Deformation Retraction and Projection/Closest Vector
$begingroup$
How do I compute the projection/closest vector to a subset? I have been thinking about this for far too long without any progress. If it helps, I am working in $Bbb{R}^2$, but I would like formalue in terms of norms and inner products, if possible.
For context, I am trying to prove that the figure eight is a deformation retract of the doubly punctured plane. And it is annoying that everything hinges on this annoyingly simple question. I have already shown that $overline{B}(0,1) setminus {p,q}$ is a deformation retraction of $Bbb{R}^2 setminus {p,q}$, where $p = (-frac{1}{2},0)$ and $q = (frac{1}{2},0)$. Now I just need to show that $overline{B}(0,1) setminus {p,q}$ deformation retracts to the union of the two discs with one centered at $p$, the other centered at $q$, but I am currently facing the obstacle discussed above.
EDIT: It just occurred to me that deformation retraction I had in mind won't be well-defined. Any point on the y-axis contained in $overline{B}(0,1)$ won't have a unique projection/closest vector in the union of the two open discs contained in $overline{B}(0,1)$...Hmm..need to rethink my approach...Of course, I wouldn't be opposed to any suggestions!
linear-algebra general-topology algebraic-topology projection retraction
$endgroup$
add a comment |
$begingroup$
How do I compute the projection/closest vector to a subset? I have been thinking about this for far too long without any progress. If it helps, I am working in $Bbb{R}^2$, but I would like formalue in terms of norms and inner products, if possible.
For context, I am trying to prove that the figure eight is a deformation retract of the doubly punctured plane. And it is annoying that everything hinges on this annoyingly simple question. I have already shown that $overline{B}(0,1) setminus {p,q}$ is a deformation retraction of $Bbb{R}^2 setminus {p,q}$, where $p = (-frac{1}{2},0)$ and $q = (frac{1}{2},0)$. Now I just need to show that $overline{B}(0,1) setminus {p,q}$ deformation retracts to the union of the two discs with one centered at $p$, the other centered at $q$, but I am currently facing the obstacle discussed above.
EDIT: It just occurred to me that deformation retraction I had in mind won't be well-defined. Any point on the y-axis contained in $overline{B}(0,1)$ won't have a unique projection/closest vector in the union of the two open discs contained in $overline{B}(0,1)$...Hmm..need to rethink my approach...Of course, I wouldn't be opposed to any suggestions!
linear-algebra general-topology algebraic-topology projection retraction
$endgroup$
add a comment |
$begingroup$
How do I compute the projection/closest vector to a subset? I have been thinking about this for far too long without any progress. If it helps, I am working in $Bbb{R}^2$, but I would like formalue in terms of norms and inner products, if possible.
For context, I am trying to prove that the figure eight is a deformation retract of the doubly punctured plane. And it is annoying that everything hinges on this annoyingly simple question. I have already shown that $overline{B}(0,1) setminus {p,q}$ is a deformation retraction of $Bbb{R}^2 setminus {p,q}$, where $p = (-frac{1}{2},0)$ and $q = (frac{1}{2},0)$. Now I just need to show that $overline{B}(0,1) setminus {p,q}$ deformation retracts to the union of the two discs with one centered at $p$, the other centered at $q$, but I am currently facing the obstacle discussed above.
EDIT: It just occurred to me that deformation retraction I had in mind won't be well-defined. Any point on the y-axis contained in $overline{B}(0,1)$ won't have a unique projection/closest vector in the union of the two open discs contained in $overline{B}(0,1)$...Hmm..need to rethink my approach...Of course, I wouldn't be opposed to any suggestions!
linear-algebra general-topology algebraic-topology projection retraction
$endgroup$
How do I compute the projection/closest vector to a subset? I have been thinking about this for far too long without any progress. If it helps, I am working in $Bbb{R}^2$, but I would like formalue in terms of norms and inner products, if possible.
