[proofplan] We use the [stable manifold theorem](/theorems/2778) for hyperbolic sets to obtain local stable and unstable plaques that vary continuously in the $C^1$ topology and remain uniformly transverse over the compact set $\Lambda$. In adapted local coordinates, an unstable plaque through $x$ and a stable plaque through $y$ are graphs over complementary hyperbolic subspaces, so their intersection is reduced to solving a uniformly transverse finite-dimensional equation. A uniform [implicit function theorem](/theorems/52) gives existence and uniqueness of the nearby intersection whenever $x$ and $y$ are close enough. If $\Lambda$ is locally maximal, an isolating neighbourhood contains a small tubular neighbourhood of $\Lambda$, and the defining stable and unstable estimates keep every forward and backward iterate of the bracket point inside that isolating neighbourhood. [/proofplan] [step:Choose uniform local stable and unstable plaques over the hyperbolic set] Because $\Lambda$ is a compact hyperbolic set for the $C^1$ diffeomorphism $f: M \to M$, the stable manifold theorem for hyperbolic sets applies (citing a result not yet in the wiki: Stable Manifold Theorem for Hyperbolic Sets). Thus there exists $\varepsilon_1 > 0$ such that for every $p \in \Lambda$ and every $\varepsilon \in (0,\varepsilon_1]$, the local stable plaque $W^s_\varepsilon(p)$ and the local unstable plaque $W^u_\varepsilon(p)$ are embedded $C^1$ disks satisfying \begin{align*} T_p W^s_\varepsilon(p) = E^s_p \end{align*} and \begin{align*} T_p W^u_\varepsilon(p) = E^u_p, \end{align*} where $T_pM = E^s_p \oplus E^u_p$ is the hyperbolic splitting over $\Lambda$. The same theorem gives continuous variation of the plaques in the $C^1$ topology with respect to the base point $p \in \Lambda$. Since $\Lambda$ is compact and the splitting $T_\Lambda M = E^s \oplus E^u$ is continuous with $E^s_p \cap E^u_p = \{0\}$ for every $p \in \Lambda$, the angle between $E^s_p$ and $E^u_p$ has a positive lower bound on $\Lambda$. After decreasing $\varepsilon_1$ if necessary, all stable plaques and unstable plaques of size at most $\varepsilon_1$ are uniformly transverse whenever their base points are sufficiently close. [guided] The only substantial input at this stage is the stable manifold theorem for hyperbolic sets (citing a result not yet in the wiki: Stable Manifold Theorem for Hyperbolic Sets). Its hypotheses are exactly the hypotheses now in force: $f: M \to M$ is a $C^1$ diffeomorphism, $\Lambda$ is compact, $f(\Lambda)=\Lambda$, and $\Lambda$ is hyperbolic. Therefore, for sufficiently small plaque size $\varepsilon_1 > 0$, each $p \in \Lambda$ has a local stable plaque $W^s_\varepsilon(p)$ and a local unstable plaque $W^u_\varepsilon(p)$ for every $\varepsilon \in (0,\varepsilon_1]$. These plaques are embedded $C^1$ disks, and at their base point their tangent spaces are the two subspaces in the hyperbolic splitting: \begin{align*} T_p W^s_\varepsilon(p) = E^s_p \end{align*} and \begin{align*} T_p W^u_\varepsilon(p) = E^u_p. \end{align*} The splitting is direct, so $E^s_p \cap E^u_p = \{0\}$ and $T_pM = E^s_p \oplus E^u_p$. The compactness of $\Lambda$ matters here because it upgrades pointwise transversality into uniform transversality: the angle between $E^s_p$ and $E^u_p$ is a positive [continuous function](/page/Continuous%20Function) of $p \in \Lambda$, hence has a positive minimum on $\Lambda$. Finally, the stable manifold theorem gives continuous $C^1$ variation of the plaques with the base point. Thus, if $x,y \in \Lambda$ are close, then the tangent spaces of $W^u_\varepsilon(x)$ and $W^s_\varepsilon(y)$ remain close to $E^u_x$ and $E^s_x$, respectively. Since $E^u_x$ and $E^s_x$ are uniformly transverse, the nearby plaques remain uniformly transverse after decreasing $\varepsilon_1$ and the allowed distance between $x$ and $y$. [/guided] [/step] [step:Represent nearby plaques as transverse graphs in one coordinate chart] Fix $\varepsilon \in (0,\varepsilon_1]$. For each $p \in \Lambda$, choose a coordinate neighbourhood $U_p \subset M$ and a $C^1$ chart \begin{align*} \psi_p: U_p \to V_p \subset E^u_p \oplus E^s_p \end{align*} with $\psi_p(p)=0$ and