Mixing
Retract of noncompact surface to its boundary?
$begingroup$
Suppose $M$ is a connected, noncompact 2-manifold, and its boundary $partial M$ is a circle. What's the simplest way to show there is a retraction $r: Mrightarrow partial M$?
Here are some examples of such surfaces:
- Probably the easiest example is the closed unit disk, minus the origin.
- A more complicated example is the closed unit disk minus a Cantor set.
- A totally different example is to join two infinite-genus tori, and attach a cylinder to that:
surfaces manifolds-with-boundary retraction
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
$endgroup$
|
show 2 more comments
$begingroup$
Suppose $M$ is a connected, noncompact 2-manifold, and its boundary $partial M$ is a circle. What's the simplest way to show there is a retraction $r: Mrightarrow partial M$?
Here are some examples of such surfaces:
- Probably the easiest example is the closed unit disk, minus the origin.
- A more complicated example is the closed unit disk minus a Cantor set.
- A totally different example is to join two infinite-genus tori, and attach a cylinder to that:
surfaces manifolds-with-boundary retraction
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
$endgroup$
$begingroup$
What is your definition of retraction? I'm familiar with the definition which requires the target be a subset of $M$. This does not happen if $M$ is, as a very simple example, the open unit disk in $mathbb{R}^2$.
$endgroup$
– Eric Towers
Jan 22 at 4:35
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@EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:46
$begingroup$
@LordSharktheUnknown : I agree, which is why I ask for clarification. Perhaps a minimal example of a connected, noncompact 2-manifold with boundary having a single boundary component would be illuminating.
$endgroup$
– Eric Towers
Jan 22 at 4:52
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@EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:54
$begingroup$
@EricTowers: yes, you can take the disk minus some points (even something like a Cantor set). By retract I mean a map that is the identity when restricted the the subspace. That is, a left inverse of the inclusion map $partial Mrightarrow M$.
$endgroup$
– Hempelicious
Jan 22 at 5:07
|
show 2 more comments
$begingroup$
Suppose $M$ is a connected, noncompact 2-manifold, and its boundary $partial M$ is a circle. What's the simplest way to show there is a retraction $r: Mrightarrow partial M$?
Here are some examples of such surfaces:
- Probably the easiest example is the closed unit disk, minus the origin.
- A more complicated example is the closed unit disk minus a Cantor set.
- A totally different example is to join two infinite-genus tori, and attach a cylinder to that:
surfaces manifolds-with-boundary retraction
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
$endgroup$
Suppose $M$ is a connected, noncompact 2-manifold, and its boundary $partial M$ is a circle. What's the simplest way to show there is a retraction $r: Mrightarrow partial M$?
Here are some examples of such surfaces:
- Probably the easiest example is the closed unit disk, minus the origin.
- A more complicated example is the closed unit disk minus a Cantor set.
- A totally different example is to join two infinite-genus tori, and attach a cylinder to that:
surfaces manifolds-with-boundary retraction
surfaces manifolds-with-boundary retraction
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
edited Jan 22 at 17:42
Hempelicious
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
asked Jan 22 at 4:17
HempeliciousHempelicious
147111
147111
$begingroup$
What is your definition of retraction? I'm familiar with the definition which requires the target be a subset of $M$. This does not happen if $M$ is, as a very simple example, the open unit disk in $mathbb{R}^2$.
$endgroup$
– Eric Towers
Jan 22 at 4:35
$begingroup$
@EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:46
$begingroup$
@LordSharktheUnknown : I agree, which is why I ask for clarification. Perhaps a minimal example of a connected, noncompact 2-manifold with boundary having a single boundary component would be illuminating.
$endgroup$
– Eric Towers
Jan 22 at 4:52
$begingroup$
@EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:54
$begingroup$
@EricTowers: yes, you can take the disk minus some points (even something like a Cantor set). By retract I mean a map that is the identity when restricted the the subspace. That is, a left inverse of the inclusion map $partial Mrightarrow M$.
$endgroup$
– Hempelicious
Jan 22 at 5:07
|
show 2 more comments
$begingroup$
What is your definition of retraction? I'm familiar with the definition which requires the target be a subset of $M$. This does not happen if $M$ is, as a very simple example, the open unit disk in $mathbb{R}^2$.
$endgroup$
– Eric Towers
Jan 22 at 4:35
$begingroup$
@EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:46
$begingroup$
@LordSharktheUnknown : I agree, which is why I ask for clarification. Perhaps a minimal example of a connected, noncompact 2-manifold with boundary having a single boundary component would be illuminating.
$endgroup$
– Eric Towers
Jan 22 at 4:52
$begingroup$
@EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:54
$begingroup$
@EricTowers: yes, you can take the disk minus some points (even something like a Cantor set). By retract I mean a map that is the identity when restricted the the subspace. That is, a left inverse of the inclusion map $partial Mrightarrow M$.
$endgroup$
– Hempelicious
Jan 22 at 5:07
$begingroup$
What is your definition of retraction? I'm familiar with the definition which requires the target be a subset of $M$. This does not happen if $M$ is, as a very simple example, the open unit disk in $mathbb{R}^2$.
