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$\mathbf{Definition:}$ 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

Badshah Khan
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  • 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 '20 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... – Badshah Khan Jan 16 '20 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... – Badshah Khan Jan 16 '20 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... – Badshah Khan Jan 16 '20 at 16:18

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