For context, I am trying to prove that the figure eight is a deformation retract of the doubly punctured plane. And it is annoying that everything hinges on this annoyingly simple question. I have already shown that $overline{B}(0,1) setminus {p,q}$ is a deformation retraction of $Bbb{R}^2 setminus {p,q}$, where $p = (-frac{1}{2},0)$ and $q = (frac{1}{2},0)$. Now I just need to show that $overline{B}(0,1) setminus {p,q}$ deformation retracts to the union of the two discs with one centered at $p$, the other centered at $q$, but I am currently facing the obstacle discussed above.
EDIT: It just occurred to me that deformation retraction I had in mind won't be well-defined. Any point on the y-axis contained in $overline{B}(0,1)$ won't have a unique projection/closest vector in the union of the two open discs contained in $overline{B}(0,1)$...Hmm..need to rethink my approach...Of course, I wouldn't be opposed to any suggestions!
linear-algebra general-topology algebraic-topology projection retraction
linear-algebra general-topology algebraic-topology projection retraction
edited Jan 10 at 14:02
Paul Frost
10.1k3933
10.1k3933
asked Jan 10 at 12:43
user193319user193319
2,3342924
2,3342924
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
I do not really understand what you mean by "the projection/closest vector to a subset".
However, I shall explain how to get the desired strong deformation retraction. Let $p_{pm1}$ denote the points $(pm1,0) in mathbb{R}^2$, let the figure eight be the space $E = S_{+1} cup S_{-1}$, where $S_{pm1}$ is the circle around $p_{pm1}$ with radius $1$ and let the doubly punctured plane be the space $P = mathbb{R}^2 setminus { p_{+1}, p_{-1} }$. Define $r : P to E$ as follows. For $p = (x,y)$ set
$$r(p) =
begin{cases}
p_{+1} + dfrac{p - p_{+1}}{lVert p - p_{+1} rVert} & lVert p - p_{+1} rVert le 1 \
p_{-1} + dfrac{p - p_{-1}}{lVert p - p_{-1} rVert} & lVert p - p_{-1} rVert le 1 \
dfrac{2xp}{lVert p rVert^2} & lVert p - p_{+1} rVert ge 1, p ne 0, x ge 0 \
-dfrac{2xp}{lVert p rVert^2} & lVert p - p_{-1} rVert ge 1, p ne 0, x le 0
end{cases}
$$
Here $lVert - rVert$ denotes the Euclidean norm $lVert (x,y) rVert = sqrt{x^2+ y^2}$. Note that the denominator $lVert p - p_{pm 1} rVert$ does not vanish on $P$.
What happen geometrically? Let $D_{pm 1}$ = closed unit disk with center $p_{pm 1}$, $H_{pm 1}$ = right/left half plane minus the interior of $D_{pm 1}$.
The first two lines describe the radial strong deformation retractions from $B_{pm 1} = D_{pm 1} setminus { p_{pm 1} }$ to $S_{pm 1}$. In fact, for $p in B_{pm 1}$ we have $lVert r(p) - p_{pm 1} rVert = lVert dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} rVert = 1$, and for $p in S_{pm 1}$ we have $p_{pm 1} + dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} = p$.
Note that $B_{+1} cap B_{-1} = { 0 }$. Both line 1 and line 2 yield $r(0) = 0$.
The last two lines (together with $r(0) = 0$) describe strong deformation retractions of $H_{pm 1}$ to $S_{pm 1}$. This is done by shifting each point $p ne 0$ along the line through $0$ and $p$ until it reaches $S_{pm 1}$. To be formal, this line is given by $l_p(t) = t p$, and for $p in H_{pm 1} setminus { 0 }$ we must find $t$ such that $lVert t p - p_{pm 1} rVert = 1$. Easy computations show $t = pm dfrac{2x}{lVert p rVert^2}$, and in fact we defined $r(p) = l_p(t) = t p$.
Note that for $p in Y = H_{+1} cap H_{-1}$ = $y$-axis = set of points with $x = 0$ we have $r(p) = 0$.
Thus all four lines give us a consistent definition on the whole space $P$. It remains to show that $r mid_{H_{pm 1}}$ is continuous in $p = 0$. We have $lVert p - p_{pm 1} rVert ge 1$, i.e. $(x mp 1)^2 + y^2 ge 1$. This is equivalent to $pm 2x le x^2 + y^2$ which means $2lvert x rvert le lVert p rVert^2$ since $p in H_{pm 1}$. Hence $lVert r(p) rVert = dfrac{2 lvert x rvert}{lVert p rVert} le lVert p rVert$ for $p ne 0$. This immediately implies continuity.