$endgroup$
– Eric Towers
Jan 22 at 4:35
$begingroup$
What is your definition of retraction? I'm familiar with the definition which requires the target be a subset of $M$. This does not happen if $M$ is, as a very simple example, the open unit disk in $mathbb{R}^2$.
$endgroup$
– Eric Towers
Jan 22 at 4:35
$begingroup$
@EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:46
$begingroup$
@EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:46
$begingroup$
@LordSharktheUnknown : I agree, which is why I ask for clarification. Perhaps a minimal example of a connected, noncompact 2-manifold with boundary having a single boundary component would be illuminating.
$endgroup$
– Eric Towers
Jan 22 at 4:52
$begingroup$
@LordSharktheUnknown : I agree, which is why I ask for clarification. Perhaps a minimal example of a connected, noncompact 2-manifold with boundary having a single boundary component would be illuminating.
$endgroup$
– Eric Towers
Jan 22 at 4:52
$begingroup$
@EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:54
$begingroup$
@EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 4:54
$begingroup$
@EricTowers: yes, you can take the disk minus some points (even something like a Cantor set). By retract I mean a map that is the identity when restricted the the subspace. That is, a left inverse of the inclusion map $partial Mrightarrow M$.
$endgroup$
– Hempelicious
Jan 22 at 5:07
$begingroup$
@EricTowers: yes, you can take the disk minus some points (even something like a Cantor set). By retract I mean a map that is the identity when restricted the the subspace. That is, a left inverse of the inclusion map $partial Mrightarrow M$.
$endgroup$
– Hempelicious
Jan 22 at 5:07
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
"Simple" depends on what you've studied. Under one valuation ...
Let $M$ be a connected, noncompact, 2-manifold with boundary having circle boundary, $partial M cong S^1$. Show there is a retraction $r: M rightarrow partial M$.
Let $E$ be the set of ideal points in the Freudenthal end point compactification of $M$ and let $M''$ be the Freudenthal compactification of $M$. Let $e in E$ and $M' = M'' smallsetminus {e}$, the Freudenthal end point compactification of $M$, ignoring the end corresponding to $e$. A quick reminder of these ideas is here. For more on this, see Raymond, Frank, The End Point Compactification of Manifolds, including details of the construction when $E$ is not countable. $M'$ is a connected, noncompact, 2-manifold with circle boundary $partial M = partial M'$, and one end. We will abuse notation and call this end $e$.
Under a nested compact exhaustion of $M'$, there is an annular neighborhood of the end $e$. (It is an open disk neighborhood of $e$ in the Freudenthal compactification.) (See O. Ya. Viro et al, Elementary Topology: Textbook in Problems, Ch. XI, 48${}^circ$2x for more.) Let $A$ be a core $S^1$ of this annulus and $p_1$ be a point of $A$. Let $B = partial M' cong S^1$ and $p_2$ be a point of $B$. There is a simple path, $p$, in $M'$ connecting $p_1$ and $p_2$ (which we would use to change the basepoint of the fundamental group from one to the other). $B$ has an open collar homotopic to an annulus closed on the $B$ boundary component and open on the other. $p$ has an open bicollar homeomorphic to $p times (0,1)$. Let $S$ be the union of these three open collars. ($S$ is a regular neighborhood of a $1$-skeleton of the boundary, the end $e$, and a path between the two in $M'$.)
Observe that $M' smallsetminus S$ is a connected, noncompact 2-manifold with circle boundary. We construct $r$ in two steps. Let $r_1$ be the continuous map holding $A$, $B$, and $p$ fixed and contracting $M' smallsetminus S$ to a point. Let $D$ be the open unit disk and $r_1(M')+D$ be $D$ injectively identified to the boundary of $r_1(M')$. The nullhomotopic homotopy classes were parallel to $A$ and $B$, and were rendered nullhomotopic by transport across $D$, so $r_1(M')+D$ is a simply connected, noncompact 2-manifold without boundary with one end. By Viro et al., 53.Ax, a simply connected non-compact manifold of dimension two without boundary is homeomorphic to $mathbb{R}^2$. Therefore, $r_1(M') +D cong mathbb{R}^2$. Constructing the retract $r_2:mathbb{R}^2 smallsetminus{(0,0)]} rightarrow B$ is a standard exercise. Then $left. r_2 right|_{r_1(M')}$ is a domain restriction of a continuous function, so is continuous, and $r = left. r_2 right|_{r_1(M)} circ left. r_1 right|_{M}$ is the desired map.
In short: find a very simple neighborhood of one end, the boundary, and a path between the two. Crush everything else to a point, yielding a half-infinite cylinder. Then telescope the cylinder onto the boundary.
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
$endgroup$
$begingroup$
How do you know $r_1$ is continuous?
$endgroup$
– Hempelicious
Jan 22 at 20:44
$begingroup$
@Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' smallsetminus S$, which is $M' smallsetminus S$ union an open regular neighborhood of $partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point.
$endgroup$
– Eric Towers
Jan 23 at 7:18
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I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'setminus S$. What does $r_1$ look like in this case?
