h^{k}(x)}$ as $\pi ^{\tau }_{k}$ , $|\cdot |_{k}$ , respectively.

Note that

$$ \begin{align*}|\widetilde{h}^{k}_x(u)|_{k}=\max\{|\pi_{k}^{u}\widetilde{h}^{k}_x(u)|_{k}, |\pi_{k}^{cs}\widetilde{h}^{k}_x(u)|_{k}\}.\end{align*} $$

If $ |\widetilde {h}^{k}_x(u)|_{k}=|\pi _{k}^{u}\widetilde {h}^{k}_x(u)|_{k}$ . By part (1) of Corollary 4.8, we have that

$$ \begin{align*} \delta\geq|\widetilde{h}^{n}_x(u)|_{n}\geq e^{2(n-k)\varepsilon} |\widetilde{h}^{k}_x(u)|_{k}.\end{align*} $$

Thus, $|\widetilde {h}^{k}_x(u)|_{k}\leq \delta e^{-2(n-k)\varepsilon }$ . If $|\widetilde {h}^{k}_x(u)|_{k}=|\pi _{k}^{cs}\widetilde {h}^{k}_x(u)|_{k}$ , by part (2) of Corollary 4.8, we have that

$$ \begin{align*}|\widetilde{h}^{k}_x(u)|_{h^k(x)}\leq e^{2k\varepsilon}|u|_{x} \leq \delta e^{-2(n-k)\varepsilon}.\end{align*} $$

This proves the first statement.

To prove the second statement, note that

$$ \begin{align*}|\widetilde{h}^{k}_x(u)|_{h^k(x)}=\max\{|\pi_{k}^{uc}\widetilde{h}^{k}_x(u)|_{k},|\pi_{k}^{s}\widetilde{h}^{k}_x(u)|_{k}\}.\end{align*} $$

If $|\widetilde {h}^{k}_x(u)|_{k}=|\pi _{k}^{uc}\widetilde {h}^{k}_x(u)|_{k}$ . By part (1) of Corollary 4.8 and Remark 4.9, we have that

$$ \begin{align*} \delta e^{-2n\varepsilon} \geq|\widetilde{h}^{n}_x(u)|_{n}\geq e^{-2(n-k)\varepsilon} |\widetilde{h}^{k}_x(u)|_{k}.\end{align*} $$

Thus, $|\widetilde {h}^{k}_x(u)|_{k}\leq \delta e^{-2k\varepsilon }$ . If $|\widetilde {h}^{k}_x(u)|_{k}=|\pi _{k}^{s}\widetilde {h}^{k}_x(u)|_{k}$ , by part (2) of Corollary 4.8 and Remark 4.9, we have that

$$ \begin{align*}|\widetilde{h}^{k}_x(u)|_{k}\leq e^{-2k\varepsilon}|u|_{x} \leq \delta e^{-2k\varepsilon}.\end{align*} $$

This completes the proof of the lemma.

Proof. For simplicity, denote $|\cdot |_{h^{k}(x)}$ by $|\cdot |_{k}$ . For each $y\in B_n(x,\rho _{\delta ,h},h)$ , let $u=\exp ^{-1}_x y$ . For every $k\in \{0,1\ldots ,n\}$ with $h^k(x)\in \Gamma _\ell $ , we have that

$$ \begin{align*}|\widetilde{h}^{k}_x(u)|_{k}\leq \ell(h^k(x))|\widetilde{h}^{k}_x(u)|\leq \ell|h^k(x)-h^k(y)| \leq \ell\delta \leq \ell^{-1}\delta_1\leq \delta_1\ell(h^k(x))^{-1}.\end{align*} $$

If $h^k(x)\notin \Gamma _\ell $ , denote by $n_k<k$ the biggest non-negative integer such that $h^{n_k}(x)\in \Gamma _\ell $ , then $\tau _{h}(h^{n_k}(x))>k-n_k$ . We have

$$ \begin{align*}|\widetilde{h}^{n_k}_x(u)|_{n_k}\leq \ell|h^{n_k}(x)-h^{n_k}(y)|\leq \ell^{-1}\delta_1 e^{-(\unicode{x3bb}+\varepsilon)\tau_{h}(h^{n_k}(x))} \leq \ell^{-1}\delta_1 e^{-(\unicode{x3bb}+\varepsilon)(k-n_k)}.\end{align*} $$