We now have constructed a retraction $r$. To see that it is a strong deformation retraction, define a homotopy
$$H : P times I to P, H(p,t) = (1-t)p + tr(p) .$$
It is readily verified that in fact $H(p,t) ne p_{pm 1}$ for all $(p,t)$ (check all 4 lines in the definition of $r$). This is a homotopy from $id_P$ to $r$ which is stationary on $E$.
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068599%2fdeformation-retraction-and-projection-closest-vector%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
I do not really understand what you mean by "the projection/closest vector to a subset".
However, I shall explain how to get the desired strong deformation retraction. Let $p_{pm1}$ denote the points $(pm1,0) in mathbb{R}^2$, let the figure eight be the space $E = S_{+1} cup S_{-1}$, where $S_{pm1}$ is the circle around $p_{pm1}$ with radius $1$ and let the doubly punctured plane be the space $P = mathbb{R}^2 setminus { p_{+1}, p_{-1} }$. Define $r : P to E$ as follows. For $p = (x,y)$ set
$$r(p) =
begin{cases}
p_{+1} + dfrac{p - p_{+1}}{lVert p - p_{+1} rVert} & lVert p - p_{+1} rVert le 1 \
p_{-1} + dfrac{p - p_{-1}}{lVert p - p_{-1} rVert} & lVert p - p_{-1} rVert le 1 \
dfrac{2xp}{lVert p rVert^2} & lVert p - p_{+1} rVert ge 1, p ne 0, x ge 0 \
-dfrac{2xp}{lVert p rVert^2} & lVert p - p_{-1} rVert ge 1, p ne 0, x le 0
end{cases}
$$
Here $lVert - rVert$ denotes the Euclidean norm $lVert (x,y) rVert = sqrt{x^2+ y^2}$. Note that the denominator $lVert p - p_{pm 1} rVert$ does not vanish on $P$.
What happen geometrically? Let $D_{pm 1}$ = closed unit disk with center $p_{pm 1}$, $H_{pm 1}$ = right/left half plane minus the interior of $D_{pm 1}$.
The first two lines describe the radial strong deformation retractions from $B_{pm 1} = D_{pm 1} setminus { p_{pm 1} }$ to $S_{pm 1}$. In fact, for $p in B_{pm 1}$ we have $lVert r(p) - p_{pm 1} rVert = lVert dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} rVert = 1$, and for $p in S_{pm 1}$ we have $p_{pm 1} + dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} = p$.
Note that $B_{+1} cap B_{-1} = { 0 }$. Both line 1 and line 2 yield $r(0) = 0$.
The last two lines (together with $r(0) = 0$) describe strong deformation retractions of $H_{pm 1}$ to $S_{pm 1}$. This is done by shifting each point $p ne 0$ along the line through $0$ and $p$ until it reaches $S_{pm 1}$. To be formal, this line is given by $l_p(t) = t p$, and for $p in H_{pm 1} setminus { 0 }$ we must find $t$ such that $lVert t p - p_{pm 1} rVert = 1$. Easy computations show $t = pm dfrac{2x}{lVert p rVert^2}$, and in fact we defined $r(p) = l_p(t) = t p$.
Note that for $p in Y = H_{+1} cap H_{-1}$ = $y$-axis = set of points with $x = 0$ we have $r(p) = 0$.
Thus all four lines give us a consistent definition on the whole space $P$. It remains to show that $r mid_{H_{pm 1}}$ is continuous in $p = 0$. We have $lVert p - p_{pm 1} rVert ge 1$, i.e. $(x mp 1)^2 + y^2 ge 1$. This is equivalent to $pm 2x le x^2 + y^2$ which means $2lvert x rvert le lVert p rVert^2$ since $p in H_{pm 1}$. Hence $lVert r(p) rVert = dfrac{2 lvert x rvert}{lVert p rVert} le lVert p rVert$ for $p ne 0$. This immediately implies continuity.
