Mixing
Relation between fixed point and retraction theorem
$begingroup$
There is this particular exercise in Lawvere/Schanuels book "Conceptual Mathematics: A first introduction to categories" that I've worked on, but I'm not entirely sure if I'm correct. Plus, I'm a little confused with regards to certain elements.
Question:
"Let $j: Cto D$ be an inclusion map from circle C into disk D. Suppose that we have two continuous maps $f: D to D$ and $g: Dto D$, and that g satisfies $g circ j = j$. Use the retraction theorem to show that there must be a point x in the disk at which f(x) = g(x). (Hint: The fixed point theorem is the special case g = $1_D$, so try to generalize the argument we used in that special case.)"
Attempted Answer:
Knowing that there is an inclusion map j and a requirement of a retraction is needed, I first state the obvious
$r: D to C$
which would create the identity $1_C$ when subject to $r circ j$.
Since a retraction cannot co-exist with fixed points, the continuous maps f and g have to be:
$f(x) neq x$ & $g(x) neq x$ respectively.
Since g satisfies $g circ j = j$, one can assume that since $g: D to D$ is an endomap yielding $1_D$ implies $g = 1_D$. (Which apparently is a special case of the fixed point theorem? I assume due to the intermediate value theorem of a fixed point theorem it is such that for a given $f(a)$ and $f(b)$ there exists an $f(c)$ which yields $f(c)-c=0$ which is fine).
However, by introducing a special case of the fixed point theorem, isnt it in violation of the rule that retractions cannot exist with fixed points?
Disregarding the concern I have for the violation, proceeding with composition:
$rcirc gcirc j = rcirc fcirc j = r circ j = 1_C$
As f and g are equivalent maps:
$g circ j = f circ j = 1_D circ j = j$
Using Associativity laws and right hand identity, then left hand inverse:
$(r circ g) = (r circ f) = r$
$g = f = 1_D $
Hence there exists an identity map whereby f and g are equivalent and that the mapping yields x. But doesnt that mean that it is a contradiction, whereby $f(x) neq x$, $g(x) neq x$? Was this a trick question..? Because I certainly feel like 12 types of stupid right now.
EDIT: Additionally, it was mentioned thereafter in the textbook that each retraction theorem is equivalent to a fixed point theorem, that the fixed point theorem was deducible from the retraction theorem and vice versa.
I understand that the contrapositive statement exists, is that what is implied by the equivalence?
category-theory
asked Jun 15 '15 at 3:52
CyrusCyrus
124
$endgroup$
add a comment |
$begingroup$
There is this particular exercise in Lawvere/Schanuels book "Conceptual Mathematics: A first introduction to categories" that I've worked on, but I'm not entirely sure if I'm correct. Plus, I'm a little confused with regards to certain elements.
Question:
"Let $j: Cto D$ be an inclusion map from circle C into disk D. Suppose that we have two continuous maps $f: D to D$ and $g: Dto D$, and that g satisfies $g circ j = j$. Use the retraction theorem to show that there must be a point x in the disk at which f(x) = g(x). (Hint: The fixed point theorem is the special case g = $1_D$, so try to generalize the argument we used in that special case.)"
Attempted Answer:
Knowing that there is an inclusion map j and a requirement of a retraction is needed, I first state the obvious
$r: D to C$
which would create the identity $1_C$ when subject to $r circ j$.
Since a retraction cannot co-exist with fixed points, the continuous maps f and g have to be:
$f(x) neq x$ & $g(x) neq x$ respectively.
Since g satisfies $g circ j = j$, one can assume that since $g: D to D$ is an endomap yielding $1_D$ implies $g = 1_D$. (Which apparently is a special case of the fixed point theorem? I assume due to the intermediate value theorem of a fixed point theorem it is such that for a given $f(a)$ and $f(b)$ there exists an $f(c)$ which yields $f(c)-c=0$ which is fine).
However, by introducing a special case of the fixed point theorem, isnt it in violation of the rule that retractions cannot exist with fixed points?
