Show that $f^n(\gamma([0,1]))\cap B_\varepsilon(x)\neq \emptyset$ for some $n>>0$.

Question

Let $f: \mathbb{T}^2\to\mathbb{T}^2$ the arnold's cat map and $\gamma:[0,1]\to \mathbb{T}^2$ a $C^1$- injective curve. Show that $f^n(\gamma([0,1]))\cap B_\varepsilon(x)\neq \emptyset$ for some $n>>0$. ( $B_\varepsilon(x)$ is the ball centered in $x\in \mathbb{T}^2$ of radius $\varepsilon>0$)

My attempt: I have already shown that all points have a stable and unstable manifold, parallel to the eigenspaces associated with the eigenvalues of the matrix $2111$ and since the eigenspaces have an irrational slope, each of these stable and unstable manifolds are dense on torus. That is, every point on the curve intersects a dense subset.

Answer 1

Modulo the obstruction I mentioned in the comment above, this follows from an inclination lemma type argument. Here is an outline (one can generalize this argument quite a bit; I'll just mention the qualitative part):

Claim: Let $f:\mathbb{T}^2\to\mathbb{T}^2$ be the automorphism defined by the matrix $ \begin{pmatrix} 2 & 1 \\ 1 & 1 \end{pmatrix} $, $\gamma:[0,1]\to \mathbb{T}^2$ be a $C^1$ embedding with image $C=\gamma([0,1])$ that is transverse to the stable foliation of $f$. Then for any $x\in \mathbb{T}^2$, and for any $\epsilon\in\mathbb{R}_{>0}$, there is an $n\in\mathbb{Z}_{>0}$ such that $f^n(C)\cap B(x,\epsilon)\neq\emptyset$.

Sketch of Proof: $f$ is a hyperbolic automorphism and as such its periodic points are dense (in this case they are exactly the rational points $\mathbb{Q}^2/\mathbb{Z}^2\leq \mathbb{R}^2/\mathbb{Z}^2=\mathbb{T}^2$; see e.g. Periodic points of hyperbolic toral automorphism are dense on the torus). For any $y\in \mathbb{T}^2$ and $r\in\mathbb{R}_{>0}$, denote by $V_{y,r}(f)=\mathcal{S}_{y,r}(f)\times \mathcal{U}_{y,r}(f)$ the product of the local stable and unstable manifolds of $f$ at $y$ of size $r$; for $r$ small enough $V_{y,r}(f)$ is a coordinate chart for $\mathbb{T}^2$ (this is the so-called local product structure of $f$). In particular for some $r$ small enough there is a periodic point $p$ of $f$ such that $C\cap V_{p,r}(f)\neq\emptyset$.

Next as you said the global stable and unstable manifolds of $f$ are dense on $\mathbb{T}^2$; in particular $\mathcal{U}_{p}(f)\cap B(x,\epsilon)\neq\emptyset$. $\mathcal{U}_p(f)$ is one dimensional, so we may choose an orientation on it and parameterize it; let $D$ be the embedded curve ($\dagger$) that corresponds to the part of $\mathcal{U}_p(f)$ from $p$ to some point in the intersection $\mathcal{U}_{p}(f)\cap B(x,\epsilon)\neq\emptyset$.

[($\dagger$) In this case the curve will be $C^\infty$ and more generally it'll be as regular as the Anosov diffeomorphism $f$.]

Then by some version of the inclination lemma (that one can write down quite explicitly in this case as the dynamics is algebraic) the sequence of embeddings $n\mapsto f^n(C)$ will converge to $D$ with respect to the $C^1$ compact-open topology on embeddings; consequently $f^n(C)\cap B(x,\epsilon)\neq\emptyset$.

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Here is a related humble interactive graph based on Exercise 5.46 on p.210 of Robinson's Dynamical Systems: Stability, Symbolic Dynamics, and Chaos: https://www.desmos.com/calculator/u4hbqnu3ma .

Let us finally remark that similarly $C$ being transverse to the unstable foliation of $f$ would result in a negative hitting time $-n$.