[proofplan] We use the uniform Hadamard-Perron graph transform theorem over the compact hyperbolic set $\Lambda$. Compactness lets us choose one system of local charts, one radius, and one contraction rate valid for every $x\in\Lambda$. Applying the stable graph transform to $f$ gives the local stable disks, and applying the same theorem to $f^{-1}$ gives the local unstable disks. The exponential estimates follow from the contraction estimates supplied by the graph transform, after comparing the chart norms with the Riemannian distance on a fixed compact neighbourhood of $\Lambda$. [/proofplan]

[step:Choose uniform hyperbolicity data on the compact invariant set]Since $\Lambda$ is hyperbolic, there are constants $A\ge 1$ and $0<\lambda<1$ such that for every $x\in\Lambda$, every integer $n\ge 0$, every $v_s\in E^s_x$, and every $v_u\in E^u_x$, \begin{align*} |Df_x^n v_s|_{g_{f^n(x)}}\le A\lambda^n |v_s|_{g_x} \end{align*} and \begin{align*} |Df_x^{-n} v_u|_{g_{f^{-n}(x)}}\le A\lambda^n |v_u|_{g_x}. \end{align*} The subbundles $E^s\to\Lambda$ and $E^u\to\Lambda$ are continuous and $Df$-invariant. Because $\Lambda$ is compact, choose an open neighbourhood $U\subset M$ of $\Lambda$ with compact closure $N:=\overline U$ and choose an adapted Riemannian metric $\tilde g$ on an open neighbourhood of $N$. Let $\tilde d:N\times N\to[0,\infty)$ denote the ambient Riemannian distance induced by $\tilde g$, restricted to pairs of points in $N$. Choose a number $\lambda_0$ with $\lambda<\lambda_0<1$ so that \begin{align*} |Df_x v_s|_{\tilde g_{f(x)}}\le \lambda_0 |v_s|_{\tilde g_x} \end{align*} for all $x\in\Lambda$ and $v_s\in E^s_x$, and \begin{align*} |Df_x^{-1} v_u|_{\tilde g_{f^{-1}(x)}}\le \lambda_0 |v_u|_{\tilde g_x} \end{align*} for all $x\in\Lambda$ and $v_u\in E^u_x$. Since $g$ and $\tilde g$ are equivalent on the compact set $N$, after shrinking the graph-transform radius so all disks and their iterates appearing in the local estimates lie in $N$, there is a constant $B\ge 1$ such that \begin{align*} B^{-1}d(p,q)\le \tilde d(p,q)\le B d(p,q) \end{align*} for all $p,q\in N$. We will apply the graph transform using $\tilde g$ and transfer the final estimates back to $d$ using this comparison.[/step]

[guided]The role of this step is to make the hyperbolicity constants uniform. The definition of a compact hyperbolic set gives a $Df$-invariant splitting \begin{align*} T_xM=E^s_x\oplus E^u_x \end{align*} for every $x\in\Lambda$, together with constants $A\ge 1$ and $0<\lambda<1$ satisfying \begin{align*} |Df_x^n v_s|_{g_{f^n(x)}}\le A\lambda^n |v_s|_{g_x} \end{align*} for all $v_s\in E^s_x$ and \begin{align*} |Df_x^{-n} v_u|_{g_{f^{-n}(x)}}\le A\lambda^n |v_u|_{g_x} \end{align*} for all $v_u\in E^u_x$. The graph transform proof is cleanest when the one-step estimates already contract by a number strictly below $1$. This is obtained by passing to an adapted Riemannian metric on an open neighbourhood of a compact neighbourhood of $\Lambda$. Choose an open neighbourhood $U\subset M$ of $\Lambda$ with compact closure $N:=\overline U$, let $\tilde g$ denote the adapted metric on an open neighbourhood of $N$, and let $\tilde d:N\times N\to[0,\infty)$ denote the ambient Riemannian distance induced by $\tilde g$, restricted to $N$. Concretely, one chooses $0<\lambda_0<1$ with $\lambda<\lambda_0$ and uses the standard adapted-metric construction for hyperbolic splittings to arrange \begin{align*} |Df_x v_s|_{\tilde g_{f(x)}}\le \lambda_0 |v_s|_{\tilde g_x} \end{align*} and \begin{align*} |Df_x^{-1} v_u|_{\tilde g_{f^{-1}(x)}}\le \lambda_0 |v_u|_{\tilde g_x}. \end{align*} This does not change the substance of the theorem. After shrinking the graph-transform radius so all local disks and the orbit segments used in the local estimates remain in $N$, the comparison of the two metrics is only needed on the compact set $N$. Since $g$ and $\tilde g$ are equivalent on $N$, there is a constant $B\ge 1$ such that \begin{align*} B^{-1}d(p,q)\le \tilde d(p,q)\le B d(p,q) \end{align*} for all $p,q\in N$. Therefore an exponential estimate in $\tilde d$ becomes an exponential estimate in the original distance $d$ after multiplying the front constant by $B^2$.[/guided]