Disregarding the concern I have for the violation, proceeding with composition:
$rcirc gcirc j = rcirc fcirc j = r circ j = 1_C$
As f and g are equivalent maps:
$g circ j = f circ j = 1_D circ j = j$
Using Associativity laws and right hand identity, then left hand inverse:
$(r circ g) = (r circ f) = r$
$g = f = 1_D $
Hence there exists an identity map whereby f and g are equivalent and that the mapping yields x. But doesnt that mean that it is a contradiction, whereby $f(x) neq x$, $g(x) neq x$? Was this a trick question..? Because I certainly feel like 12 types of stupid right now.
EDIT: Additionally, it was mentioned thereafter in the textbook that each retraction theorem is equivalent to a fixed point theorem, that the fixed point theorem was deducible from the retraction theorem and vice versa.
I understand that the contrapositive statement exists, is that what is implied by the equivalence?
category-theory
asked Jun 15 '15 at 3:52
CyrusCyrus
124
$endgroup$
add a comment |
$begingroup$
There is this particular exercise in Lawvere/Schanuels book "Conceptual Mathematics: A first introduction to categories" that I've worked on, but I'm not entirely sure if I'm correct. Plus, I'm a little confused with regards to certain elements.
Question:
"Let $j: Cto D$ be an inclusion map from circle C into disk D. Suppose that we have two continuous maps $f: D to D$ and $g: Dto D$, and that g satisfies $g circ j = j$. Use the retraction theorem to show that there must be a point x in the disk at which f(x) = g(x). (Hint: The fixed point theorem is the special case g = $1_D$, so try to generalize the argument we used in that special case.)"
Attempted Answer:
Knowing that there is an inclusion map j and a requirement of a retraction is needed, I first state the obvious
$r: D to C$
which would create the identity $1_C$ when subject to $r circ j$.
Since a retraction cannot co-exist with fixed points, the continuous maps f and g have to be:
$f(x) neq x$ & $g(x) neq x$ respectively.
Since g satisfies $g circ j = j$, one can assume that since $g: D to D$ is an endomap yielding $1_D$ implies $g = 1_D$. (Which apparently is a special case of the fixed point theorem? I assume due to the intermediate value theorem of a fixed point theorem it is such that for a given $f(a)$ and $f(b)$ there exists an $f(c)$ which yields $f(c)-c=0$ which is fine).
However, by introducing a special case of the fixed point theorem, isnt it in violation of the rule that retractions cannot exist with fixed points?
Disregarding the concern I have for the violation, proceeding with composition:
$rcirc gcirc j = rcirc fcirc j = r circ j = 1_C$
As f and g are equivalent maps:
$g circ j = f circ j = 1_D circ j = j$
Using Associativity laws and right hand identity, then left hand inverse:
$(r circ g) = (r circ f) = r$
$g = f = 1_D $
Hence there exists an identity map whereby f and g are equivalent and that the mapping yields x. But doesnt that mean that it is a contradiction, whereby $f(x) neq x$, $g(x) neq x$? Was this a trick question..? Because I certainly feel like 12 types of stupid right now.
EDIT: Additionally, it was mentioned thereafter in the textbook that each retraction theorem is equivalent to a fixed point theorem, that the fixed point theorem was deducible from the retraction theorem and vice versa.
I understand that the contrapositive statement exists, is that what is implied by the equivalence?
category-theory
asked Jun 15 '15 at 3:52
CyrusCyrus
124
$endgroup$
There is this particular exercise in Lawvere/Schanuels book "Conceptual Mathematics: A first introduction to categories" that I've worked on, but I'm not entirely sure if I'm correct. Plus, I'm a little confused with regards to certain elements.
Question:
"Let $j: Cto D$ be an inclusion map from circle C into disk D. Suppose that we have two continuous maps $f: D to D$ and $g: Dto D$, and that g satisfies $g circ j = j$. Use the retraction theorem to show that there must be a point x in the disk at which f(x) = g(x). (Hint: The fixed point theorem is the special case g = $1_D$, so try to generalize the argument we used in that special case.)"