We now have constructed a retraction $r$. To see that it is a strong deformation retraction, define a homotopy
$$H : P times I to P, H(p,t) = (1-t)p + tr(p) .$$
It is readily verified that in fact $H(p,t) ne p_{pm 1}$ for all $(p,t)$ (check all 4 lines in the definition of $r$). This is a homotopy from $id_P$ to $r$ which is stationary on $E$.
$endgroup$
add a comment |
$begingroup$
I do not really understand what you mean by "the projection/closest vector to a subset".
However, I shall explain how to get the desired strong deformation retraction. Let $p_{pm1}$ denote the points $(pm1,0) in mathbb{R}^2$, let the figure eight be the space $E = S_{+1} cup S_{-1}$, where $S_{pm1}$ is the circle around $p_{pm1}$ with radius $1$ and let the doubly punctured plane be the space $P = mathbb{R}^2 setminus { p_{+1}, p_{-1} }$. Define $r : P to E$ as follows. For $p = (x,y)$ set
$$r(p) =
begin{cases}
p_{+1} + dfrac{p - p_{+1}}{lVert p - p_{+1} rVert} & lVert p - p_{+1} rVert le 1 \
p_{-1} + dfrac{p - p_{-1}}{lVert p - p_{-1} rVert} & lVert p - p_{-1} rVert le 1 \
dfrac{2xp}{lVert p rVert^2} & lVert p - p_{+1} rVert ge 1, p ne 0, x ge 0 \
-dfrac{2xp}{lVert p rVert^2} & lVert p - p_{-1} rVert ge 1, p ne 0, x le 0
end{cases}
$$
Here $lVert - rVert$ denotes the Euclidean norm $lVert (x,y) rVert = sqrt{x^2+ y^2}$. Note that the denominator $lVert p - p_{pm 1} rVert$ does not vanish on $P$.
What happen geometrically? Let $D_{pm 1}$ = closed unit disk with center $p_{pm 1}$, $H_{pm 1}$ = right/left half plane minus the interior of $D_{pm 1}$.
The first two lines describe the radial strong deformation retractions from $B_{pm 1} = D_{pm 1} setminus { p_{pm 1} }$ to $S_{pm 1}$. In fact, for $p in B_{pm 1}$ we have $lVert r(p) - p_{pm 1} rVert = lVert dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} rVert = 1$, and for $p in S_{pm 1}$ we have $p_{pm 1} + dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} = p$.
Note that $B_{+1} cap B_{-1} = { 0 }$. Both line 1 and line 2 yield $r(0) = 0$.
The last two lines (together with $r(0) = 0$) describe strong deformation retractions of $H_{pm 1}$ to $S_{pm 1}$. This is done by shifting each point $p ne 0$ along the line through $0$ and $p$ until it reaches $S_{pm 1}$. To be formal, this line is given by $l_p(t) = t p$, and for $p in H_{pm 1} setminus { 0 }$ we must find $t$ such that $lVert t p - p_{pm 1} rVert = 1$. Easy computations show $t = pm dfrac{2x}{lVert p rVert^2}$, and in fact we defined $r(p) = l_p(t) = t p$.
Note that for $p in Y = H_{+1} cap H_{-1}$ = $y$-axis = set of points with $x = 0$ we have $r(p) = 0$.
Thus all four lines give us a consistent definition on the whole space $P$. It remains to show that $r mid_{H_{pm 1}}$ is continuous in $p = 0$. We have $lVert p - p_{pm 1} rVert ge 1$, i.e. $(x mp 1)^2 + y^2 ge 1$. This is equivalent to $pm 2x le x^2 + y^2$ which means $2lvert x rvert le lVert p rVert^2$ since $p in H_{pm 1}$. Hence $lVert r(p) rVert = dfrac{2 lvert x rvert}{lVert p rVert} le lVert p rVert$ for $p ne 0$. This immediately implies continuity.