By equation (4.12), we have that

$$ \begin{align*} |\widetilde{h}^{k}_x(u)|_{k}\leq e^{\unicode{x3bb}(k-n_k)}|\widetilde{h}^{n_k}_x(u)|_{n_k}\leq \ell^{-1}e^{-(k-n_k)\varepsilon}\delta_1\leq \delta_1\ell(h^k(x))^{-1}\end{align*} $$

since $\ell (h^{n_k}(x))=\ell $ . Therefore, we have that

$$ \begin{align*}|\widetilde{h}^{k}_x(u)|_{k}\leq \delta_1\ell(h^k(x))^{-1}\end{align*} $$

for every $k=0,1\ldots ,n$ . Moreover, if $y\in B(x,\delta e^{-2n\varepsilon })$ , then $|u|_{x}\leq \ell (x)|u|\leq \ell \delta e^{-2n\varepsilon }$ . Similarly, $h^n(y)\in B(h^n(x),\delta e^{-2n\varepsilon })$ implies $|\widetilde {h}^{n}_x(u)|_{n}\leq \ell \delta e^{-2n\varepsilon }$ . Statements (1) and (2) follow immediately from Lemma 5.5.

Proof. For each $y\in \Delta $ , let $u=\exp ^{-1}_x y$ . Since $x\in \Gamma _\ell \cap f^{-n}\Gamma _\ell $ and $y\in B_n(x,\rho _{\delta ,f},f)\cap B(x,\delta e^{-2n\varepsilon })$ , by part (1) of Lemma 5.6, we have that

(5.1) $$ \begin{align} \widetilde{f}^{n-k}_{x}u\in \widetilde{B}_{f^{n-k}(x)}(\delta \ell e^{-2k\varepsilon}) \end{align} $$

for every $0\leq k\leq n$ . Similarly, since $f^n(x)\in \Gamma _\ell \cap g^{-n}\Gamma _\ell $ and $f^{n}(y)\in B_n(f^n(x), \rho _{\delta ,g},g) ) \cap g^{-n}(B((fg)^n(x), \delta e^{-2n\varepsilon }))$ , by part (2) of Lemma 5.6, we have that

(5.2) $$ \begin{align} \widetilde{(fg)}^{k}_{f^{n-k}(x)}\circ \widetilde{f}^{n-k}_{x}u\in \widetilde{B}_{f^ng^k(x)}(\delta \ell e^{-2k\varepsilon}) \end{align} $$

for every $0\leq k\leq n$ .

For each $0\leq k\leq n$ , we claim that

(5.3) $$ \begin{align} \widetilde{(fg)}^{i}_{f^{n-k}(x)}\circ \widetilde{f}^{n-k}_{x}u \in \widetilde{B}_{(fg)^i(f^{n-k}(x))}(\delta \ell e^{-2\max \{i,k-i\}\varepsilon}) \ \text{for} \, i=0,1,\ldots, k. \end{align} $$

We will show the above claim by induction. For $k=0$ , the claim follows from equation (5.1). Suppose that the claim is true for $k-1$ , that is,

$$ \begin{align*}\widetilde{(fg)}^{i-1}_{f^{n-k+1}(x)}\circ \widetilde{f}^{n-k+1}_{x}u \in \widetilde{B}_{(fg)^{i-1}(f^{n-k+1}(x))}(\delta \ell e^{-2\max \{i-1,k-i\}\varepsilon})\end{align*} $$

for every $1\leq i \leq k$ . Note that

$$ \begin{align*} (\widetilde{g}\circ\widetilde{(fg)}^{i-1})_{f^{n-k+1}x}\circ \widetilde{f}^{n-k+1}_{x} u = \widetilde{(fg)}^{i}_{f^{n-k}(x)}\circ \widetilde{f}^{n-k}_{x}u. \end{align*} $$

By equation (4.12) and the induction assumption, for every $1\leq i \leq k$ , we have that

$$ \begin{align*} |\widetilde{(fg)}^{i}\circ \widetilde{f}^{n-k}_{x}u|_{(fg)^{i}(f^{n-k}(x))} &\leq e^\unicode{x3bb}|\widetilde{(fg)}^{i-1}_{f^{n-k+1}x}\circ \widetilde{f}^{n-k+1}_{x}u|_{(fg)^{i-1}(f^{n-k+1}(x))}\\ &\leq e^\unicode{x3bb} \cdot \delta \ell e^{-2\max \{i-1,k-i\}\varepsilon}\\ &\leq \delta_1 \ell^{-1}e^{-k\varepsilon}. \end{align*} $$