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– Hempelicious
Jan 23 at 20:26
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@Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk.
$endgroup$
– Eric Towers
Jan 23 at 20:48
$begingroup$
Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me!
$endgroup$
– Hempelicious
Jan 23 at 21:23
|
show 1 more comment
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Here is the approach I know, which is used in Hubbard's book Teichmuller Theory vol. 1.
One thing I don't like about this proof is you either need some smooth structure on $M$, or you need to verify the "supposes" in the next paragraph.
Suppose we have an embedding $rho:mathbb{H}rightarrow M$, where $mathbb{H}=[0,infty)$. Let $xinpartial M$ and further suppose this embedding is such that
begin{equation*}
rho(mathbb{H})cappartial M = rho(0)=x
end{equation*}
Finally, suppose that there's a neighborhood of $rho(mathbb{H})$ that looks like $mathbb{H}times (-1,1)$.
Then we can cut $M$ along $rho(mathbb{H})$ to get a new manifold $N$, where $partial Ncongmathbb{R}$. We can think of this boundary as divided into three pieces: left, middle, and right. Here left and right come from the two sides of $rho(mathbb{H})$, and middle is $partial M$ cut at $x$. Note that middle is homemorphic to $[0,1]$.
We define a map $f:partial Nrightarrow [0,1]$ on each piece:
begin{equation*}
f(z) = begin{cases}
0 & zintextit{left}\
z & zintextit{middle}\
1 & zintextit{right}\
end{cases}
end{equation*}
By the Tietze extension theorem, this extends to a map $F: Nrightarrow [0,1]$. If we compose that with the quotient map $pi:[0,1]rightarrow S^1$, we get a map $pi F:Nrightarrow S^1$. Because this map agrees on left and right, it descends to a map
begin{equation*}
widetilde{pi F}:Mrightarrow S^1
end{equation*}
which is the identity on $partial M$. That means it is our desired retraction.
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
$endgroup$
$begingroup$
This $rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x in partial M$ exists. My argument replaces putting $rho(infty)$ on the end with putting $rho(text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $rho$ has an open tubular neighborhood.
$endgroup$
– Eric Towers
Jan 24 at 3:08
$begingroup$
@EricTowers: yes, $rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful.
$endgroup$
– Hempelicious
Jan 24 at 4:09
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2 Answers
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2 Answers
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$begingroup$
"Simple" depends on what you've studied. Under one valuation ...
Let $M$ be a connected, noncompact, 2-manifold with boundary having circle boundary, $partial M cong S^1$. Show there is a retraction $r: M rightarrow partial M$.
Let $E$ be the set of ideal points in the Freudenthal end point compactification of $M$ and let $M''$ be the Freudenthal compactification of $M$. Let $e in E$ and $M' = M'' smallsetminus {e}$, the Freudenthal end point compactification of $M$, ignoring the end corresponding to $e$. A quick reminder of these ideas is here. For more on this, see Raymond, Frank, The End Point Compactification of Manifolds, including details of the construction when $E$ is not countable. $M'$ is a connected, noncompact, 2-manifold with circle boundary $partial M = partial M'$, and one end. We will abuse notation and call this end $e$.
Under a nested compact exhaustion of $M'$, there is an annular neighborhood of the end $e$. (It is an open disk neighborhood of $e$ in the Freudenthal compactification.) (See O. Ya. Viro et al, Elementary Topology: Textbook in Problems, Ch. XI, 48${}^circ$2x for more.) Let $A$ be a core $S^1$ of this annulus and $p_1$ be a point of $A$. Let $B = partial M' cong S^1$ and $p_2$ be a point of $B$. There is a simple path, $p$, in $M'$ connecting $p_1$ and $p_2$ (which we would use to change the basepoint of the fundamental group from one to the other). $B$ has an open collar homotopic to an annulus closed on the $B$ boundary component and open on the other. $p$ has an open bicollar homeomorphic to $p times (0,1)$. Let $S$ be the union of these three open collars. ($S$ is a regular neighborhood of a $1$-skeleton of the boundary, the end $e$, and a path between the two in $M'$.)
Observe that $M' smallsetminus S$ is a connected, noncompact 2-manifold with circle boundary. We construct $r$ in two steps. Let $r_1$ be the continuous map holding $A$, $B$, and $p$ fixed and contracting $M' smallsetminus S$ to a point. Let $D$ be the open unit disk and $r_1(M')+D$ be $D$ injectively identified to the boundary of $r_1(M')$. The nullhomotopic homotopy classes were parallel to $A$ and $B$, and were rendered nullhomotopic by transport across $D$, so $r_1(M')+D$ is a simply connected, noncompact 2-manifold without boundary with one end. By Viro et al., 53.Ax, a simply connected non-compact manifold of dimension two without boundary is homeomorphic to $mathbb{R}^2$. Therefore, $r_1(M') +D cong mathbb{R}^2$. Constructing the retract $r_2:mathbb{R}^2 smallsetminus{(0,0)]} rightarrow B$ is a standard exercise. Then $left. r_2 right|_{r_1(M')}$ is a domain restriction of a continuous function, so is continuous, and $r = left. r_2 right|_{r_1(M)} circ left. r_1 right|_{M}$ is the desired map.
