Mixing
Orientation preserving homeomorphism on a torus
$begingroup$
Let $Τ^2=S^1times S^1$ be the standard torus, $f:T^2 to T^2$ an orientation preserving homeomorphism that is the identity on a meridian $m$ and $A$ an $ε$-neighbourhood of $m$.
My question is, if we cut $f(T^2)$ along $f(m)$ do points in $A$ on the left of $m$ are still mapped on the left of $f(m)$?
I believe they do but I can't rigorously see why. If so, can someone please explain why?
EDIT: I have given it some thought and came up with something, I would like to check if it's correct.
Let's say, without loss of generality that a cylindrical neighbourhood of $m$ is mapped homeomorphically to another cylindrical neighbouhood of $m$. For connectivity reasons the left region will either be mapped homeomorphically on the left or right region of the image. Let's say it's mapped on the right region, then the right side will be mapped on the left side. Let's take a circle inside our neighbourhood crossing $m$ on points $a$ and $b$ and choose points $p$ and $q$ on our circle, $p$ on the left region and $q$ on the right region. If $overrightarrow {pq}$ is the arc containing $a$ then it will be mapped to the arc $overleftarrow {f(q)f(p)}$ that contains $a=f(a)$. But that would contradict $f$ being orientation preserving.
proof-verification geometric-topology
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
$endgroup$
add a comment |
$begingroup$
Let $Τ^2=S^1times S^1$ be the standard torus, $f:T^2 to T^2$ an orientation preserving homeomorphism that is the identity on a meridian $m$ and $A$ an $ε$-neighbourhood of $m$.
My question is, if we cut $f(T^2)$ along $f(m)$ do points in $A$ on the left of $m$ are still mapped on the left of $f(m)$?
I believe they do but I can't rigorously see why. If so, can someone please explain why?
EDIT: I have given it some thought and came up with something, I would like to check if it's correct.
Let's say, without loss of generality that a cylindrical neighbourhood of $m$ is mapped homeomorphically to another cylindrical neighbouhood of $m$. For connectivity reasons the left region will either be mapped homeomorphically on the left or right region of the image. Let's say it's mapped on the right region, then the right side will be mapped on the left side. Let's take a circle inside our neighbourhood crossing $m$ on points $a$ and $b$ and choose points $p$ and $q$ on our circle, $p$ on the left region and $q$ on the right region. If $overrightarrow {pq}$ is the arc containing $a$ then it will be mapped to the arc $overleftarrow {f(q)f(p)}$ that contains $a=f(a)$. But that would contradict $f$ being orientation preserving.
proof-verification geometric-topology
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
$endgroup$
$begingroup$
How much algebraic topology do you know? Do you know what $H_2(T^2, T^2-p)$ means?
$endgroup$
– Moishe Cohen
Jan 10 at 23:57
$begingroup$
Yes, that's the relative homology group. I am also aware of the definition of orientation preserving map in regards to the relative homology groups, i.e. $f_*:H_2(T^2,T^2-p)to H_2(T^2,T^2-f(p))$ should be the identity
$endgroup$
– Amontillado
Jan 11 at 3:17
add a comment |
$begingroup$
Let $Τ^2=S^1times S^1$ be the standard torus, $f:T^2 to T^2$ an orientation preserving homeomorphism that is the identity on a meridian $m$ and $A$ an $ε$-neighbourhood of $m$.
My question is, if we cut $f(T^2)$ along $f(m)$ do points in $A$ on the left of $m$ are still mapped on the left of $f(m)$?
I believe they do but I can't rigorously see why. If so, can someone please explain why?
EDIT: I have given it some thought and came up with something, I would like to check if it's correct.
Let's say, without loss of generality that a cylindrical neighbourhood of $m$ is mapped homeomorphically to another cylindrical neighbouhood of $m$. For connectivity reasons the left region will either be mapped homeomorphically on the left or right region of the image. Let's say it's mapped on the right region, then the right side will be mapped on the left side. Let's take a circle inside our neighbourhood crossing $m$ on points $a$ and $b$ and choose points $p$ and $q$ on our circle, $p$ on the left region and $q$ on the right region. If $overrightarrow {pq}$ is the arc containing $a$ then it will be mapped to the arc $overleftarrow {f(q)f(p)}$ that contains $a=f(a)$. But that would contradict $f$ being orientation preserving.
proof-verification geometric-topology
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
$endgroup$
Let $Τ^2=S^1times S^1$ be the standard torus, $f:T^2 to T^2$ an orientation preserving homeomorphism that is the identity on a meridian $m$ and $A$ an $ε$-neighbourhood of $m$.