By the definition of the function $\ell (x)$ , we have that

$$ \begin{align*}\ell((fg)^{i}(f^{n-k}(x)))=\ell(f^{-(k-i)}g^{i}f^n(x))\leq \ell(f^n(x))e^{(k-i)\varepsilon+i\varepsilon}\leq \ell e^{k\varepsilon} \end{align*} $$

since $f^n(x)\in \Gamma _\ell $ . Therefore, we have that

$$ \begin{align*}\widetilde{(fg)}^{i}\circ \widetilde{f}^{n-k}_{x}u \in \widetilde{B}_{(fg)^if^{n-k}(x)}(\delta_1 \ell((fg)^{i}f^{n-k}(x))^{-1}) \quad \text{ for all } i=1,2,\ldots, k.\end{align*} $$

Applying Lemma 5.5 to vector $\widetilde {f}^{n-k}_{x}u$ and map $\widetilde {fg}$ with equations (5.1) and (5.2), we have that

$$ \begin{align*}\widetilde{(fg)}^{i}_{f^{n-k}(x)}\circ \widetilde{f}^{n-k}_{x}u \in \widetilde{B}_{(fg)^i(f^{n-k}(x))}(\delta \ell e^{-2\max \{k,k-i\}\varepsilon})\quad \text{ for all } i=0,1,\ldots, k.\end{align*} $$

This completes the proof of the claim.

By the claim, for every $0 \leq i \leq n$ , we have that

$$ \begin{align*}|\widetilde{(fg)}^{i}_x u|_{(fg)^i(x)}\leq \delta \ell e^{-2\max \{i,n-i\}\varepsilon}\leq \delta \ell.\end{align*} $$

Since $ \widetilde {(fg)}^{i}_x u=(fg)^i(y)-(fg)^i(x)$ , it follows from equation (4.10) that

$$ \begin{align*} |(fg)^i(x)-(fg)^i(y)|=|\widetilde{(fg)}^{i}_x u|\leq K_0 |\widetilde{(fg)}^{i}_x u|_{(fg)^i(x)} \leq \delta \ell K_0 \end{align*} $$

for every $0 \leq i \leq n$ . Hence, $y\in B_n(x,\delta \ell K_0,fg).$ This completes the proof of the lemma.

Proof of Theorem A

Take $r\in (0,1)$ . Fix $\ell>1$ such that $\mu (\Gamma _\ell )>1-r/5$ . Let $\rho _{\delta ,f}$ and $\rho _{\delta ,g}$ be two families of functions as in §5.2.

For $h=f$ or g, $n\in \mathbb {N}$ and sufficiently small number $\varepsilon>0$ and $\delta>0$ , let

$$ \begin{align*}A^{h}_{n,\delta,\varepsilon}=\{x\in \Gamma:\mu(B_k(x,\rho_{\delta,h},h))\geq e^{-k(h_{\mu}(h)+\varepsilon)} \text{ for all } k\geq n\}.\end{align*} $$

It follows from Proposition 5.2 that

$$ \begin{align*}h_{\mu}(h)\geq \limsup_{n\rightarrow \infty} -\frac{1}{n} \log \mu(B_n(x,\rho_{\delta,h},h)) \quad \mu\text{-almost everywhere}.\end{align*} $$

Thus, for every $\delta>0$ , we have $\mu (A^{h}_{n,\delta ,\varepsilon })\rightarrow 1$ as $n\rightarrow \infty $ . Choose $n_h(\delta )>0$ , such that $\mu (A^{h}_{n,\delta ,\varepsilon })>1-r/5$ for every $n>n_h(\delta )$ .

By the definition of $A^{h}_{n,\delta ,\varepsilon }$ , there are at most $e^{n(h_{\mu }(h)+\varepsilon )}$ disjoint $(n,\rho _{\delta ,h},h)$ -balls centred at points in $A^{h}_{n,\delta ,\varepsilon }$ . So the same number of $(n,2\rho _{\delta ,h},h)$ -balls centred at points in $A^{h}_{n,\delta ,\varepsilon }$ can cover $A^{h}_{n,\delta ,\varepsilon }$ . That is, there exists a set $S_h\subset A^{h}_{n,\delta ,\varepsilon }$ with $|S_h|\leq e^{n(h_{\mu }(h)+\varepsilon )}$ such that

$$ \begin{align*}A^{h}_{n,\delta,\varepsilon} \subset \bigcup_{x\in S_h}B_n(x,2\rho_{\delta,h},h),\end{align*} $$

where $|S|$ denotes the cardinality of the set S.