In short: find a very simple neighborhood of one end, the boundary, and a path between the two. Crush everything else to a point, yielding a half-infinite cylinder. Then telescope the cylinder onto the boundary.
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
$endgroup$
$begingroup$
How do you know $r_1$ is continuous?
$endgroup$
– Hempelicious
Jan 22 at 20:44
$begingroup$
@Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' smallsetminus S$, which is $M' smallsetminus S$ union an open regular neighborhood of $partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point.
$endgroup$
– Eric Towers
Jan 23 at 7:18
$begingroup$
I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'setminus S$. What does $r_1$ look like in this case?
$endgroup$
– Hempelicious
Jan 23 at 20:26
$begingroup$
@Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk.
$endgroup$
– Eric Towers
Jan 23 at 20:48
$begingroup$
Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me!
$endgroup$
– Hempelicious
Jan 23 at 21:23
|
show 1 more comment
$begingroup$
"Simple" depends on what you've studied. Under one valuation ...
Let $M$ be a connected, noncompact, 2-manifold with boundary having circle boundary, $partial M cong S^1$. Show there is a retraction $r: M rightarrow partial M$.
Let $E$ be the set of ideal points in the Freudenthal end point compactification of $M$ and let $M''$ be the Freudenthal compactification of $M$. Let $e in E$ and $M' = M'' smallsetminus {e}$, the Freudenthal end point compactification of $M$, ignoring the end corresponding to $e$. A quick reminder of these ideas is here. For more on this, see Raymond, Frank, The End Point Compactification of Manifolds, including details of the construction when $E$ is not countable. $M'$ is a connected, noncompact, 2-manifold with circle boundary $partial M = partial M'$, and one end. We will abuse notation and call this end $e$.
Under a nested compact exhaustion of $M'$, there is an annular neighborhood of the end $e$. (It is an open disk neighborhood of $e$ in the Freudenthal compactification.) (See O. Ya. Viro et al, Elementary Topology: Textbook in Problems, Ch. XI, 48${}^circ$2x for more.) Let $A$ be a core $S^1$ of this annulus and $p_1$ be a point of $A$. Let $B = partial M' cong S^1$ and $p_2$ be a point of $B$. There is a simple path, $p$, in $M'$ connecting $p_1$ and $p_2$ (which we would use to change the basepoint of the fundamental group from one to the other). $B$ has an open collar homotopic to an annulus closed on the $B$ boundary component and open on the other. $p$ has an open bicollar homeomorphic to $p times (0,1)$. Let $S$ be the union of these three open collars. ($S$ is a regular neighborhood of a $1$-skeleton of the boundary, the end $e$, and a path between the two in $M'$.)
Observe that $M' smallsetminus S$ is a connected, noncompact 2-manifold with circle boundary. We construct $r$ in two steps. Let $r_1$ be the continuous map holding $A$, $B$, and $p$ fixed and contracting $M' smallsetminus S$ to a point. Let $D$ be the open unit disk and $r_1(M')+D$ be $D$ injectively identified to the boundary of $r_1(M')$. The nullhomotopic homotopy classes were parallel to $A$ and $B$, and were rendered nullhomotopic by transport across $D$, so $r_1(M')+D$ is a simply connected, noncompact 2-manifold without boundary with one end. By Viro et al., 53.Ax, a simply connected non-compact manifold of dimension two without boundary is homeomorphic to $mathbb{R}^2$. Therefore, $r_1(M') +D cong mathbb{R}^2$. Constructing the retract $r_2:mathbb{R}^2 smallsetminus{(0,0)]} rightarrow B$ is a standard exercise. Then $left. r_2 right|_{r_1(M')}$ is a domain restriction of a continuous function, so is continuous, and $r = left. r_2 right|_{r_1(M)} circ left. r_1 right|_{M}$ is the desired map.
In short: find a very simple neighborhood of one end, the boundary, and a path between the two. Crush everything else to a point, yielding a half-infinite cylinder. Then telescope the cylinder onto the boundary.
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
$endgroup$
$begingroup$
How do you know $r_1$ is continuous?
$endgroup$
– Hempelicious
Jan 22 at 20:44
$begingroup$
@Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' smallsetminus S$, which is $M' smallsetminus S$ union an open regular neighborhood of $partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point.
$endgroup$
– Eric Towers
Jan 23 at 7:18
$begingroup$
I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'setminus S$. What does $r_1$ look like in this case?
$endgroup$
– Hempelicious
Jan 23 at 20:26
$begingroup$
@Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk.
$endgroup$
– Eric Towers
Jan 23 at 20:48
$begingroup$
Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me!
$endgroup$
– Hempelicious
Jan 23 at 21:23
|
show 1 more comment
$begingroup$
"Simple" depends on what you've studied. Under one valuation ...
Let $M$ be a connected, noncompact, 2-manifold with boundary having circle boundary, $partial M cong S^1$. Show there is a retraction $r: M rightarrow partial M$.