Attempted Answer:
Knowing that there is an inclusion map j and a requirement of a retraction is needed, I first state the obvious
$r: D to C$
which would create the identity $1_C$ when subject to $r circ j$.
Since a retraction cannot co-exist with fixed points, the continuous maps f and g have to be:
$f(x) neq x$ & $g(x) neq x$ respectively.
Since g satisfies $g circ j = j$, one can assume that since $g: D to D$ is an endomap yielding $1_D$ implies $g = 1_D$. (Which apparently is a special case of the fixed point theorem? I assume due to the intermediate value theorem of a fixed point theorem it is such that for a given $f(a)$ and $f(b)$ there exists an $f(c)$ which yields $f(c)-c=0$ which is fine).
However, by introducing a special case of the fixed point theorem, isnt it in violation of the rule that retractions cannot exist with fixed points?
Disregarding the concern I have for the violation, proceeding with composition:
$rcirc gcirc j = rcirc fcirc j = r circ j = 1_C$
As f and g are equivalent maps:
$g circ j = f circ j = 1_D circ j = j$
Using Associativity laws and right hand identity, then left hand inverse:
$(r circ g) = (r circ f) = r$
$g = f = 1_D $
Hence there exists an identity map whereby f and g are equivalent and that the mapping yields x. But doesnt that mean that it is a contradiction, whereby $f(x) neq x$, $g(x) neq x$? Was this a trick question..? Because I certainly feel like 12 types of stupid right now.
EDIT: Additionally, it was mentioned thereafter in the textbook that each retraction theorem is equivalent to a fixed point theorem, that the fixed point theorem was deducible from the retraction theorem and vice versa.
I understand that the contrapositive statement exists, is that what is implied by the equivalence?
category-theory
category-theory
asked Jun 15 '15 at 3:52
CyrusCyrus
124
asked Jun 15 '15 at 3:52
CyrusCyrus
124
asked Jun 15 '15 at 3:52
CyrusCyrus
124
asked Jun 15 '15 at 3:52
CyrusCyrus
124
asked Jun 15 '15 at 3:52
CyrusCyrus
124
124
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
I imagine you are studying on your own, and have not done a full introductory course in point set topology. There is a standard picture that you would probably have seen...
Assume that $f(x) neq g(x)$ for every $x in D.$ For every $x,$ draw the line segment that begins at $f(x),$ goes through $g(x),$ and continues on until some point on $C.$ Note that, when $x in C,$ we already have $g(x) = x in C,$ so the line segment stops at $x$ itself. Let us name $h(x)$ the point in $C$ where the line segment ends. There is a fair amount of justification needed for this step: $h$ is a continuous mapping. Since $h$ takes all of $D$ to $C$ and fixes $C$ pointwise, it is a retraction. However, we have found out that there is no retraction of the disk onto the circle. This contradicts the assumption that $f(x) neq g(x)$ for every $x in D.$
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
I definitely have not started learning about point set topology and I do lack a great deal of mathematical maturity - I've just started purely on category theory and a long way to go. Ok, to wrap my idea around what you've said, the maps $f(x)$ and $g(x)$ do not connect except via line segment to substantiate the claim $f(x) neq g(x)$. $g(x)=x$ will occur when x becomes an element of C, whereby $h(x)$, presumably will be a fixed point. $h(x)$ as apparent from the line segments D to C is negated, hence no fixed point, hence $g(x) neq x$ and thus $f(x) = g(x)$?
$endgroup$
– Cyrus
Jun 15 '15 at 4:56
$begingroup$
@Cyrus, let me see if I can find a picture online somewhere.
$endgroup$
– Will Jagy
Jun 15 '15 at 4:59
$begingroup$
Ouch, sounds like I didnt get it right. Thanks in advance for literally giving me a clearer picture.