My question is, if we cut $f(T^2)$ along $f(m)$ do points in $A$ on the left of $m$ are still mapped on the left of $f(m)$?
I believe they do but I can't rigorously see why. If so, can someone please explain why?
EDIT: I have given it some thought and came up with something, I would like to check if it's correct.
Let's say, without loss of generality that a cylindrical neighbourhood of $m$ is mapped homeomorphically to another cylindrical neighbouhood of $m$. For connectivity reasons the left region will either be mapped homeomorphically on the left or right region of the image. Let's say it's mapped on the right region, then the right side will be mapped on the left side. Let's take a circle inside our neighbourhood crossing $m$ on points $a$ and $b$ and choose points $p$ and $q$ on our circle, $p$ on the left region and $q$ on the right region. If $overrightarrow {pq}$ is the arc containing $a$ then it will be mapped to the arc $overleftarrow {f(q)f(p)}$ that contains $a=f(a)$. But that would contradict $f$ being orientation preserving.
proof-verification geometric-topology
proof-verification geometric-topology
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
edited Jan 13 at 15:27
Amontillado
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
asked Jan 10 at 2:35
AmontilladoAmontillado
449313
449313
$begingroup$
How much algebraic topology do you know? Do you know what $H_2(T^2, T^2-p)$ means?
$endgroup$
– Moishe Cohen
Jan 10 at 23:57
$begingroup$
Yes, that's the relative homology group. I am also aware of the definition of orientation preserving map in regards to the relative homology groups, i.e. $f_*:H_2(T^2,T^2-p)to H_2(T^2,T^2-f(p))$ should be the identity
$endgroup$
– Amontillado
Jan 11 at 3:17
add a comment |
$begingroup$
How much algebraic topology do you know? Do you know what $H_2(T^2, T^2-p)$ means?
$endgroup$
– Moishe Cohen
Jan 10 at 23:57
$begingroup$
Yes, that's the relative homology group. I am also aware of the definition of orientation preserving map in regards to the relative homology groups, i.e. $f_*:H_2(T^2,T^2-p)to H_2(T^2,T^2-f(p))$ should be the identity
$endgroup$
– Amontillado
Jan 11 at 3:17
$begingroup$
How much algebraic topology do you know? Do you know what $H_2(T^2, T^2-p)$ means?
$endgroup$
– Moishe Cohen
Jan 10 at 23:57
$begingroup$
How much algebraic topology do you know? Do you know what $H_2(T^2, T^2-p)$ means?
$endgroup$
– Moishe Cohen
Jan 10 at 23:57
$begingroup$
Yes, that's the relative homology group. I am also aware of the definition of orientation preserving map in regards to the relative homology groups, i.e. $f_*:H_2(T^2,T^2-p)to H_2(T^2,T^2-f(p))$ should be the identity
$endgroup$
– Amontillado
Jan 11 at 3:17
$begingroup$
Yes, that's the relative homology group. I am also aware of the definition of orientation preserving map in regards to the relative homology groups, i.e. $f_*:H_2(T^2,T^2-p)to H_2(T^2,T^2-f(p))$ should be the identity
$endgroup$
– Amontillado
Jan 11 at 3:17
add a comment |
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$begingroup$
How much algebraic topology do you know? Do you know what $H_2(T^2, T^2-p)$ means?
$endgroup$
– Moishe Cohen
Jan 10 at 23:57
$begingroup$
Yes, that's the relative homology group. I am also aware of the definition of orientation preserving map in regards to the relative homology groups, i.e. $f_*:H_2(T^2,T^2-p)to H_2(T^2,T^2-f(p))$ should be the identity
$endgroup$
– Amontillado
Jan 11 at 3:17