Let $E$ be the set of ideal points in the Freudenthal end point compactification of $M$ and let $M''$ be the Freudenthal compactification of $M$. Let $e in E$ and $M' = M'' smallsetminus {e}$, the Freudenthal end point compactification of $M$, ignoring the end corresponding to $e$. A quick reminder of these ideas is here. For more on this, see Raymond, Frank, The End Point Compactification of Manifolds, including details of the construction when $E$ is not countable. $M'$ is a connected, noncompact, 2-manifold with circle boundary $partial M = partial M'$, and one end. We will abuse notation and call this end $e$.
Under a nested compact exhaustion of $M'$, there is an annular neighborhood of the end $e$. (It is an open disk neighborhood of $e$ in the Freudenthal compactification.) (See O. Ya. Viro et al, Elementary Topology: Textbook in Problems, Ch. XI, 48${}^circ$2x for more.) Let $A$ be a core $S^1$ of this annulus and $p_1$ be a point of $A$. Let $B = partial M' cong S^1$ and $p_2$ be a point of $B$. There is a simple path, $p$, in $M'$ connecting $p_1$ and $p_2$ (which we would use to change the basepoint of the fundamental group from one to the other). $B$ has an open collar homotopic to an annulus closed on the $B$ boundary component and open on the other. $p$ has an open bicollar homeomorphic to $p times (0,1)$. Let $S$ be the union of these three open collars. ($S$ is a regular neighborhood of a $1$-skeleton of the boundary, the end $e$, and a path between the two in $M'$.)
Observe that $M' smallsetminus S$ is a connected, noncompact 2-manifold with circle boundary. We construct $r$ in two steps. Let $r_1$ be the continuous map holding $A$, $B$, and $p$ fixed and contracting $M' smallsetminus S$ to a point. Let $D$ be the open unit disk and $r_1(M')+D$ be $D$ injectively identified to the boundary of $r_1(M')$. The nullhomotopic homotopy classes were parallel to $A$ and $B$, and were rendered nullhomotopic by transport across $D$, so $r_1(M')+D$ is a simply connected, noncompact 2-manifold without boundary with one end. By Viro et al., 53.Ax, a simply connected non-compact manifold of dimension two without boundary is homeomorphic to $mathbb{R}^2$. Therefore, $r_1(M') +D cong mathbb{R}^2$. Constructing the retract $r_2:mathbb{R}^2 smallsetminus{(0,0)]} rightarrow B$ is a standard exercise. Then $left. r_2 right|_{r_1(M')}$ is a domain restriction of a continuous function, so is continuous, and $r = left. r_2 right|_{r_1(M)} circ left. r_1 right|_{M}$ is the desired map.
In short: find a very simple neighborhood of one end, the boundary, and a path between the two. Crush everything else to a point, yielding a half-infinite cylinder. Then telescope the cylinder onto the boundary.
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
$endgroup$
"Simple" depends on what you've studied. Under one valuation ...
Let $M$ be a connected, noncompact, 2-manifold with boundary having circle boundary, $partial M cong S^1$. Show there is a retraction $r: M rightarrow partial M$.
Let $E$ be the set of ideal points in the Freudenthal end point compactification of $M$ and let $M''$ be the Freudenthal compactification of $M$. Let $e in E$ and $M' = M'' smallsetminus {e}$, the Freudenthal end point compactification of $M$, ignoring the end corresponding to $e$. A quick reminder of these ideas is here. For more on this, see Raymond, Frank, The End Point Compactification of Manifolds, including details of the construction when $E$ is not countable. $M'$ is a connected, noncompact, 2-manifold with circle boundary $partial M = partial M'$, and one end. We will abuse notation and call this end $e$.
Under a nested compact exhaustion of $M'$, there is an annular neighborhood of the end $e$. (It is an open disk neighborhood of $e$ in the Freudenthal compactification.) (See O. Ya. Viro et al, Elementary Topology: Textbook in Problems, Ch. XI, 48${}^circ$2x for more.) Let $A$ be a core $S^1$ of this annulus and $p_1$ be a point of $A$. Let $B = partial M' cong S^1$ and $p_2$ be a point of $B$. There is a simple path, $p$, in $M'$ connecting $p_1$ and $p_2$ (which we would use to change the basepoint of the fundamental group from one to the other). $B$ has an open collar homotopic to an annulus closed on the $B$ boundary component and open on the other. $p$ has an open bicollar homeomorphic to $p times (0,1)$. Let $S$ be the union of these three open collars. ($S$ is a regular neighborhood of a $1$-skeleton of the boundary, the end $e$, and a path between the two in $M'$.)
Observe that $M' smallsetminus S$ is a connected, noncompact 2-manifold with circle boundary. We construct $r$ in two steps. Let $r_1$ be the continuous map holding $A$, $B$, and $p$ fixed and contracting $M' smallsetminus S$ to a point. Let $D$ be the open unit disk and $r_1(M')+D$ be $D$ injectively identified to the boundary of $r_1(M')$. The nullhomotopic homotopy classes were parallel to $A$ and $B$, and were rendered nullhomotopic by transport across $D$, so $r_1(M')+D$ is a simply connected, noncompact 2-manifold without boundary with one end. By Viro et al., 53.Ax, a simply connected non-compact manifold of dimension two without boundary is homeomorphic to $mathbb{R}^2$. Therefore, $r_1(M') +D cong mathbb{R}^2$. Constructing the retract $r_2:mathbb{R}^2 smallsetminus{(0,0)]} rightarrow B$ is a standard exercise. Then $left. r_2 right|_{r_1(M')}$ is a domain restriction of a continuous function, so is continuous, and $r = left. r_2 right|_{r_1(M)} circ left. r_1 right|_{M}$ is the desired map.