$endgroup$
– Cyrus
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, please see the picture on the right in en.wikipedia.org/wiki/… where their line segment starts at $f(x)$ and goes back through $x$ until the boundary circle is reached, and they call that point $F(x).$
$endgroup$
– Will Jagy
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, meanwhile, your summary above seems sort of correct, but i cannot really tell; it might be different if we were speaking in person. Rather late for me; I will chack this tomorrow to see if you had more to say.
$endgroup$
– Will Jagy
Jun 15 '15 at 5:08
|
show 1 more comment
$begingroup$
However, by introducing a special case of the fixed point theorem,
isn't it in violation of the rule that retractions cannot exist with fixed points?
Retractions and fixed points can coexist. The book says there is no retraction between "boundary" and "disk", namely no $B^{n+1} to S^n$. This is not true for all $A subset X$, namely there exists $A subset X$ and A is a retract of X. In this case, if X has fixed point property then A has fixed point property. One can refer here for a brief proof. https://topospaces.subwiki.org/wiki/Fixed-point_property_is_retract-hereditary
answered Jan 17 at 9:03
XingtaoXingtao
11
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add a comment |
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2 Answers
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2 Answers
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active
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$begingroup$
I imagine you are studying on your own, and have not done a full introductory course in point set topology. There is a standard picture that you would probably have seen...
Assume that $f(x) neq g(x)$ for every $x in D.$ For every $x,$ draw the line segment that begins at $f(x),$ goes through $g(x),$ and continues on until some point on $C.$ Note that, when $x in C,$ we already have $g(x) = x in C,$ so the line segment stops at $x$ itself. Let us name $h(x)$ the point in $C$ where the line segment ends. There is a fair amount of justification needed for this step: $h$ is a continuous mapping. Since $h$ takes all of $D$ to $C$ and fixes $C$ pointwise, it is a retraction. However, we have found out that there is no retraction of the disk onto the circle. This contradicts the assumption that $f(x) neq g(x)$ for every $x in D.$
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
I definitely have not started learning about point set topology and I do lack a great deal of mathematical maturity - I've just started purely on category theory and a long way to go. Ok, to wrap my idea around what you've said, the maps $f(x)$ and $g(x)$ do not connect except via line segment to substantiate the claim $f(x) neq g(x)$. $g(x)=x$ will occur when x becomes an element of C, whereby $h(x)$, presumably will be a fixed point. $h(x)$ as apparent from the line segments D to C is negated, hence no fixed point, hence $g(x) neq x$ and thus $f(x) = g(x)$?
$endgroup$
– Cyrus
Jun 15 '15 at 4:56
$begingroup$
@Cyrus, let me see if I can find a picture online somewhere.
$endgroup$
– Will Jagy
Jun 15 '15 at 4:59
$begingroup$
Ouch, sounds like I didnt get it right. Thanks in advance for literally giving me a clearer picture.
$endgroup$
– Cyrus
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, please see the picture on the right in en.wikipedia.org/wiki/… where their line segment starts at $f(x)$ and goes back through $x$ until the boundary circle is reached, and they call that point $F(x).$
$endgroup$
– Will Jagy
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, meanwhile, your summary above seems sort of correct, but i cannot really tell; it might be different if we were speaking in person. Rather late for me; I will chack this tomorrow to see if you had more to say.
$endgroup$
– Will Jagy
Jun 15 '15 at 5:08
|
show 1 more comment
$begingroup$
I imagine you are studying on your own, and have not done a full introductory course in point set topology. There is a standard picture that you would probably have seen...