We now have constructed a retraction $r$. To see that it is a strong deformation retraction, define a homotopy
$$H : P times I to P, H(p,t) = (1-t)p + tr(p) .$$
It is readily verified that in fact $H(p,t) ne p_{pm 1}$ for all $(p,t)$ (check all 4 lines in the definition of $r$). This is a homotopy from $id_P$ to $r$ which is stationary on $E$.
$endgroup$
add a comment |
$begingroup$
I do not really understand what you mean by "the projection/closest vector to a subset".
However, I shall explain how to get the desired strong deformation retraction. Let $p_{pm1}$ denote the points $(pm1,0) in mathbb{R}^2$, let the figure eight be the space $E = S_{+1} cup S_{-1}$, where $S_{pm1}$ is the circle around $p_{pm1}$ with radius $1$ and let the doubly punctured plane be the space $P = mathbb{R}^2 setminus { p_{+1}, p_{-1} }$. Define $r : P to E$ as follows. For $p = (x,y)$ set
$$r(p) =
begin{cases}
p_{+1} + dfrac{p - p_{+1}}{lVert p - p_{+1} rVert} & lVert p - p_{+1} rVert le 1 \
p_{-1} + dfrac{p - p_{-1}}{lVert p - p_{-1} rVert} & lVert p - p_{-1} rVert le 1 \
dfrac{2xp}{lVert p rVert^2} & lVert p - p_{+1} rVert ge 1, p ne 0, x ge 0 \
-dfrac{2xp}{lVert p rVert^2} & lVert p - p_{-1} rVert ge 1, p ne 0, x le 0
end{cases}
$$
Here $lVert - rVert$ denotes the Euclidean norm $lVert (x,y) rVert = sqrt{x^2+ y^2}$. Note that the denominator $lVert p - p_{pm 1} rVert$ does not vanish on $P$.
What happen geometrically? Let $D_{pm 1}$ = closed unit disk with center $p_{pm 1}$, $H_{pm 1}$ = right/left half plane minus the interior of $D_{pm 1}$.
The first two lines describe the radial strong deformation retractions from $B_{pm 1} = D_{pm 1} setminus { p_{pm 1} }$ to $S_{pm 1}$. In fact, for $p in B_{pm 1}$ we have $lVert r(p) - p_{pm 1} rVert = lVert dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} rVert = 1$, and for $p in S_{pm 1}$ we have $p_{pm 1} + dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} = p$.
Note that $B_{+1} cap B_{-1} = { 0 }$. Both line 1 and line 2 yield $r(0) = 0$.
The last two lines (together with $r(0) = 0$) describe strong deformation retractions of $H_{pm 1}$ to $S_{pm 1}$. This is done by shifting each point $p ne 0$ along the line through $0$ and $p$ until it reaches $S_{pm 1}$. To be formal, this line is given by $l_p(t) = t p$, and for $p in H_{pm 1} setminus { 0 }$ we must find $t$ such that $lVert t p - p_{pm 1} rVert = 1$. Easy computations show $t = pm dfrac{2x}{lVert p rVert^2}$, and in fact we defined $r(p) = l_p(t) = t p$.
Note that for $p in Y = H_{+1} cap H_{-1}$ = $y$-axis = set of points with $x = 0$ we have $r(p) = 0$.
Thus all four lines give us a consistent definition on the whole space $P$. It remains to show that $r mid_{H_{pm 1}}$ is continuous in $p = 0$. We have $lVert p - p_{pm 1} rVert ge 1$, i.e. $(x mp 1)^2 + y^2 ge 1$. This is equivalent to $pm 2x le x^2 + y^2$ which means $2lvert x rvert le lVert p rVert^2$ since $p in H_{pm 1}$. Hence $lVert r(p) rVert = dfrac{2 lvert x rvert}{lVert p rVert} le lVert p rVert$ for $p ne 0$. This immediately implies continuity.
We now have constructed a retraction $r$. To see that it is a strong deformation retraction, define a homotopy
$$H : P times I to P, H(p,t) = (1-t)p + tr(p) .$$
It is readily verified that in fact $H(p,t) ne p_{pm 1}$ for all $(p,t)$ (check all 4 lines in the definition of $r$). This is a homotopy from $id_P$ to $r$ which is stationary on $E$.