In short: find a very simple neighborhood of one end, the boundary, and a path between the two. Crush everything else to a point, yielding a half-infinite cylinder. Then telescope the cylinder onto the boundary.
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
edited Jan 24 at 2:39
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
answered Jan 22 at 20:29
Eric TowersEric Towers
32.8k22370
32.8k22370
$begingroup$
How do you know $r_1$ is continuous?
$endgroup$
– Hempelicious
Jan 22 at 20:44
$begingroup$
@Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' smallsetminus S$, which is $M' smallsetminus S$ union an open regular neighborhood of $partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point.
$endgroup$
– Eric Towers
Jan 23 at 7:18
$begingroup$
I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'setminus S$. What does $r_1$ look like in this case?
$endgroup$
– Hempelicious
Jan 23 at 20:26
$begingroup$
@Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk.
$endgroup$
– Eric Towers
Jan 23 at 20:48
$begingroup$
Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me!
$endgroup$
– Hempelicious
Jan 23 at 21:23
|
show 1 more comment
$begingroup$
How do you know $r_1$ is continuous?
$endgroup$
– Hempelicious
Jan 22 at 20:44
$begingroup$
@Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' smallsetminus S$, which is $M' smallsetminus S$ union an open regular neighborhood of $partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point.
$endgroup$
– Eric Towers
Jan 23 at 7:18
$begingroup$
I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'setminus S$. What does $r_1$ look like in this case?
$endgroup$
– Hempelicious
Jan 23 at 20:26
$begingroup$
@Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk.
$endgroup$
– Eric Towers
Jan 23 at 20:48
$begingroup$
Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me!
$endgroup$
– Hempelicious
Jan 23 at 21:23
$begingroup$
How do you know $r_1$ is continuous?
$endgroup$
– Hempelicious
Jan 22 at 20:44
$begingroup$
How do you know $r_1$ is continuous?
$endgroup$
– Hempelicious
Jan 22 at 20:44
$begingroup$
@Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' smallsetminus S$, which is $M' smallsetminus S$ union an open regular neighborhood of $partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point.
$endgroup$
– Eric Towers
Jan 23 at 7:18
$begingroup$
@Hempelicious : Any constant map is continuous, so $r_1$ is continuous on $M' smallsetminus S$. Any (small) open neighborhood of that point pulls back to a regular neighborhood of $M' smallsetminus S$, which is $M' smallsetminus S$ union an open regular neighborhood of $partial S$, which is open. Any open set excluding the point pulls back to an open set. It is perhaps easiest to think of $r_1$ telescoping $p$ until only one point of $p$ is outside the union of the collars of $A$ and $B$, then stretching the $(0,1)$ bicollar of that point around to meet at the point.
$endgroup$
– Eric Towers
Jan 23 at 7:18
$begingroup$
I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'setminus S$. What does $r_1$ look like in this case?
$endgroup$
– Hempelicious
Jan 23 at 20:26
$begingroup$
I'm sorry, I don't follow. I'm not sure what it means to "telescope" a path. You say that the preimage of a nbhd of the point should be a regular nbhd of $M'setminus S$, but I don't see how that can be true without being more explicit about what $r_1$ does. For example, here is the "simplest" case: $M'$ is a punctured disk. I've drawn what you call $A$ in red, and the shaded region is $M'setminus S$. What does $r_1$ look like in this case?
$endgroup$
– Hempelicious
Jan 23 at 20:26
$begingroup$
@Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk.
$endgroup$
– Eric Towers
Jan 23 at 20:48
$begingroup$
@Hempelicious : $r_1$ deformation retracts the gray disk to a point in its interior, dragging the blue boundary curve to that point. This is equivalent to quotienting by the closure of the gray disk.
$endgroup$
– Eric Towers
Jan 23 at 20:48
$begingroup$
Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me!
$endgroup$
– Hempelicious
Jan 23 at 21:23
$begingroup$
Ah ok, now I get it! You are deforming the boundary of $S$, but leaving the "core" graph fixed. OK I think I now understand your argument. Thanks for sticking with me!
$endgroup$
– Hempelicious
Jan 23 at 21:23
|
show 1 more comment
$begingroup$
Here is the approach I know, which is used in Hubbard's book Teichmuller Theory vol. 1.
One thing I don't like about this proof is you either need some smooth structure on $M$, or you need to verify the "supposes" in the next paragraph.
Suppose we have an embedding $rho:mathbb{H}rightarrow M$, where $mathbb{H}=[0,infty)$. Let $xinpartial M$ and further suppose this embedding is such that
begin{equation*}
rho(mathbb{H})cappartial M = rho(0)=x
end{equation*}
Finally, suppose that there's a neighborhood of $rho(mathbb{H})$ that looks like $mathbb{H}times (-1,1)$.