Assume that $f(x) neq g(x)$ for every $x in D.$ For every $x,$ draw the line segment that begins at $f(x),$ goes through $g(x),$ and continues on until some point on $C.$ Note that, when $x in C,$ we already have $g(x) = x in C,$ so the line segment stops at $x$ itself. Let us name $h(x)$ the point in $C$ where the line segment ends. There is a fair amount of justification needed for this step: $h$ is a continuous mapping. Since $h$ takes all of $D$ to $C$ and fixes $C$ pointwise, it is a retraction. However, we have found out that there is no retraction of the disk onto the circle. This contradicts the assumption that $f(x) neq g(x)$ for every $x in D.$
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
$endgroup$
$begingroup$
I definitely have not started learning about point set topology and I do lack a great deal of mathematical maturity - I've just started purely on category theory and a long way to go. Ok, to wrap my idea around what you've said, the maps $f(x)$ and $g(x)$ do not connect except via line segment to substantiate the claim $f(x) neq g(x)$. $g(x)=x$ will occur when x becomes an element of C, whereby $h(x)$, presumably will be a fixed point. $h(x)$ as apparent from the line segments D to C is negated, hence no fixed point, hence $g(x) neq x$ and thus $f(x) = g(x)$?
$endgroup$
– Cyrus
Jun 15 '15 at 4:56
$begingroup$
@Cyrus, let me see if I can find a picture online somewhere.
$endgroup$
– Will Jagy
Jun 15 '15 at 4:59
$begingroup$
Ouch, sounds like I didnt get it right. Thanks in advance for literally giving me a clearer picture.
$endgroup$
– Cyrus
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, please see the picture on the right in en.wikipedia.org/wiki/… where their line segment starts at $f(x)$ and goes back through $x$ until the boundary circle is reached, and they call that point $F(x).$
$endgroup$
– Will Jagy
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, meanwhile, your summary above seems sort of correct, but i cannot really tell; it might be different if we were speaking in person. Rather late for me; I will chack this tomorrow to see if you had more to say.
$endgroup$
– Will Jagy
Jun 15 '15 at 5:08
|
show 1 more comment
$begingroup$
I imagine you are studying on your own, and have not done a full introductory course in point set topology. There is a standard picture that you would probably have seen...
Assume that $f(x) neq g(x)$ for every $x in D.$ For every $x,$ draw the line segment that begins at $f(x),$ goes through $g(x),$ and continues on until some point on $C.$ Note that, when $x in C,$ we already have $g(x) = x in C,$ so the line segment stops at $x$ itself. Let us name $h(x)$ the point in $C$ where the line segment ends. There is a fair amount of justification needed for this step: $h$ is a continuous mapping. Since $h$ takes all of $D$ to $C$ and fixes $C$ pointwise, it is a retraction. However, we have found out that there is no retraction of the disk onto the circle. This contradicts the assumption that $f(x) neq g(x)$ for every $x in D.$
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
$endgroup$
I imagine you are studying on your own, and have not done a full introductory course in point set topology. There is a standard picture that you would probably have seen...
Assume that $f(x) neq g(x)$ for every $x in D.$ For every $x,$ draw the line segment that begins at $f(x),$ goes through $g(x),$ and continues on until some point on $C.$ Note that, when $x in C,$ we already have $g(x) = x in C,$ so the line segment stops at $x$ itself. Let us name $h(x)$ the point in $C$ where the line segment ends. There is a fair amount of justification needed for this step: $h$ is a continuous mapping. Since $h$ takes all of $D$ to $C$ and fixes $C$ pointwise, it is a retraction. However, we have found out that there is no retraction of the disk onto the circle. This contradicts the assumption that $f(x) neq g(x)$ for every $x in D.$
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
answered Jun 15 '15 at 4:10
Will JagyWill Jagy
103k5102200
103k5102200
$begingroup$
I definitely have not started learning about point set topology and I do lack a great deal of mathematical maturity - I've just started purely on category theory and a long way to go. Ok, to wrap my idea around what you've said, the maps $f(x)$ and $g(x)$ do not connect except via line segment to substantiate the claim $f(x) neq g(x)$. $g(x)=x$ will occur when x becomes an element of C, whereby $h(x)$, presumably will be a fixed point. $h(x)$ as apparent from the line segments D to C is negated, hence no fixed point, hence $g(x) neq x$ and thus $f(x) = g(x)$?