[step:Apply the uniform stable graph transform over $\Lambda$]We now invoke the uniform Hadamard-Perron graph transform theorem for compact hyperbolic sets, stated here as an external result not yet in the wiki: if a $C^r$ diffeomorphism has a compact hyperbolic invariant set with adapted one-step contraction constant $\lambda_0<1$, then there exist a radius $\rho_s>0$, a constant $K_s\ge 1$, and a number $\mu_s$ with $\lambda_0<\mu_s<1$ such that, for each $x\in\Lambda$, there is a $C^r$ embedded disk $W^s_{\rho_s}(x)\subset M$ satisfying \begin{align*} x\in W^s_{\rho_s}(x) \end{align*} and \begin{align*} T_xW^s_{\rho_s}(x)=E^s_x. \end{align*} The same theorem gives local forward invariance: \begin{align*} f(W^s_{\rho_s}(x))\subset W^s_{\rho_s}(f(x)), \end{align*} and the stable contraction estimate in the adapted distance \begin{align*} \tilde d(f^n(y),f^n(x))\le K_s\mu_s^n \tilde d(y,x) \end{align*} for every $y\in W^s_{\rho_s}(x)$ and every integer $n\ge 0$. We set $\varepsilon_s:=\rho_s$ and retain the stable disks $W^s_{\varepsilon_s}(x):=W^s_{\rho_s}(x)$.[/step]

[guided]We apply the uniform Hadamard-Perron graph transform theorem for compact hyperbolic sets. This result is not yet available as a linked theorem in the wiki, so we state exactly the part being used. Its hypotheses are: a $C^r$ diffeomorphism, a compact invariant hyperbolic set, a continuous invariant splitting, and uniform one-step contraction in an adapted metric. These hypotheses have been verified in the previous step: $f$ is a $C^r$ diffeomorphism by assumption, $\Lambda$ is compact and $f$-invariant by the meaning of compact hyperbolic set, and the adapted metric gives the one-step estimates with contraction constant $\lambda_0<1$. The graph transform works in local charts centered at points $x\in\Lambda$, writing nearby points as graphs over $E^s_x$ with values in $E^u_x$. The stable directions contract under $Df$, while the unstable directions expand forward; this imbalance makes the graph transform a contraction on a suitable space of small $C^r$ graphs. Compactness of $\Lambda$ is essential because it gives one chart radius, one graph-size bound, and one contraction rate valid for all base points $x\in\Lambda$. The theorem therefore supplies a radius $\rho_s>0$, a constant $K_s\ge 1$, and a number $\mu_s$ with $\lambda_0<\mu_s<1$ such that each $x\in\Lambda$ has a $C^r$ embedded disk $W^s_{\rho_s}(x)\subset M$ through $x$. The disk is tangent to the stable subspace: \begin{align*} T_xW^s_{\rho_s}(x)=E^s_x. \end{align*} It is also locally forward invariant: \begin{align*} f(W^s_{\rho_s}(x))\subset W^s_{\rho_s}(f(x)). \end{align*} Finally, for every $y\in W^s_{\rho_s}(x)$ and every integer $n\ge 0$, the graph transform contraction estimate gives \begin{align*} \tilde d(f^n(y),f^n(x))\le K_s\mu_s^n \tilde d(y,x). \end{align*} We define $\varepsilon_s:=\rho_s$ and write $W^s_{\varepsilon_s}(x):=W^s_{\rho_s}(x)$.[/guided]

[step:Apply the same construction to $f^{-1}$ to obtain unstable disks]The map \begin{align*} f^{-1}:M\to M \end{align*} is also a $C^r$ diffeomorphism, and $\Lambda$ is compact and $f^{-1}$-invariant. For $f^{-1}$, the stable bundle over $\Lambda$ is precisely $E^u$. Applying the same uniform stable graph transform theorem to $f^{-1}$ gives a radius $\rho_u>0$, a constant $K_u\ge 1$, and a number $\mu_u\in(0,1)$ such that for each $x\in\Lambda$ there is a $C^r$ embedded disk $W^u_{\rho_u}(x)\subset M$ with \begin{align*} T_xW^u_{\rho_u}(x)=E^u_x, \end{align*} with local invariance under $f^{-1}$: \begin{align*} f^{-1}(W^u_{\rho_u}(x))\subset W^u_{\rho_u}(f^{-1}(x)), \end{align*} and with the backward estimate in the adapted distance \begin{align*} \tilde d(f^{-n}(y),f^{-n}(x))\le K_u\mu_u^n \tilde d(y,x) \end{align*} for every $y\in W^u_{\rho_u}(x)$ and every integer $n\ge 0$.[/step]