Then we can cut $M$ along $rho(mathbb{H})$ to get a new manifold $N$, where $partial Ncongmathbb{R}$. We can think of this boundary as divided into three pieces: left, middle, and right. Here left and right come from the two sides of $rho(mathbb{H})$, and middle is $partial M$ cut at $x$. Note that middle is homemorphic to $[0,1]$.
We define a map $f:partial Nrightarrow [0,1]$ on each piece:
begin{equation*}
f(z) = begin{cases}
0 & zintextit{left}\
z & zintextit{middle}\
1 & zintextit{right}\
end{cases}
end{equation*}
By the Tietze extension theorem, this extends to a map $F: Nrightarrow [0,1]$. If we compose that with the quotient map $pi:[0,1]rightarrow S^1$, we get a map $pi F:Nrightarrow S^1$. Because this map agrees on left and right, it descends to a map
begin{equation*}
widetilde{pi F}:Mrightarrow S^1
end{equation*}
which is the identity on $partial M$. That means it is our desired retraction.
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
$endgroup$
$begingroup$
This $rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x in partial M$ exists. My argument replaces putting $rho(infty)$ on the end with putting $rho(text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $rho$ has an open tubular neighborhood.
$endgroup$
– Eric Towers
Jan 24 at 3:08
$begingroup$
@EricTowers: yes, $rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful.
$endgroup$
– Hempelicious
Jan 24 at 4:09
add a comment |
$begingroup$
Here is the approach I know, which is used in Hubbard's book Teichmuller Theory vol. 1.
One thing I don't like about this proof is you either need some smooth structure on $M$, or you need to verify the "supposes" in the next paragraph.
Suppose we have an embedding $rho:mathbb{H}rightarrow M$, where $mathbb{H}=[0,infty)$. Let $xinpartial M$ and further suppose this embedding is such that
begin{equation*}
rho(mathbb{H})cappartial M = rho(0)=x
end{equation*}
Finally, suppose that there's a neighborhood of $rho(mathbb{H})$ that looks like $mathbb{H}times (-1,1)$.
Then we can cut $M$ along $rho(mathbb{H})$ to get a new manifold $N$, where $partial Ncongmathbb{R}$. We can think of this boundary as divided into three pieces: left, middle, and right. Here left and right come from the two sides of $rho(mathbb{H})$, and middle is $partial M$ cut at $x$. Note that middle is homemorphic to $[0,1]$.
We define a map $f:partial Nrightarrow [0,1]$ on each piece:
begin{equation*}
f(z) = begin{cases}
0 & zintextit{left}\
z & zintextit{middle}\
1 & zintextit{right}\
end{cases}
end{equation*}
By the Tietze extension theorem, this extends to a map $F: Nrightarrow [0,1]$. If we compose that with the quotient map $pi:[0,1]rightarrow S^1$, we get a map $pi F:Nrightarrow S^1$. Because this map agrees on left and right, it descends to a map
begin{equation*}
widetilde{pi F}:Mrightarrow S^1
end{equation*}
which is the identity on $partial M$. That means it is our desired retraction.
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
$endgroup$
$begingroup$
This $rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x in partial M$ exists. My argument replaces putting $rho(infty)$ on the end with putting $rho(text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $rho$ has an open tubular neighborhood.
$endgroup$
– Eric Towers
Jan 24 at 3:08
$begingroup$
@EricTowers: yes, $rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful.
$endgroup$
– Hempelicious
Jan 24 at 4:09
add a comment |
$begingroup$
Here is the approach I know, which is used in Hubbard's book Teichmuller Theory vol. 1.
One thing I don't like about this proof is you either need some smooth structure on $M$, or you need to verify the "supposes" in the next paragraph.
Suppose we have an embedding $rho:mathbb{H}rightarrow M$, where $mathbb{H}=[0,infty)$. Let $xinpartial M$ and further suppose this embedding is such that
begin{equation*}
rho(mathbb{H})cappartial M = rho(0)=x
end{equation*}
Finally, suppose that there's a neighborhood of $rho(mathbb{H})$ that looks like $mathbb{H}times (-1,1)$.
Then we can cut $M$ along $rho(mathbb{H})$ to get a new manifold $N$, where $partial Ncongmathbb{R}$. We can think of this boundary as divided into three pieces: left, middle, and right. Here left and right come from the two sides of $rho(mathbb{H})$, and middle is $partial M$ cut at $x$. Note that middle is homemorphic to $[0,1]$.
We define a map $f:partial Nrightarrow [0,1]$ on each piece:
begin{equation*}
f(z) = begin{cases}
0 & zintextit{left}\
z & zintextit{middle}\
1 & zintextit{right}\
end{cases}
end{equation*}
By the Tietze extension theorem, this extends to a map $F: Nrightarrow [0,1]$. If we compose that with the quotient map $pi:[0,1]rightarrow S^1$, we get a map $pi F:Nrightarrow S^1$. Because this map agrees on left and right, it descends to a map
begin{equation*}
widetilde{pi F}:Mrightarrow S^1
end{equation*}
which is the identity on $partial M$. That means it is our desired retraction.