$endgroup$
– Cyrus
Jun 15 '15 at 4:56
$begingroup$
@Cyrus, let me see if I can find a picture online somewhere.
$endgroup$
– Will Jagy
Jun 15 '15 at 4:59
$begingroup$
Ouch, sounds like I didnt get it right. Thanks in advance for literally giving me a clearer picture.
$endgroup$
– Cyrus
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, please see the picture on the right in en.wikipedia.org/wiki/… where their line segment starts at $f(x)$ and goes back through $x$ until the boundary circle is reached, and they call that point $F(x).$
$endgroup$
– Will Jagy
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, meanwhile, your summary above seems sort of correct, but i cannot really tell; it might be different if we were speaking in person. Rather late for me; I will chack this tomorrow to see if you had more to say.
$endgroup$
– Will Jagy
Jun 15 '15 at 5:08
|
show 1 more comment
$begingroup$
I definitely have not started learning about point set topology and I do lack a great deal of mathematical maturity - I've just started purely on category theory and a long way to go. Ok, to wrap my idea around what you've said, the maps $f(x)$ and $g(x)$ do not connect except via line segment to substantiate the claim $f(x) neq g(x)$. $g(x)=x$ will occur when x becomes an element of C, whereby $h(x)$, presumably will be a fixed point. $h(x)$ as apparent from the line segments D to C is negated, hence no fixed point, hence $g(x) neq x$ and thus $f(x) = g(x)$?
$endgroup$
– Cyrus
Jun 15 '15 at 4:56
$begingroup$
@Cyrus, let me see if I can find a picture online somewhere.
$endgroup$
– Will Jagy
Jun 15 '15 at 4:59
$begingroup$
Ouch, sounds like I didnt get it right. Thanks in advance for literally giving me a clearer picture.
$endgroup$
– Cyrus
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, please see the picture on the right in en.wikipedia.org/wiki/… where their line segment starts at $f(x)$ and goes back through $x$ until the boundary circle is reached, and they call that point $F(x).$
$endgroup$
– Will Jagy
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, meanwhile, your summary above seems sort of correct, but i cannot really tell; it might be different if we were speaking in person. Rather late for me; I will chack this tomorrow to see if you had more to say.
$endgroup$
– Will Jagy
Jun 15 '15 at 5:08
$begingroup$
I definitely have not started learning about point set topology and I do lack a great deal of mathematical maturity - I've just started purely on category theory and a long way to go. Ok, to wrap my idea around what you've said, the maps $f(x)$ and $g(x)$ do not connect except via line segment to substantiate the claim $f(x) neq g(x)$. $g(x)=x$ will occur when x becomes an element of C, whereby $h(x)$, presumably will be a fixed point. $h(x)$ as apparent from the line segments D to C is negated, hence no fixed point, hence $g(x) neq x$ and thus $f(x) = g(x)$?
$endgroup$
– Cyrus
Jun 15 '15 at 4:56
$begingroup$
I definitely have not started learning about point set topology and I do lack a great deal of mathematical maturity - I've just started purely on category theory and a long way to go. Ok, to wrap my idea around what you've said, the maps $f(x)$ and $g(x)$ do not connect except via line segment to substantiate the claim $f(x) neq g(x)$. $g(x)=x$ will occur when x becomes an element of C, whereby $h(x)$, presumably will be a fixed point. $h(x)$ as apparent from the line segments D to C is negated, hence no fixed point, hence $g(x) neq x$ and thus $f(x) = g(x)$?
$endgroup$
– Cyrus
Jun 15 '15 at 4:56
$begingroup$
@Cyrus, let me see if I can find a picture online somewhere.
$endgroup$
– Will Jagy
Jun 15 '15 at 4:59
$begingroup$
@Cyrus, let me see if I can find a picture online somewhere.
$endgroup$
– Will Jagy
Jun 15 '15 at 4:59
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Ouch, sounds like I didnt get it right. Thanks in advance for literally giving me a clearer picture.
