# Topological Conjugacy of Arnalds Cat map

$$\mathbf{Definition:}$$ Let $$X,Y$$ be two metric spaces and $$f:X\to X$$ and $$g:Y \to Y$$ be two mappings, $$f$$ and $$g$$ are said to be Topollogically Conjugated (denoted by $$f\sim g$$) if there exist $$h:X\to Y$$ homeomorphism s.t $$h\circ f = g\circ h$$

$$\mathbf{Question:}$$ I am given with two mappings, $$f$$ and $$g$$, both from $$\Bbb T^2 \to \Bbb T^2$$, s.t $$f\begin{pmatrix}x\\y\\\end{pmatrix} = \begin{pmatrix}2&1\\1&1\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix} + \begin{pmatrix}a\\b\\\end{pmatrix}$$ Where $$a,b\in \Bbb R$$. And the second is "Arnold's cat map" given as $$g\begin{pmatrix}x\\y\\\end{pmatrix} = \begin{pmatrix}2&1\\1&1\\\end{pmatrix}\begin{pmatrix}x\\y\\\end{pmatrix}$$ And I have asked to prove $$f\sim g$$. I tried a lot, but I didn't succeeded. Please if any one help me regarding finding that homeomorphism '$$h$$'. Thanks

• You should not expect something so immediate. Note however that there are dense orbits. So it suffices to take one and extend the map by continuity, which gives easily a semiconjugacy. Then you prove that it is invertible. – John B Jan 16 at 16:00
• Yes Sir... You are right. I do the same thing for many other proofs of Conjugacy. But it didn't work here... – Mohammad Shoaib Jan 16 at 16:15
• Like in many other questions, i find 'h' for just positive end, and than take the negation with possible way to make it homeo.... But here, i even not able to find atleast one way map... – Mohammad Shoaib Jan 16 at 16:17
• Actually there is some properties of this homeo map 'h', that is, fixed point of h is fixed point of g, and periodic point of h is also the periodic point of g... – Mohammad Shoaib Jan 16 at 16:18