[proofplan] Work in local coordinates around $p$ in which the local stable and unstable manifolds are straightened to coordinate axes. First follow a sufficiently small transverse subarc through $x$ until it enters the local chart near $p$ as an admissible graph. Then apply the local graph transform: the height and slope of the iterated graphs tend to zero while the unstable-coordinate domains expand. The expanding-domain property ensures that, for all sufficiently large iterates, one can restrict to a subarc whose projection covers the prescribed compact unstable subarc $J$ and whose $C^1$ distance from $J$ is as small as required. [/proofplan]

[step:Represent the transverse arc as a local graph]Let $(U,\varphi)$ be a chart around $p$ with $\varphi(p)=0$ and write $\varphi(U)\subset \mathbb R^2$ with coordinates $(u,v)$. By the [Stable Manifold Theorem](/theorems/2778) for the hyperbolic fixed point $p$, after shrinking $U$ we may assume that $\varphi(W^s_{\mathrm{loc}}(p))=\{u=0\}$ and $\varphi(W^u_{\mathrm{loc}}(p))=\{v=0\}$. Since $\gamma$ meets $W^s_{\mathrm{loc}}(p)$ transversely at $x$, after shrinking $\gamma$ to a smaller subarc through $x$ its image under $\varphi$ is the graph of a $C^r$ function over an interval in the $u$-axis.[/step]

[guided]The transversality hypothesis is used exactly here: it lets us solve locally for one coordinate as a function of the other. In the chosen chart, the stable manifold is the vertical axis and the unstable manifold is the horizontal axis, so we have \begin{align*} \varphi(p) = 0, \end{align*} \begin{align*} \varphi(W^s_{\mathrm{loc}}(p)) = \{(u,v) \in \mathbb R^2 : u = 0\}, \end{align*} \begin{align*} \varphi(W^u_{\mathrm{loc}}(p)) = \{(u,v) \in \mathbb R^2 : v = 0\}. \end{align*} Since $\gamma$ meets $W^s_{\mathrm{loc}}(p)$ transversely at $x$, the tangent line $T_x\gamma$ is not vertical in these coordinates. After shrinking $\gamma$ to a smaller subarc through $x$, there is an interval $I \subset \mathbb R$ and a $C^r$ function $\psi: I \to \mathbb R$ such that \begin{align*} \varphi(\gamma) = \{(u,\psi(u)) : u \in I\}. \end{align*} Equivalently, the chosen subarc of $\gamma$ is a graph over the unstable coordinate $u$, and this is the local input that turns the dynamical statement into a statement about the evolution of graphs under iteration.[/guided]

[step:Enter the local chart and apply the expanding graph transform]Define the local coordinate representation of $f$ by \begin{align*} F: \varphi(U) \to \mathbb R^2, \qquad F = \varphi \circ f \circ \varphi^{-1}. \end{align*} Because $x \in W^s_{\mathrm{loc}}(p)$, the forward orbit $f^m(x)$ converges to $p$ inside the chosen local stable manifold. Hence, after replacing the transverse subarc of $\gamma$ through $x$ by a smaller subarc and choosing $m_0 \in \mathbb N$, the arc $f^{m_0}(\gamma)$ enters $U$ as an admissible $C^1$ graph over the unstable coordinate. The graph-transform hypothesis in the formalized statement applies to this graph: for all $k \geq 0$, the arc $f^{m_0+k}(\gamma)$ contains a graph over an interval $I_k$ in the unstable coordinate, its height and slope tend to $0$, and the intervals $I_k$ eventually contain every compact subinterval corresponding to a compact subarc of $W^u_{\mathrm{loc}}(p)$.[/step]

[guided]Why is the graph transform not applied immediately at $x$? The point $x$ lies on the local stable manifold, but it need not equal $p$, and the original arc need not initially be an admissible graph in the chart where the stable and unstable manifolds have been straightened. Since $x \in W^s_{\mathrm{loc}}(p)$, its forward orbit satisfies $f^m(x) \to p$. Therefore a sufficiently small subarc of $\gamma$ through $x$ can be chosen so that, for some $m_0 \in \mathbb N$, its image under $f^{m_0}$ lies in $U$ and is an admissible $C^1$ graph transverse to the stable cone. In the chart $(U,\varphi)$ define \begin{align*} F: \varphi(U) \to \mathbb R^2, \end{align*} with \begin{align*} F = \varphi \circ f \circ \varphi^{-1}. \end{align*} The graph-transform hypothesis from the statement is now applicable. If the entered graph is written as \begin{align*} \Gamma_0 = \{(u,\psi_0(u)) : u \in I_0\}, \end{align*} then for each $k \geq 0$ there is a subarc of $F^k(\Gamma_0)$ of the form \begin{align*} \Gamma_k = \{(u,\psi_k(u)) : u \in I_k\}, \end{align*} where $I_k$ is an interval in the unstable coordinate. The graph transform gives two pieces of information, both of which are needed for the theorem: the height and slope of $\psi_k$ tend to $0$ in the $C^1$ topology, and the intervals $I_k$ expand so that they eventually contain every fixed compact subinterval of the local unstable axis. The first assertion gives $C^1$ closeness to the unstable manifold; the second assertion ensures that the relevant graph actually covers the prescribed compact unstable subarc rather than merely lying close to the unstable axis somewhere else.[/guided]

[step:Choose an iterate whose subarc lies in the prescribed neighborhood] Let $\mathcal U$ be a $C^1$ neighborhood of the compact subarc $J \subset W^u_{\mathrm{loc}}(p)$. By definition of the $C^1$ topology on embedded arcs, there exists $\varepsilon > 0$ such that any embedded $C^1$ arc whose image lies in a sufficiently small tubular neighborhood of $J$ and whose tangent lines are $\varepsilon$-close in angle to those of $J$ belongs to $\mathcal U$. The previous step shows more than closeness to the unstable axis: after the entry time $m_0$, the graph domains eventually contain the unstable-coordinate interval corresponding to $J$, while the graph functions and their first derivatives tend to $0$. Therefore, for all sufficiently large $n$, we may restrict the graph contained in $f^n(\gamma)$ to the coordinate interval over $J$; the resulting embedded subarc has image in the required tubular neighborhood of $J$ and tangent lines $\varepsilon$-close to those of $J$. Hence there exists $N \in \mathbb N$ such that for every $n \ge N$, the curve $f^n(\gamma)$ contains a subarc $\gamma_n$ with $\gamma_n \in \mathcal U$. This proves the lambda lemma. [/step]