$endgroup$
I do not really understand what you mean by "the projection/closest vector to a subset".
However, I shall explain how to get the desired strong deformation retraction. Let $p_{pm1}$ denote the points $(pm1,0) in mathbb{R}^2$, let the figure eight be the space $E = S_{+1} cup S_{-1}$, where $S_{pm1}$ is the circle around $p_{pm1}$ with radius $1$ and let the doubly punctured plane be the space $P = mathbb{R}^2 setminus { p_{+1}, p_{-1} }$. Define $r : P to E$ as follows. For $p = (x,y)$ set
$$r(p) =
begin{cases}
p_{+1} + dfrac{p - p_{+1}}{lVert p - p_{+1} rVert} & lVert p - p_{+1} rVert le 1 \
p_{-1} + dfrac{p - p_{-1}}{lVert p - p_{-1} rVert} & lVert p - p_{-1} rVert le 1 \
dfrac{2xp}{lVert p rVert^2} & lVert p - p_{+1} rVert ge 1, p ne 0, x ge 0 \
-dfrac{2xp}{lVert p rVert^2} & lVert p - p_{-1} rVert ge 1, p ne 0, x le 0
end{cases}
$$
Here $lVert - rVert$ denotes the Euclidean norm $lVert (x,y) rVert = sqrt{x^2+ y^2}$. Note that the denominator $lVert p - p_{pm 1} rVert$ does not vanish on $P$.
What happen geometrically? Let $D_{pm 1}$ = closed unit disk with center $p_{pm 1}$, $H_{pm 1}$ = right/left half plane minus the interior of $D_{pm 1}$.
The first two lines describe the radial strong deformation retractions from $B_{pm 1} = D_{pm 1} setminus { p_{pm 1} }$ to $S_{pm 1}$. In fact, for $p in B_{pm 1}$ we have $lVert r(p) - p_{pm 1} rVert = lVert dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} rVert = 1$, and for $p in S_{pm 1}$ we have $p_{pm 1} + dfrac{p - p_{pm 1}}{lVert p - p_{pm 1} rVert} = p$.
Note that $B_{+1} cap B_{-1} = { 0 }$. Both line 1 and line 2 yield $r(0) = 0$.
The last two lines (together with $r(0) = 0$) describe strong deformation retractions of $H_{pm 1}$ to $S_{pm 1}$. This is done by shifting each point $p ne 0$ along the line through $0$ and $p$ until it reaches $S_{pm 1}$. To be formal, this line is given by $l_p(t) = t p$, and for $p in H_{pm 1} setminus { 0 }$ we must find $t$ such that $lVert t p - p_{pm 1} rVert = 1$. Easy computations show $t = pm dfrac{2x}{lVert p rVert^2}$, and in fact we defined $r(p) = l_p(t) = t p$.
Note that for $p in Y = H_{+1} cap H_{-1}$ = $y$-axis = set of points with $x = 0$ we have $r(p) = 0$.
Thus all four lines give us a consistent definition on the whole space $P$. It remains to show that $r mid_{H_{pm 1}}$ is continuous in $p = 0$. We have $lVert p - p_{pm 1} rVert ge 1$, i.e. $(x mp 1)^2 + y^2 ge 1$. This is equivalent to $pm 2x le x^2 + y^2$ which means $2lvert x rvert le lVert p rVert^2$ since $p in H_{pm 1}$. Hence $lVert r(p) rVert = dfrac{2 lvert x rvert}{lVert p rVert} le lVert p rVert$ for $p ne 0$. This immediately implies continuity.
We now have constructed a retraction $r$. To see that it is a strong deformation retraction, define a homotopy
$$H : P times I to P, H(p,t) = (1-t)p + tr(p) .$$
It is readily verified that in fact $H(p,t) ne p_{pm 1}$ for all $(p,t)$ (check all 4 lines in the definition of $r$). This is a homotopy from $id_P$ to $r$ which is stationary on $E$.
edited Jan 11 at 9:38
answered Jan 10 at 16:51
Paul FrostPaul Frost
10.1k3933
10.1k3933
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068599%2fdeformation-retraction-and-projection-closest-vector%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown