Real plane with two holes and circumference are not homotopy equivalent Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalentSpaces homotopy equivalent to $mathbbR^2$ minus two pointsAre these two spaces homotopy equivalent?Non-homotopy equivalent spaces with isomorphic fundamental groupsWhich of these are homotopy equivalent? $S^1, mathbbR, *$$mathbbR^2$ and $mathbbR times [0, +infty]$ are homotopy equivalent, but not homeomorphicProve that the Torus is not homotopy equivalent to $S^1vee S^1vee S^2$Infinite wedge sum of circles and a space homotopy equivalentExample of two spaces with same cohomology ring but different homotopy groupInfinite dimensional CW complexes with nontrivial reduced homology are not homotopy equivalent to finite dimensional ones?
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Real plane with two holes and circumference are not homotopy equivalent
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Two CW complexes with isomorphic homotopy groups and homology, yet not homotopy equivalentSpaces homotopy equivalent to $mathbbR^2$ minus two pointsAre these two spaces homotopy equivalent?Non-homotopy equivalent spaces with isomorphic fundamental groupsWhich of these are homotopy equivalent? $S^1, mathbbR, *$$mathbbR^2$ and $mathbbR times [0, +infty]$ are homotopy equivalent, but not homeomorphicProve that the Torus is not homotopy equivalent to $S^1vee S^1vee S^2$Infinite wedge sum of circles and a space homotopy equivalentExample of two spaces with same cohomology ring but different homotopy groupInfinite dimensional CW complexes with nontrivial reduced homology are not homotopy equivalent to finite dimensional ones?
$begingroup$
I´d like to obtain an argument to prove that the real plane with two holes, for example $mathbbR setminus p,q$ is not homotopy equivalent to the circumference $S^1$.
I know they have different number of holes, but I´d like an argument (without using homology, but homotopy or fundamental group is valid!).
Thanks and regards!
algebraic-topology homotopy-theory fundamental-groups
asked Mar 27 at 17:49
user183002user183002
312
$endgroup$
add a comment |
$begingroup$
I´d like to obtain an argument to prove that the real plane with two holes, for example $mathbbR setminus p,q$ is not homotopy equivalent to the circumference $S^1$.
I know they have different number of holes, but I´d like an argument (without using homology, but homotopy or fundamental group is valid!).
Thanks and regards!
algebraic-topology homotopy-theory fundamental-groups
asked Mar 27 at 17:49
user183002user183002
312
$endgroup$
add a comment |
$begingroup$
I´d like to obtain an argument to prove that the real plane with two holes, for example $mathbbR setminus p,q$ is not homotopy equivalent to the circumference $S^1$.
I know they have different number of holes, but I´d like an argument (without using homology, but homotopy or fundamental group is valid!).
Thanks and regards!
algebraic-topology homotopy-theory fundamental-groups
asked Mar 27 at 17:49
user183002user183002
312
$endgroup$
I´d like to obtain an argument to prove that the real plane with two holes, for example $mathbbR setminus p,q$ is not homotopy equivalent to the circumference $S^1$.
I know they have different number of holes, but I´d like an argument (without using homology, but homotopy or fundamental group is valid!).
Thanks and regards!
algebraic-topology homotopy-theory fundamental-groups
algebraic-topology homotopy-theory fundamental-groups
asked Mar 27 at 17:49
user183002user183002
312
asked Mar 27 at 17:49
user183002user183002
312
asked Mar 27 at 17:49
user183002user183002
312
asked Mar 27 at 17:49
user183002user183002
312
asked Mar 27 at 17:49
user183002user183002
312
312
add a comment |
add a comment |
1 Answer
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I got it! (I think). This is an argument:
$mathbbR setminus p,q$ is homotopy equivalent to the eight space (indeed, it is a deformation retract). So they have same fundamental group, that is $mathbbZ * mathbbZ$, the free group with two generators.
Since the $pi(S^1) = mathbbZ$, they can´t be homotopy equivalent.
Now, just a question for the comments. You know $mathbbZ = <x : emptyset >$, but how you write $mathbbZ * mathbbZ$?
answered Mar 27 at 20:21
user183002user183002
312
$endgroup$
1
$begingroup$
$mathbbZastmathbbZ = langle x,y: emptysetrangle,$ or, just $mathbbZastmathbbZ = langle x,yrangle.$ (Also, note the use of langle and rangle to make $langle$ and $rangle$, which is prettier than using $<$ and $>$.)
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– Jason DeVito
Mar 27 at 21:05
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I got it! (I think). This is an argument:
$mathbbR setminus p,q$ is homotopy equivalent to the eight space (indeed, it is a deformation retract). So they have same fundamental group, that is $mathbbZ * mathbbZ$, the free group with two generators.
Since the $pi(S^1) = mathbbZ$, they can´t be homotopy equivalent.
Now, just a question for the comments. You know $mathbbZ = <x : emptyset >$, but how you write $mathbbZ * mathbbZ$?
answered Mar 27 at 20:21
user183002user183002
312
$endgroup$
1
$begingroup$
$mathbbZastmathbbZ = langle x,y: emptysetrangle,$ or, just $mathbbZastmathbbZ = langle x,yrangle.$ (Also, note the use of langle and rangle to make $langle$ and $rangle$, which is prettier than using $<$ and $>$.)
$endgroup$
– Jason DeVito
Mar 27 at 21:05
add a comment |
$begingroup$
I got it! (I think). This is an argument:
$mathbbR setminus p,q$ is homotopy equivalent to the eight space (indeed, it is a deformation retract). So they have same fundamental group, that is $mathbbZ * mathbbZ$, the free group with two generators.
Since the $pi(S^1) = mathbbZ$, they can´t be homotopy equivalent.
Now, just a question for the comments. You know $mathbbZ = <x : emptyset >$, but how you write $mathbbZ * mathbbZ$?
answered Mar 27 at 20:21
user183002user183002
312
$endgroup$
1
$begingroup$
$mathbbZastmathbbZ = langle x,y: emptysetrangle,$ or, just $mathbbZastmathbbZ = langle x,yrangle.$ (Also, note the use of langle and rangle to make $langle$ and $rangle$, which is prettier than using $<$ and $>$.)
$endgroup$
– Jason DeVito
Mar 27 at 21:05
add a comment |
$begingroup$
I got it! (I think). This is an argument:
$mathbbR setminus p,q$ is homotopy equivalent to the eight space (indeed, it is a deformation retract). So they have same fundamental group, that is $mathbbZ * mathbbZ$, the free group with two generators.
Since the $pi(S^1) = mathbbZ$, they can´t be homotopy equivalent.
Now, just a question for the comments. You know $mathbbZ = <x : emptyset >$, but how you write $mathbbZ * mathbbZ$?
answered Mar 27 at 20:21
user183002user183002
312
$endgroup$
I got it! (I think). This is an argument:
$mathbbR setminus p,q$ is homotopy equivalent to the eight space (indeed, it is a deformation retract). So they have same fundamental group, that is $mathbbZ * mathbbZ$, the free group with two generators.
Since the $pi(S^1) = mathbbZ$, they can´t be homotopy equivalent.
Now, just a question for the comments. You know $mathbbZ = <x : emptyset >$, but how you write $mathbbZ * mathbbZ$?
answered Mar 27 at 20:21
user183002user183002
312
answered Mar 27 at 20:21
user183002user183002
312
answered Mar 27 at 20:21
user183002user183002
312
answered Mar 27 at 20:21
user183002user183002
312
312
1
$begingroup$
$mathbbZastmathbbZ = langle x,y: emptysetrangle,$ or, just $mathbbZastmathbbZ = langle x,yrangle.$ (Also, note the use of langle and rangle to make $langle$ and $rangle$, which is prettier than using $<$ and $>$.)
$endgroup$
– Jason DeVito
Mar 27 at 21:05
add a comment |
1
$begingroup$
$mathbbZastmathbbZ = langle x,y: emptysetrangle,$ or, just $mathbbZastmathbbZ = langle x,yrangle.$ (Also, note the use of langle and rangle to make $langle$ and $rangle$, which is prettier than using $<$ and $>$.)
$endgroup$
– Jason DeVito
Mar 27 at 21:05
1
1
$begingroup$
$mathbbZastmathbbZ = langle x,y: emptysetrangle,$ or, just $mathbbZastmathbbZ = langle x,yrangle.$ (Also, note the use of langle and rangle to make $langle$ and $rangle$, which is prettier than using $<$ and $>$.)
$endgroup$
– Jason DeVito
Mar 27 at 21:05
$begingroup$
$mathbbZastmathbbZ = langle x,y: emptysetrangle,$ or, just $mathbbZastmathbbZ = langle x,yrangle.$ (Also, note the use of langle and rangle to make $langle$ and $rangle$, which is prettier than using $<$ and $>$.)
$endgroup$
– Jason DeVito
Mar 27 at 21:05
add a comment |
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