with derivative $d(\psi_p)_p: T_pM \to E^u_p \oplus E^s_p$ equal to the splitting identification. Since $\Lambda$ is compact, finitely many such neighbourhoods cover $\Lambda$. By the graph form of the local stable manifold theorem, compactness of $\Lambda$ allows the following uniform choice. After decreasing $\varepsilon_1$, choose for each fixed $\varepsilon \in (0,\varepsilon_1]$ a number $r=r(\varepsilon)>0$ and then choose $\delta_1=\delta_1(\varepsilon)>0$ such that whenever $x,y \in \Lambda$ and $d(x,y)<\delta_1$, both full plaques $W^u_\varepsilon(x)$ and $W^s_\varepsilon(y)$ lie in one adapted chart centred at some $p \in \Lambda$. In that chart their entire coordinate images are $C^1$ graphs over closed coordinate balls \begin{align*} \overline{B}^u_r := \{a \in E^u_p : |a| \le r\} \end{align*} and \begin{align*} \overline{B}^s_r := \{b \in E^s_p : |b| \le r\}. \end{align*} More precisely, \begin{align*} \Gamma^u_x = \{a + \varphi^u_x(a) : a \in \overline{B}^u_r\} \end{align*} and \begin{align*} \Gamma^s_y = \{\varphi^s_y(b) + b : b \in \overline{B}^s_r\}, \end{align*} where \begin{align*} \varphi^u_x: \overline{B}^u_r \to \overline{B}^s_r \end{align*} and \begin{align*} \varphi^s_y: \overline{B}^s_r \to \overline{B}^u_r \end{align*} are $C^1$ maps. The $C^1$ variation of the plaques and the uniform transversality of $E^u_p$ and $E^s_p$ allow the chart size and $\delta_1$ to be chosen so that both graph maps have Lipschitz constant at most $1/4$ on their domains. Because the displayed graphs represent the entire local plaques of size $\varepsilon$, any intersection of $W^u_\varepsilon(x)$ and $W^s_\varepsilon(y)$ must occur inside this product of closed coordinate balls. [/step] [step:Solve the graph intersection equation by a contraction argument] In the common chart, an intersection point of $W^u_\varepsilon(x)$ and $W^s_\varepsilon(y)$ is equivalent to a pair $(a,b) \in \overline{B}^u_r \times \overline{B}^s_r$ satisfying \begin{align*} a + \varphi^u_x(a) = \varphi^s_y(b) + b. \end{align*} Because $E^u_p \oplus E^s_p$ is a direct sum, this equality is equivalent to the coupled equations \begin{align*} a = \varphi^s_y(b) \end{align*} and \begin{align*} b = \varphi^u_x(a). \end{align*} Define the map \begin{align*} T_{x,y}: \overline{B}^u_r \times \overline{B}^s_r \to \overline{B}^u_r \times \overline{B}^s_r \end{align*} by \begin{align*} T_{x,y}(a,b) = (\varphi^s_y(b),\varphi^u_x(a)). \end{align*} The domain is a [complete metric space](/page/Complete%20Metric%20Space) with the product metric induced by the Euclidean norms on $E^u_p$ and $E^s_p$. Since both graph maps have Lipschitz constant at most $1/4$, the map $T_{x,y}$ is a contraction with Lipschitz constant at most $1/4$. By the [contraction mapping theorem](/theorems/71), $T_{x,y}$ has a unique fixed point $(a_*,b_*) \in \overline{B}^u_r \times \overline{B}^s_r$. The fixed point identities give $a_* = \varphi^s_y(b_*)$ and $b_* = \varphi^u_x(a_*)$, hence \begin{align*} a_* + \varphi^u_x(a_*) = \varphi^s_y(b_*) + b_*. \end{align*} Thus the corresponding point of the common graph is an intersection point. Pulling it back through the chart gives \begin{align*} [x,y] \in W^u_\varepsilon(x) \cap W^s_\varepsilon(y). \end{align*} Conversely, every intersection of the full plaques is represented by some pair in $\overline{B}^u_r \times \overline{B}^s_r$ because the previous step represented the entire plaques in this chart. Such a pair is a fixed point of $T_{x,y}$, and the fixed point is unique. Therefore $W^u_\varepsilon(x) \cap W^s_\varepsilon(y)=\{[x,y]\}$. [guided] We now turn the geometric intersection problem into a fixed point problem on a precisely defined product box. In the adapted chart, the whole unstable plaque through $x$ is the graph \begin{align*} \Gamma^u_x = \{a + \varphi^u_x(a) : a \in \overline{B}^u_r\}, \end{align*} where $\varphi^u_x: \overline{B}^u_r \to \overline{B}^s_r$ is a $C^1$ map with Lipschitz constant at most $1/4$. The whole stable plaque through $y$ is the graph \begin{align*} \Gamma^s_y = \{\varphi^s_y(b) + b : b \in \overline{B}^s_r\}, \end{align*} where $\varphi^s_y: \overline{B}^s_r \to \overline{B}^u_r$ is a $C^1$ map with Lipschitz constant at most $1/4$. A point belongs to both graphs exactly when there are vectors $a \in \overline{B}^u_r$ and $b \in \overline{B}^s_r$ such that \begin{align*} a + \varphi^u_x(a) = \varphi^s_y(b) + b. \end{align*} The decomposition $E^u_p \oplus E^s_p$ is direct, so the $E^u_p$ and $E^s_p$ components must agree separately. Therefore the intersection equation is equivalent to \begin{align*} a = \varphi^s_y(b) \end{align*} and \begin{align*} b = \varphi^u_x(a). \end{align*} This is why a fixed point argument is the correct tool: define \begin{align*} T_{x,y}: \overline{B}^u_r \times \overline{B}^s_r \to \overline{B}^u_r \times \overline{B}^s_r \end{align*} by \begin{align*} T_{x,y}(a,b) = (\varphi^s_y(b),\varphi^u_x(a)). \end{align*} The codomain is the same product box because $\varphi^s_y$ takes values in $\overline{B}^u_r$ and $\varphi^u_x$ takes values in $\overline{B}^s_r$. We verify the hypotheses of the contraction mapping theorem. The closed balls $\overline{B}^u_r \subset E^u_p$ and $\overline{B}^s_r \subset E^s_p$ are complete because they are closed subsets of finite-dimensional normed vector spaces. Hence their product is complete for the product metric. For two pairs $(a_1,b_1)$ and $(a_2,b_2)$ in the product box, the Lipschitz bounds on the graph maps give \begin{align*} |\varphi^s_y(b_1)-\varphi^s_y(b_2)| \leq \frac{1}{4}|b_1-b_2| \end{align*} and \begin{align*} |\varphi^u_x(a_1)-\varphi^u_x(a_2)| \leq \frac{1}{4}|a_1-a_2|. \end{align*} Thus $T_{x,y}$ is a contraction, with contraction constant at most $1/4$ for the product metric. The contraction mapping theorem gives a unique fixed point $(a_*,b_*)$ in the whole product box. The fixed point condition means \begin{align*} a_* = \varphi^s_y(b_*) \end{align*} and \begin{align*} b_* = \varphi^u_x(a_*). \end{align*} Substituting these identities gives \begin{align*} a_* + \varphi^u_x(a_*) = \varphi^s_y(b_*) + b_*. \end{align*} Therefore the point represented by this vector in the common chart lies on both plaques. Pulling it back through the chart defines the bracket point \begin{align*} [x,y] \in W^u_\varepsilon(x) \cap W^s_\varepsilon(y). \end{align*} Finally, uniqueness is global for the local plaques, not merely local inside a smaller box: the previous step represented the entire plaques $W^u_\varepsilon(x)$ and $W^s_\varepsilon(y)$ as the two displayed graphs over $\overline{B}^u_r$ and $\overline{B}^s_r$. Hence any other plaque intersection would give another fixed point of $T_{x,y}$ in the same complete product box. Since the fixed point is unique, no second intersection exists. [/guided] [/step] [step:Use local maximality to keep the bracket orbit inside an isolating neighbourhood] Assume now that $\Lambda$ is locally maximal. By definition, there exists an open neighbourhood $U \subset M$ of $\Lambda$ such that \begin{align*} \Lambda = \bigcap_{n \in \mathbb{Z}} f^n(U). \end{align*} Since $\Lambda$ is compact and $U$ is open, choose $\rho>0$ such that \begin{align*} \{q \in M : d(q,\Lambda)<\rho\} \subset U. \end{align*} Decrease $\varepsilon_1$ so that $\varepsilon_1<\rho$. Let $\varepsilon \in (0,\varepsilon_1]$, choose $\delta=\delta(\varepsilon)$ as above, and let $x,y \in \Lambda$ satisfy $d(x,y)<\delta$. Put $z=[x,y]$. Since $z \in W^s_\varepsilon(y)$, the defining property of the local stable plaque gives \begin{align*} d(f^n(z),f^n(y))<\varepsilon \end{align*} for every integer $n \ge 0$. Since $y \in \Lambda$ and $\Lambda$ is $f$-invariant, $f^n(y)\in\Lambda$ for every $n\ge 0$, hence $d(f^n(z),\Lambda)<\rho$ and $f^n(z)\in U$ for every $n\ge 0$. Similarly, since $z \in W^u_\varepsilon(x)$, the defining property of the local unstable plaque gives \begin{align*} d(f^{-n}(z),f^{-n}(x))<\varepsilon \end{align*} for every integer $n \ge 0$. Since $x \in \Lambda$ and $\Lambda$ is $f$-invariant, $f^{-n}(x)\in\Lambda$ for every $n\ge 0$, hence $f^{-n}(z)\in U$ for every $n\ge 0$. Combining the forward and backward inclusions gives $f^k(z)\in U$ for every $k\in\mathbb{Z}$. Therefore \begin{align*} z \in \bigcap_{k \in \mathbb{Z}} f^k(U) = \Lambda. \end{align*} Since $z=[x,y]$, this proves $[x,y]\in\Lambda$ in the locally maximal case. [/step]