Take $0<\delta <\min \{\tfrac 14\delta _1\ell ^{-2} e^{-\unicode{x3bb} -\varepsilon } ,\delta _2\}$ . By equation (5.4), there exists a set $S_0\subset A$ with $|S_0|=N(2\delta e^{-2n\varepsilon })\leq (2\delta e^{-2n\varepsilon })^{-D_0}$ such that

$$ \begin{align*}A\subset \bigcup_{x\in S_0}B(x,2\delta e^{-2n\varepsilon}).\end{align*} $$

For each $n>\max \{n_f(\delta ),n_g(\delta )\}$ , let

$$ \begin{align*}A_n=A^{f}_{n,\delta,\varepsilon}\cup f^{-n} A^{g}_{n,\delta,\varepsilon} \cap \Gamma_\ell \cap f^{-n}\Gamma_\ell \cap (fg)^{-n} \Gamma_\ell.\end{align*} $$

Obviously, $\mu (A_n)\geq 1-r$ . For every $x_f\in S_f, x_g\in S_g,{x}'\in S_0$ and ${x}"\in S_0$ , if the intersection

$$ \begin{align*} &A_n\cap B_n(x_f,2\rho_{\delta,f},f)\cap B({x}',2\delta e^{-2n\varepsilon}) \\ &\quad\cap f^{-n}(B_n(x_g,2\rho_{\delta,g},g) ) \cap (fg)^{-n}(B({x}", 2\delta e^{-2n\varepsilon})) \end{align*} $$

is not empty, then for each x in it, the intersection is contained in the set

$$ \begin{align*} &B_n(x,4\rho_{\delta,f},f)\cap B(x,4\delta e^{-2n\varepsilon}) \\ &\quad\cap f^{-n}(B_n(f^n(x),4\rho_{\delta,g},g) ) \cap (fg)^{-n}(B((fg)^n(x), 4\delta e^{-2n\varepsilon})). \end{align*} $$

By Lemma 5.7, the above set is contained in $B_n(x,4\delta \ell K_0,fg)$ . There are at most $|S_f|\cdot |S_g|\cdot |S_0|^2 $ different such intersections. Note that $\mu (A_n)>1-r$ . These intersections cover $A_n$ and each one is contained in an $(n,4\delta \ell K_0,fg)$ -ball. Thus, we have that

$$ \begin{align*} N_n(4\delta \ell K_0,r,fg)&\leq |S_f|\cdot|S_g|\cdot|S_0|^2\\ &\leq e^{n(h_{\mu}(f)+h_{\mu}(g)+2\varepsilon)}\cdot (2\delta e^{-2n\varepsilon})^{-2D_0}\\ &\leq (2\delta)^{-2D_0}\cdot e^{n(h_{\mu}(f)+h_{\mu}(g)+(4D_0+2)\varepsilon)}. \end{align*} $$

Therefore, by Lemma 5.3,

$$ \begin{align*}h_{\mu}(fg)\leq h_{\mu}(f)+h_{\mu}(g)+(4D_0+2)\varepsilon.\end{align*} $$

Since $\varepsilon $ is arbitrary, one has that

$$ \begin{align*} h_{\mu}(fg)\leq h_{\mu}(f)+h_{\mu}(g).\end{align*} $$

This completes the proof of Theorem A.

Proof. Fix $x\in \Gamma $ . Since $E^{u}(x,f)=E^{u}(x,g)$ , we have that $E^{cs}(x,f)=E^{cs}(x,g)$ by equation (4.1). Hence, if there is no confusion caused, we will simply write $E^{\tau }(x,h)$ as $E^{\tau }(x)$ for $\tau =u,cs$ and $h=f,g$ . This also implies that $\mathcal {W}(x,f)=\mathcal {W}(x,g)$ , and we denote the common set by $\mathcal {W}(x)$ .

In the following, we shall show that $\widetilde {f}_x(W_{\delta }^{u}(x,g))\supset W_{\delta }^{u}(fx,g) $ , the other one can be proven in a similar fashion.