[guided]We now repeat the stable construction for the inverse map. The map \begin{align*} f^{-1}:M\to M \end{align*} is a $C^r$ diffeomorphism because $f$ is a $C^r$ diffeomorphism, and $\Lambda$ is invariant under $f^{-1}$ because $f(\Lambda)=\Lambda$. For $f^{-1}$, the directions contracted in forward time are the original unstable directions $E^u_x$, since the hyperbolicity hypothesis gives \begin{align*} |Df_x^{-1}v_u|_{\tilde g_{f^{-1}(x)}}\le \lambda_0|v_u|_{\tilde g_x} \end{align*} for every $x\in\Lambda$ and every $v_u\in E^u_x$. Thus the hypotheses of the same uniform Hadamard-Perron graph transform theorem are satisfied for $f^{-1}$ with stable bundle $E^u$. The theorem supplies a radius $\rho_u>0$, a constant $K_u\ge 1$, and a number $\mu_u\in(0,1)$ such that for each $x\in\Lambda$ there is a $C^r$ embedded disk $W^u_{\rho_u}(x)\subset M$ tangent to $E^u_x$: \begin{align*} T_xW^u_{\rho_u}(x)=E^u_x. \end{align*} The local forward invariance for $f^{-1}$ is exactly the local backward invariance for $f$: \begin{align*} f^{-1}(W^u_{\rho_u}(x))\subset W^u_{\rho_u}(f^{-1}(x)). \end{align*} The contraction estimate for the inverse dynamics gives, for every $y\in W^u_{\rho_u}(x)$ and every integer $n\ge 0$, \begin{align*} \tilde d(f^{-n}(y),f^{-n}(x))\le K_u\mu_u^n \tilde d(y,x). \end{align*} This is the desired unstable estimate before transferring from the adapted distance $\tilde d$ back to the original distance $d$.[/guided]

[step:Choose common constants and conclude the theorem] The uniform graph transform theorem may be applied with the common radius \begin{align*} \varepsilon:=\min\{\varepsilon_s,\rho_u\} \end{align*} because its radius conclusion is monotone: shrinking the chart radius and graph domain preserves the graph-transform invariance estimates. We therefore take the disks $W^s_\varepsilon(x)$ and $W^u_\varepsilon(x)$ supplied by the stable construction for $f$ and the stable construction for $f^{-1}$ at this common radius. Their tangency identities and the inclusions \begin{align*} f(W^s_\varepsilon(x))\subset W^s_\varepsilon(f(x)) \end{align*} and \begin{align*} f^{-1}(W^u_\varepsilon(x))\subset W^u_\varepsilon(f^{-1}(x)) \end{align*} are therefore part of the graph-transform conclusion at radius $\varepsilon$, not consequences of arbitrarily cutting down already-chosen disks. Set \begin{align*} \mu:=\max\{\mu_s,\mu_u\} \end{align*} and set \begin{align*} C':=B^2\max\{K_s,K_u,1\}. \end{align*} Then $0<\mu<1$ and $C'\ge 1$. For every $y\in W^s_\varepsilon(x)$ and every integer $n\ge 0$, the adapted-distance estimate and the comparison $B^{-1}d\le \tilde d\le Bd$ give \begin{align*} d(f^n(y),f^n(x))\le B\tilde d(f^n(y),f^n(x))\le B K_s\mu_s^n\tilde d(y,x)\le B^2K_s\mu^n d(y,x)\le C'\mu^n d(y,x). \end{align*} For every $y\in W^u_\varepsilon(x)$ and every integer $n\ge 0$, the same comparison gives \begin{align*} d(f^{-n}(y),f^{-n}(x))\le B\tilde d(f^{-n}(y),f^{-n}(x))\le B K_u\mu_u^n\tilde d(y,x)\le B^2K_u\mu^n d(y,x)\le C'\mu^n d(y,x). \end{align*} The tangent identities are exactly those obtained from the stable graph transform for $f$ and for $f^{-1}$. Hence the disks $W^s_\varepsilon(x)$ and $W^u_\varepsilon(x)$ satisfy all asserted properties. [/step]