[proofplan] Choose an isolating neighbourhood $U$ for $\Lambda$ and shrink it so that its closure is compact. Hyperbolicity gives invariant stable and unstable cone fields on a neighbourhood of $\Lambda$, and these cone estimates persist for every diffeomorphism $g$ sufficiently close to $f$ in the $C^1$ topology; this proves that the maximal invariant set $\Lambda_g$ is again hyperbolic and locally maximal. The conjugacy is built by shadowing: the $f$-orbit of each $x \in \Lambda$ is a small two-sided $g$-pseudo-orbit, so the hyperbolic shadowing lemma gives a unique nearby $g$-orbit. Defining $h_g(x)$ to be the initial point of that orbit gives the semiconjugacy relation, and expansivity together with the symmetric shadowing construction gives bijectivity and continuity. [/proofplan] [step:Choose a compact isolating neighbourhood for the original hyperbolic set] Since $\Lambda$ is locally maximal for $f$, there exists an open neighbourhood $U_0 \subset M$ of $\Lambda$ such that \begin{align*} \Lambda = \bigcap_{n \in \mathbb{Z}} f^n(U_0). \end{align*} Because $\Lambda$ is compact and $M$ is a smooth manifold, choose an open neighbourhood $U \subset U_0$ of $\Lambda$ such that $\overline{U}$ is compact and such that \begin{align*} \Lambda = \bigcap_{n \in \mathbb{Z}} f^n(U). \end{align*} This is possible by shrinking the isolating neighbourhood inside a relatively compact coordinate neighbourhood cover of the compact set $\Lambda$. Define the maximal invariant set of a diffeomorphism $g: M \to M$ in $U$ by \begin{align*} \operatorname{Inv}(U,g) := \{x \in U : g^n(x) \in U \text{ for every } n \in \mathbb{Z}\}. \end{align*} With this notation, $\Lambda = \operatorname{Inv}(U,f)$ and, for each perturbation $g$, $\Lambda_g = \operatorname{Inv}(U,g)$. [/step] [step:Persist the invariant cone fields under small $C^1$ perturbations] Let \begin{align*} T_\Lambda M = E^s \oplus E^u \end{align*} be the hyperbolic splitting for $f$ over $\Lambda$. Thus $E^s$ and $E^u$ are continuous subbundles of $T_\Lambda M$, invariant under $df$, and there exist constants $C \geq 1$ and $\lambda \in (0,1)$ such that, for every $x \in \Lambda$ and every $n \in \mathbb{N}$, \begin{align*} |df_x^n(v)| \leq C\lambda^n |v| \end{align*} for all $v \in E_x^s$, and \begin{align*} |df_x^{-n}(w)| \leq C\lambda^n |w| \end{align*} for all $w \in E_x^u$. Choose $\theta \in (\lambda,1)$. By the standard persistence of invariant cone fields under $C^1$ perturbations, applied to the hyperbolic splitting on the compact set $\Lambda$, there exist an open neighbourhood $W \subset M$ of $\Lambda$, cone fields $C^s$ and $C^u$ on $W$, an integer $N \in \mathbb{N}$, and a $C^1$ neighbourhood $\mathcal{V}_1$ of $f$ in $\operatorname{Diff}^1(M)$ such that, for every $g \in \mathcal{V}_1$, the derivative maps $dg^N$ and $dg^{-N}$ preserve the corresponding unstable and stable cones on every orbit segment staying in $W$, and satisfy the uniform estimates \begin{align*} |dg_x^N(v)| \leq \theta^N |v| \end{align*} for every $v \in C_x^s$, and \begin{align*} |dg_x^{-N}(w)| \leq \theta^N |w| \end{align*} for every $w \in C_x^u$. We shrink $U$ if necessary so that $\overline{U} \subset W$. For every $g \in \mathcal{V}_1$, every full orbit in $\Lambda_g$ stays in $U$, hence in $W$. The cone estimates therefore define a continuous $dg$-invariant splitting \begin{align*} T_{\Lambda_g}M = E_g^s \oplus E_g^u \end{align*} with uniform contraction on $E_g^s$ and uniform contraction for $g^{-1}$ on $E_g^u$. Therefore $\Lambda_g$ is hyperbolic for $g$. The same cone-field theorem also gives an isolating property after shrinking $\mathcal{V}_1$: there exists a $C^1$ neighbourhood $\mathcal{V}_2 \subset \mathcal{V}_1$ of $f$ such that, for every $g \in \mathcal{V}_2$, \begin{align*} \operatorname{Inv}(\overline{U},g) \subset U. \end{align*} Since $\operatorname{Inv}(\overline{U},g)$ is closed in the compact set $\overline{U}$, it is compact. The displayed inclusion implies \begin{align*} \operatorname{Inv}(U,g) = \operatorname{Inv}(\overline{U},g), \end{align*} so $\Lambda_g$ is compact and locally maximal for $g$ with isolating neighbourhood $U$. [/step] [step:Construct the continuation map by shadowing the old orbit with a new orbit] Fix a number $\rho > 0$ smaller than the expansivity radius supplied by the standard expansivity theorem for hyperbolic sets, and also small enough that the closed $\rho$-neighbourhood of $\Lambda$ is contained in $U$. By the two-sided [shadowing lemma for hyperbolic sets](/theorems/7755), there are numbers $\delta > 0$ and a $C^1$ neighbourhood $\mathcal{V}_3 \subset \mathcal{V}_2$ of $f$ such that, for every $g \in \mathcal{V}_3$, every two-sided $\delta$-pseudo-orbit in $U$ is $\rho$-shadowed by a unique full $g$-orbit in $U$. For $x \in \Lambda$, define the sequence \begin{align*} a_x: \mathbb{Z} \to U \end{align*} by \begin{align*} a_x(n) := f^n(x). \end{align*} Since $g$ is $C^0$-close to $f$ on the compact set $\overline{U}$ after shrinking $\mathcal{V}_3$, we have \begin{align*} d(g(a_x(n)),a_x(n+1)) = d(g(f^n(x)),f(f^n(x))) < \delta \end{align*} for every $n \in \mathbb{Z}$. Thus $a_x$ is a two-sided $\delta$-pseudo-orbit for $g$. By shadowing, there exists a unique point $y \in U$ such that \begin{align*} d(g^n(y),f^n(x)) < \rho \end{align*} for every $n \in \mathbb{Z}$. Since the full $g$-orbit of $y$ remains in the $\rho$-neighbourhood of $\Lambda \subset U$, it lies in $U$ for all $n \in \mathbb{Z}$; hence $y \in \Lambda_g$. Define \begin{align*} h_g: \Lambda \to \Lambda_g \end{align*} by assigning to $x$ this unique initial point $y$. [guided] The goal is to attach to each old orbit of $f$ one nearby new orbit of $g$. Fix $x \in \Lambda$ and define the full orbit sequence \begin{align*} a_x: \mathbb{Z} \to U \end{align*} by \begin{align*} a_x(n) := f^n(x). \end{align*} This sequence is an exact orbit for $f$, but we want to regard it as an approximate orbit for $g$. The defect at time $n$ is the distance between applying $g$ to the current point and moving to the next point of the sequence: \begin{align*} d(g(a_x(n)),a_x(n+1)) = d(g(f^n(x)),f(f^n(x))). \end{align*} Because every point $f^n(x)$ lies in the compact set $\Lambda \subset U$, and because $g$ is chosen sufficiently close to $f$ in the $C^1$ topology, it is in particular uniformly close to $f$ on $\overline{U}$. After shrinking the neighbourhood of $f$, this defect is less than the shadowing threshold $\delta$ for every $n \in \mathbb{Z}$. Therefore $a_x$ is a two-sided $\delta$-pseudo-orbit for $g$. Now apply the two-sided shadowing lemma for hyperbolic sets. Its hypotheses are satisfied because the pseudo-orbit lies in the isolating neighbourhood $U$, and because $g$ is inside the perturbation neighbourhood for which the hyperbolic shadowing constants are valid. The conclusion gives a point $y \in U$ such that \begin{align*} d(g^n(y),a_x(n)) < \rho \end{align*} for every $n \in \mathbb{Z}$. Substituting the definition of $a_x(n)$ gives \begin{align*} d(g^n(y),f^n(x)) < \rho \end{align*} for every $n \in \mathbb{Z}$. The choice of $\rho$ ensures that the $\rho$-neighbourhood of $\Lambda$ is contained in $U$. Since $f^n(x) \in \Lambda$ for every $n \in \mathbb{Z}$, the estimate above implies $g^n(y) \in U$ for every $n \in \mathbb{Z}$. Thus $y$ belongs to \begin{align*} \Lambda_g = \bigcap_{n \in \mathbb{Z}} g^n(U). \end{align*} The shadowing lemma also gives uniqueness of this $y$ among full $g$-orbits staying in the prescribed neighbourhood. We may therefore define $h_g(x)$ unambiguously to be this unique point $y$. [/guided] [/step] [step:Derive the conjugacy equation from uniqueness of the shadowing orbit] Let $x \in \Lambda$, and write $y := h_g(x)$. By construction, \begin{align*} d(g^n(y),f^n(x)) < \rho \end{align*} for every $n \in \mathbb{Z}$. The point $g(y)$ shadows the $f$-orbit of $f(x)$, because for every $n \in \mathbb{Z}$, \begin{align*} d(g^n(g(y)),f^n(f(x))) = d(g^{n+1}(y),f^{n+1}(x)) < \rho. \end{align*} By uniqueness in the shadowing lemma, the unique initial point shadowing the orbit of $f(x)$ is $g(y)$. Hence \begin{align*} h_g(f(x)) = g(h_g(x)). \end{align*} Since $x \in \Lambda$ was arbitrary, \begin{align*} h_g \circ f = g \circ h_g. \end{align*} [/step] [step:Use symmetric shadowing to prove bijectivity] After shrinking the neighbourhood $\mathcal{V}_3$ once more, the same shadowing construction applies with the roles of $f$ and $g$ reversed, because $g$ is $C^1$-close to $f$ and $\Lambda_g$ is a compact locally maximal hyperbolic set for $g$ in the same isolating neighbourhood $U$. Thus for each $z \in \Lambda_g$ there is a unique point $k_g(z) \in \Lambda$ whose $f$-orbit $\rho$-shadows the $g$-orbit of $z$. This defines a map \begin{align*} k_g: \Lambda_g \to \Lambda. \end{align*} Let $x \in \Lambda$ and set $y := h_g(x)$. The $g$-orbit of $y$ $\rho$-shadows the $f$-orbit of $x$, and therefore the $f$-orbit of $x$ $\rho$-shadows the $g$-orbit of $y$. By uniqueness in the reverse shadowing construction, \begin{align*} k_g(h_g(x)) = x. \end{align*} Hence $k_g \circ h_g = \operatorname{id}_\Lambda$. Similarly, let $z \in \Lambda_g$ and set $x := k_g(z)$. The $f$-orbit of $x$ $\rho$-shadows the $g$-orbit of $z$, so by uniqueness in the original shadowing construction, \begin{align*} h_g(k_g(z)) = z. \end{align*} Thus $h_g \circ k_g = \operatorname{id}_{\Lambda_g}$. Therefore $h_g$ is bijective with inverse $k_g$. [/step] [step:Prove continuity of the continuation map and finish the conjugacy] Let $(x_j)_{j \in \mathbb{N}}$ be a sequence in $\Lambda$ converging to $x \in \Lambda$, and define \begin{align*} y_j := h_g(x_j) \end{align*} for each $j \in \mathbb{N}$. Since $\Lambda_g$ is compact, every subsequence of $(y_j)_{j \in \mathbb{N}}$ has a further subsequence converging to some point $y_* \in \Lambda_g$. Fix such a convergent subsequence, still denoted $(y_j)_{j \in \mathbb{N}}$, with $y_j \to y_*$. For each fixed $n \in \mathbb{Z}$, continuity of $f^n$ and $g^n$ gives \begin{align*} g^n(y_j) \to g^n(y_*) \end{align*} and \begin{align*} f^n(x_j) \to f^n(x). \end{align*} Since $y_j = h_g(x_j)$, we have \begin{align*} d(g^n(y_j),f^n(x_j)) < \rho \end{align*} for every $j \in \mathbb{N}$ and every $n \in \mathbb{Z}$. Passing to the limit for each fixed $n$ yields \begin{align*} d(g^n(y_*),f^n(x)) \leq \rho. \end{align*} Choosing the shadowing radius strictly below the expansivity threshold and shrinking it at the start if necessary, uniqueness of the shadowing orbit gives \begin{align*} y_* = h_g(x). \end{align*} Every convergent subsequence of $(y_j)$ therefore has limit $h_g(x)$, and compactness of $\Lambda_g$ implies $y_j \to h_g(x)$. Thus $h_g$ is continuous. The same argument applied to the inverse map $k_g: \Lambda_g \to \Lambda$ proves that $k_g$ is continuous. Hence $h_g$ is a homeomorphism. The identity \begin{align*} h_g \circ f = g \circ h_g \end{align*} proved above shows that $h_g$ conjugates $f|_\Lambda$ to $g|_{\Lambda_g}$. This completes the proof. [/step]