Given $\sigma _{x}^{g}\in \mathcal {W}(x)$ with $\operatorname {graph}(\sigma _{x}^{g})=W_{\delta }^{u}(x,g)$ as in Theorem 6.1, by Lemma 7.1, there exists $\tilde {\sigma }_{fx}\in \mathcal {W}(fx)$ such that

(7.1) $$ \begin{align} \widetilde{f}_x W_{\delta}^{u}(x,g)\supset \operatorname{graph}(\tilde{\sigma}_{fx}). \end{align} $$

It suffices to prove that $\tilde {\sigma }_{fx}=\sigma ^{g}_{fx}$ . For every $v \in \tilde {B}^{u}_{fx}(\delta \ell (fx)^{-1})$ , let $u:=v+\tilde {\sigma }_{fx} v\in \operatorname {graph}(\tilde {\sigma }_{fx}).$ By equation (7.1), there exists ${u}_0' \in W_{\delta }^{u}(x,g)$ such that $\widetilde {f}_x ({u}^{\prime }_0)=u$ . Since ${u}^{\prime }_0 \in W_{\delta }^{u}(x,g)$ , there exists ${u}^{\prime }_n\in W_{\delta }^{u}(g^{-n}x,g)$ such that $\widetilde {g}^n_{g^{-n}x} ({u}^{\prime }_n)={u}^{\prime }_0$ and $|\pi _{g^{-n}(x),f}^{u} {u}^{\prime }_n|_{g^{-n}(x)}=|{u}^{\prime }_n|_{g^{-n}(x)}$ for every $n\geq 0$ . Let $u_{n}=\widetilde {f}_{g^{-n}x} {u}^{\prime }_n$ , since ${u}^{\prime }_n\in \tilde {B}_{g^{-n}x}(\delta \ell (g^{-n}x)^{-1})$ . By equation (4.13) and part (1) of Corollary 4.8, we have that

$$ \begin{align*}u_{n}\in \tilde{B}_{g^{-n}(fx)}(\delta_1 \ell(g^{-n}fx)^{-1}),\quad |\pi_{g^{-n}(fx),f}^{u} u_n|_{g^{-n}(fx)}=|u_n|_{g^{-n}(fx)}.\end{align*} $$

By construction of $u_n$ , we have that $\widetilde {g}^n_{g^{-n}(fx)} u_n=u$ for every $n\geq 1$ . It follows from Lemma 6.3 that $u\in W_{\delta }^{u}(fx,g).$ This yields that $\tilde {\sigma }_{fx}= \sigma ^{g}_{fx}$ . This completes the proof of the lemma.

Proof. Given $x\in \Gamma $ and $u\in W^{u}_{\delta }(x,f)$ , it follows from Lemma 7.2 that

$$ \begin{align*}\widetilde{g}^{n}_{g^{-n}x}(W_{\delta}^{u}(g^{-n}x,f))\supset W_{\delta}^{u}(x,f)\ \ \text{ for all } \, n\geq 0. \end{align*} $$

Thus, there exists $u_n\in W^{u}_{\delta }(g^{-n}x,f)$ such that $\widetilde {g}^{n}_{g^{-n}x} u_n=u$ for every $n\geq 0$ . Since $u_n\in W^{u}_{\delta }(g^{-n}x,f)\subset \tilde {B}_{g^{-n}x}(\delta \ell (g^{-n}x)^{-1})$ and $E^{u}(g^{-n}x,f)=E^{u}(g^{-n}x,g)$ , we have that

$$ \begin{align*}|\pi_{g^{-n}(x),f}^{u} u_n|_{g^{-n}x}=|\pi_{g^{-n}(x),g}^{u} u_n|_{g^{-n}x}=|u_n|_{g^{-n}x}.\end{align*} $$

It follows from Lemma 6.3 that $u\in W^{u}_{\delta }(x,g).$ Thus, we have $W^{u}_{\delta }(x,f) \subset W^{u}_{\delta }(x,g).$

Similarly, one can show that $W^{u}_{\delta }(x,g) \subset W^{u}_{\delta }(x,f).$ Therefore, $W^{u}_{\delta }(x,g)= W^{u}_{\delta }(x,f)$ . So we conclude that $ w^{u}_{\delta }(x,f)=w^{u}_{\delta }(x,g) $ for every $x\in \Gamma $ .