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
$endgroup$
Here is the approach I know, which is used in Hubbard's book Teichmuller Theory vol. 1.
One thing I don't like about this proof is you either need some smooth structure on $M$, or you need to verify the "supposes" in the next paragraph.
Suppose we have an embedding $rho:mathbb{H}rightarrow M$, where $mathbb{H}=[0,infty)$. Let $xinpartial M$ and further suppose this embedding is such that
begin{equation*}
rho(mathbb{H})cappartial M = rho(0)=x
end{equation*}
Finally, suppose that there's a neighborhood of $rho(mathbb{H})$ that looks like $mathbb{H}times (-1,1)$.
Then we can cut $M$ along $rho(mathbb{H})$ to get a new manifold $N$, where $partial Ncongmathbb{R}$. We can think of this boundary as divided into three pieces: left, middle, and right. Here left and right come from the two sides of $rho(mathbb{H})$, and middle is $partial M$ cut at $x$. Note that middle is homemorphic to $[0,1]$.
We define a map $f:partial Nrightarrow [0,1]$ on each piece:
begin{equation*}
f(z) = begin{cases}
0 & zintextit{left}\
z & zintextit{middle}\
1 & zintextit{right}\
end{cases}
end{equation*}
By the Tietze extension theorem, this extends to a map $F: Nrightarrow [0,1]$. If we compose that with the quotient map $pi:[0,1]rightarrow S^1$, we get a map $pi F:Nrightarrow S^1$. Because this map agrees on left and right, it descends to a map
begin{equation*}
widetilde{pi F}:Mrightarrow S^1
end{equation*}
which is the identity on $partial M$. That means it is our desired retraction.
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
edited Jan 23 at 23:38
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
answered Jan 23 at 21:42
HempeliciousHempelicious
147111
147111
$begingroup$
This $rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x in partial M$ exists. My argument replaces putting $rho(infty)$ on the end with putting $rho(text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $rho$ has an open tubular neighborhood.
$endgroup$
– Eric Towers
Jan 24 at 3:08
$begingroup$
@EricTowers: yes, $rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful.
$endgroup$
– Hempelicious
Jan 24 at 4:09
add a comment |
$begingroup$
This $rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x in partial M$ exists. My argument replaces putting $rho(infty)$ on the end with putting $rho(text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $rho$ has an open tubular neighborhood.
$endgroup$
– Eric Towers
Jan 24 at 3:08
$begingroup$
@EricTowers: yes, $rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful.
$endgroup$
– Hempelicious
Jan 24 at 4:09
$begingroup$
This $rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x in partial M$ exists. My argument replaces putting $rho(infty)$ on the end with putting $rho(text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $rho$ has an open tubular neighborhood.
$endgroup$
– Eric Towers
Jan 24 at 3:08
$begingroup$
This $rho$ is my $p$. I don't force $p$ to go all the way out to the end. Checking supposes: $x in partial M$ exists. My argument replaces putting $rho(infty)$ on the end with putting $rho(text{big})$ in an annulus that is a neighborhood of the end, concatenated with a radius of the punctured disk homeomorphic to that neighborhood. There is a path traced by $rho$ because connected implies path connected. Topological manifolds admit smooth structures in dimensions 1, 2, 3, so $rho$ has an open tubular neighborhood.
$endgroup$
– Eric Towers
Jan 24 at 3:08
$begingroup$
@EricTowers: yes, $rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful.
$endgroup$
– Hempelicious
Jan 24 at 4:09
$begingroup$
@EricTowers: yes, $rho$ is easy to build using an exhaustion of $M$. It's pretty easy to show it can be made a smooth embedding, so one gets a trivial tubular nbhd, allowing the cut. But I still don't think "every surface has a smooth structure" is the simplest. But maybe I'm being wishful.
$endgroup$
– Hempelicious
Jan 24 at 4:09
add a comment |
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What is your definition of retraction? I'm familiar with the definition which requires the target be a subset of $M$. This does not happen if $M$ is, as a very simple example, the open unit disk in $mathbb{R}^2$.
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– Eric Towers
Jan 22 at 4:35
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@EricTowers That doesn't have boundary (in the sense of "manifold-with-boundary") a circle.
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– Lord Shark the Unknown
Jan 22 at 4:46
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@LordSharktheUnknown : I agree, which is why I ask for clarification. Perhaps a minimal example of a connected, noncompact 2-manifold with boundary having a single boundary component would be illuminating.
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– Eric Towers
Jan 22 at 4:52
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@EricTowers How about a closed disc with its centre removed; that obviously does retract to it boundary.
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– Lord Shark the Unknown
Jan 22 at 4:54
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@EricTowers: yes, you can take the disk minus some points (even something like a Cantor set). By retract I mean a map that is the identity when restricted the the subspace. That is, a left inverse of the inclusion map $partial Mrightarrow M$.
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– Hempelicious
Jan 22 at 5:07