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Consider dynamical systems with dicrete time $f:X\to X$. Given a function $\tau:X\to\mathbb{R}^+$ and consider the set $$Z=\{(x,t)\in X\times \mathbb{R}; 0\leq t\leq \tau(x)\}.$$ Then the set $Y=Z/\sim$, where $\sim$ is equivalence relation define as $$(x,t)\sim (y,s) \iff y=f(x), t=\tau(x), s=0.$$

I am quite confused why it is equivalence relation and how the equivalence classes look like.

Any help will be appreciated. Thanks.

Waney
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  • Try looking at the simple example of $X={0,1}$ and $f(x) = 1-x$, with $\tau(x) =1$. What does the space $Z$ look like? How about $Y$? – Dan Rust Sep 27 '21 at 13:01
  • $Z={(0,t), (1,t); 0\leq t\leq 1}$. For example point $(0,1)$ is in relation with $(1-0,0)=(1,0)$ and $(1,1)$ is in relation with $(1-1,0)=(0,0)$. But what about the point for example $(0,\frac{1}{2})$ which obviously is from $Z$? Moreover the each point from $Z$ should be in relation with itself. Am I correct, or have been I doing some mistakes? Thanks. – Waney Sep 27 '21 at 13:17
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    Ah, there's a bit of confusion here between an equivalence relation and then a 'generator' of an equivalence relation. You should thing that the relation given is in addition to the relations $(x,t)\sim (x,t)$ for all $x,t$. That is, for all other values, they are not glued to anything else except themselves. You should also extend by reflexivity, so if $(x,t)\sim(y,s)$ and $(x',t')\sim (y,s)$, then $(x,t)\sim (x',t')$. See here for more on generating equivalence relations. – Dan Rust Sep 27 '21 at 13:34
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    So the way you should think about $Z$ is that you're extending $X$ by an interval at every point, and then when you take the quotient to get $Y$, you're gluing the end points together according to how $f$ maps them. So if you follow a path along these glued intervals, you're just following an orbit of $f$ in a `smooth' way, rather than discrete. In the example I gave, you should be able to see that you get a single circle with 'circumference' $2$. If $f$ were instead the identity map, then you'd get two disjoint circles each with circumference $1$. – Dan Rust Sep 27 '21 at 13:40

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