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Let $M$ be $S^1 \times [-1,1]$, $f$ a baker map on $M$ and for $p, q \in M$ consider $W^s_p$ the stable manifold in $p$ (i.e. the set of points whose forward orbit tend to the forward orbit of $p$) and $W^u_q$ the unstable manifold in $q$ (i.e. the set of points whose backward orbit tends to backwards orbit of $q$). Let us assume, we have transversality. Question:

Let $x\in W^u_p$ and $y\in W^u_q$. Show that there exists a point $t$ so that $t\in W^u_x \cap W^u_y$, if $x$ and $y$ be small close to each other. I want to say that if two point be close to each other their unstable manifold will intersect each other at a point.

 

Under which assumption we can prove it the intersection point is unique?

When we draw baker map we easily see they intersect a point but i do not know how can i prove it?

Thanks in advance.

Let $M$ be $S^1 \times [-1,1]$, $f$ a baker map on $M$ and for $p, q \in M$ consider $W^s_p$ the stable manifold in $p$ (i.e. the set of points whose forward orbit tend to the forward orbit of $p$) and $W^u_q$ the unstable manifold in $q$ (i.e. the set of points whose backward orbit tends to backwards orbit of $q$). Let us assume, we have transversality. Question:

Let $x\in W^u_p$ and $y\in W^u_q$. Show that there exists a point $t$ so that $t\in W^u_x \cap W^u_y$, if $x$ and $y$ be small close to each other. I want to say that if two point be close to each other their unstable manifold will intersect each other at a point.

Under which assumption we can prove it the intersection point is unique?

When we draw baker map we easily see they intersect a point but i do not know how can i prove it?

Thanks in advance.

Let $M$ be $S^1 \times [-1,1]$, $f$ a Baker map on $M$ and for $p, q \in M$ consider $W^s_p$ the stable manifold in $p$ (i.e. the set of points whose forward orbit tend to the forward orbit of $p$) and $W^u_q$ the unstable manifold in $q$ (i.e. the set of points whose backward orbit tends to backwards orbit of $q$). Let us assume, we have transversality. Question:

Let $x\in W^u_p$ and $y\in W^u_q$. Show that there exists a point $t$ so that $t\in W^u_x \cap W^u_y$, if $x$ and $y$ be small close to each other. I want to say that if two point be close to each other their unstable manifold will intersect each other at a point.

Under which assumption we can prove it the intersection point is unique?

When we draw Baker map we easily see they intersect a point but i do not know how can i prove it?

Thanks in advance.

Let $M$ be $S^1 \times [-1,1]$, $f$ a baker map on $M$ and for $p, q \in M$ consider $W^s_p$ the stable manifold in $p$ (i.e. the set of points whose forward orbit tend to the forward orbit of $p$) and $W^u_q$ the unstable manifold in $q$ (i.e. the set of points whose backward orbit tends to backwards orbit of $q$). Let us assume, we have transversality. Question:

Let $x\in W^u_p$ and $y\in W^u_q$. Show that there exists a point $t$ so that $t\in W^u_x \cap W^u_y$, if $x$ and $y$ be small close to each other. I want to say that if two point be close to each other their unstable manifold will intersect each other at a point.

Under which assumption we can prove it the intersection point is unique?

When we draw baker map we easily see they intersect a point but i do not know how can i prove it?

Thanks in advance.

Let $M$ be a compact smooth manifold, $f$ a Baker map on $M$ and for $p, q \in M$ consider $W^s_p$ the stable manifold in $p$ (i.e. the set of points whose forward orbit tend to the forward orbit of $p$) and $W^u_q$ the unstable manifold in $q$ (i.e. the set of points whose backward orbit tends to backwards orbit of $q$). Let us assume, we have transversality. Question:

Let $x\in W^u_p$ and $y\in W^u_q$. Show that there exists a point $t$ so that $t\in W^u_x \cap W^u_y$, if $x$ and $y$ be small close to each other. I want to say that if two point be close to each other their unstable manifold will intersect each other at a point.

Under which assumption we can prove it the intersection point is unique?

When we draw Baker map we easily see they intersect a point but i do not know how can i prove it?

Thanks in advance.

Let $M$ be $S^1 \times [-1,1]$, $f$ a Baker map on $M$ and for $p, q \in M$ consider $W^s_p$ the stable manifold in $p$ (i.e. the set of points whose forward orbit tend to the forward orbit of $p$) and $W^u_q$ the unstable manifold in $q$ (i.e. the set of points whose backward orbit tends to backwards orbit of $q$). Let us assume, we have transversality. Question:

Let $x\in W^u_p$ and $y\in W^u_q$. Show that there exists a point $t$ so that $t\in W^u_x \cap W^u_y$, if $x$ and $y$ be small close to each other. I want to say that if two point be close to each other their unstable manifold will intersect each other at a point.

Under which assumption we can prove it the intersection point is unique?

When we draw Baker map we easily see they intersect a point but i do not know how can i prove it?

Thanks in advance.