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– Cyrus
Jun 15 '15 at 5:03
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Ouch, sounds like I didnt get it right. Thanks in advance for literally giving me a clearer picture.
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– Cyrus
Jun 15 '15 at 5:03
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@Cyrus, please see the picture on the right in en.wikipedia.org/wiki/… where their line segment starts at $f(x)$ and goes back through $x$ until the boundary circle is reached, and they call that point $F(x).$
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– Will Jagy
Jun 15 '15 at 5:03
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@Cyrus, please see the picture on the right in en.wikipedia.org/wiki/… where their line segment starts at $f(x)$ and goes back through $x$ until the boundary circle is reached, and they call that point $F(x).$
$endgroup$
– Will Jagy
Jun 15 '15 at 5:03
$begingroup$
@Cyrus, meanwhile, your summary above seems sort of correct, but i cannot really tell; it might be different if we were speaking in person. Rather late for me; I will chack this tomorrow to see if you had more to say.
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– Will Jagy
Jun 15 '15 at 5:08
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@Cyrus, meanwhile, your summary above seems sort of correct, but i cannot really tell; it might be different if we were speaking in person. Rather late for me; I will chack this tomorrow to see if you had more to say.
$endgroup$
– Will Jagy
Jun 15 '15 at 5:08
|
show 1 more comment
$begingroup$
However, by introducing a special case of the fixed point theorem,
isn't it in violation of the rule that retractions cannot exist with fixed points?
Retractions and fixed points can coexist. The book says there is no retraction between "boundary" and "disk", namely no $B^{n+1} to S^n$. This is not true for all $A subset X$, namely there exists $A subset X$ and A is a retract of X. In this case, if X has fixed point property then A has fixed point property. One can refer here for a brief proof. https://topospaces.subwiki.org/wiki/Fixed-point_property_is_retract-hereditary
answered Jan 17 at 9:03
XingtaoXingtao
11
$endgroup$
add a comment |
$begingroup$
However, by introducing a special case of the fixed point theorem,
isn't it in violation of the rule that retractions cannot exist with fixed points?
Retractions and fixed points can coexist. The book says there is no retraction between "boundary" and "disk", namely no $B^{n+1} to S^n$. This is not true for all $A subset X$, namely there exists $A subset X$ and A is a retract of X. In this case, if X has fixed point property then A has fixed point property. One can refer here for a brief proof. https://topospaces.subwiki.org/wiki/Fixed-point_property_is_retract-hereditary
answered Jan 17 at 9:03
XingtaoXingtao
11
$endgroup$
add a comment |
$begingroup$
However, by introducing a special case of the fixed point theorem,
isn't it in violation of the rule that retractions cannot exist with fixed points?
Retractions and fixed points can coexist. The book says there is no retraction between "boundary" and "disk", namely no $B^{n+1} to S^n$. This is not true for all $A subset X$, namely there exists $A subset X$ and A is a retract of X. In this case, if X has fixed point property then A has fixed point property. One can refer here for a brief proof. https://topospaces.subwiki.org/wiki/Fixed-point_property_is_retract-hereditary
answered Jan 17 at 9:03
XingtaoXingtao
11
$endgroup$
However, by introducing a special case of the fixed point theorem,
isn't it in violation of the rule that retractions cannot exist with fixed points?
Retractions and fixed points can coexist. The book says there is no retraction between "boundary" and "disk", namely no $B^{n+1} to S^n$. This is not true for all $A subset X$, namely there exists $A subset X$ and A is a retract of X. In this case, if X has fixed point property then A has fixed point property. One can refer here for a brief proof. https://topospaces.subwiki.org/wiki/Fixed-point_property_is_retract-hereditary
answered Jan 17 at 9:03
XingtaoXingtao
11
edited Jan 17 at 9:19
answered Jan 17 at 9:03
XingtaoXingtao
11
answered Jan 17 at 9:03
XingtaoXingtao
11
answered Jan 17 at 9:03
XingtaoXingtao
11
11
add a comment |
add a comment |
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