Mixing
Are the letters $O$ and $infty $ homeomorphic?
$begingroup$
I feel they are homeomorphic
But if we remove the intersection point from $infty $ we get two disconnected components, right ?
general-topology
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
$endgroup$
add a comment |
$begingroup$
I feel they are homeomorphic
But if we remove the intersection point from $infty $ we get two disconnected components, right ?
general-topology
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
$endgroup$
7
$begingroup$
This shows that feelings may be inaccurate
$endgroup$
– Hagen von Eitzen
Jan 20 at 18:06
add a comment |
$begingroup$
I feel they are homeomorphic
But if we remove the intersection point from $infty $ we get two disconnected components, right ?
general-topology
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
$endgroup$
I feel they are homeomorphic
But if we remove the intersection point from $infty $ we get two disconnected components, right ?
general-topology
general-topology
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
asked Jan 20 at 18:05
Pedro AlvarèsPedro Alvarès
636
636
7
$begingroup$
This shows that feelings may be inaccurate
$endgroup$
– Hagen von Eitzen
Jan 20 at 18:06
add a comment |
7
$begingroup$
This shows that feelings may be inaccurate
$endgroup$
– Hagen von Eitzen
Jan 20 at 18:06
7
7
$begingroup$
This shows that feelings may be inaccurate
$endgroup$
– Hagen von Eitzen
Jan 20 at 18:06
$begingroup$
This shows that feelings may be inaccurate
$endgroup$
– Hagen von Eitzen
Jan 20 at 18:06
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
You have correctly observed that your feeling is wrong: since removing a point from a figure-eight may result in a disconnected space, but removing a point from a circle never will, the two spaces are not homeomorphic.
But this isn't quite the end of the story. Why did you feel that they should be homeomorphic? And, what do the facts leading to that feeling actually imply?
I can think of two ways one might be led to believe that they are "essentially the same shape:
Artistic license. The process of drawing a figure-eight seems the same as the process of drawing a circle (you just "slightly change" the details of how the pencil moves around). What's going on here is that this process of drawing is really a continuous function (your pencil doesn't "jump") from TIME - which we can think of as $[0,1]$, in the sense that there's a start time $0$, a stop time $1$, and time is "line-like" - to the plane $mathbb{R}^2$, and the resulting curve is the image of this function. That is: both the circle and the figure-eight are the continuous image of the same space. Moreover, the continuous functions "drawing" each shape - taking the codomains as the curves being drawn, not $mathbb{R}^2$ - are "almost bijective:" each is trivially surjective, and going from $[0,1]$ to the circle we only have one non-injectivity (we wind up where we start, so $0$ and $1$ get "glued together") and going from $[0,1]$ to the figure eight we only have two non-injectivities (we wind up where we start and along the way we cross our path). But this kind of "almost homeomorphism" is quite deceptive: e.g. there is no almost-injective continuous surjection from the figure-eight to the circle, or from the circle to the line (but there is one from the circle to the figure-eight)!
- Incidentally, drawing pictures in spaces is really important, but that's a more advanced topic.
Accidentally using scissors We often think of topology as "ignoring small distortions," and so one might feel that we should be able to "unpinch" the x-point in the figure-eight to get a circle(-ish shape). However, this is not in fact a "topologically innocent" transformation - there is a sense in which it is "small," but it does change the homeomorphism type of the space.
None of the ideas above are silly - "small deviations" from nice behavior are often interesting, and later on you may see e.g. the notion of an immersion - but they do not correspond to genuine homeomorphicitude.
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
As you have realized, a useful topological invariant is the set of path components $pi_0$. A continuous map $fcolon Xrightarrow Y$ must induce a map on path components $fcolon pi_0(X)rightarrow pi_0(Y)$. This is because it respects the equivalence relation which defines path components; if $betacolon Irightarrow X$ is a path from $x$ to $x'$ (both points in $X$), then $fcirc beta colon Irightarrow Y$ is a path from $f(x)$ to $f(x')$. Importantly, homeomorphisms induce bijections on $pi_0$ (this is an easy exercise).
This is often useful in "local" considerations like the one you make in your question. Indeed, if $xin X$ then $f$ must induce a homeomorphism from $Xsetminus x$ to $Ysetminus f(x)$. But this will then induce a bijection between $pi_0 (Xsetminus x)$ and $pi_0 (Ysetminus f(x))$.
answered Jan 20 at 18:13
o.h.o.h.
4717
$endgroup$
add a comment |
$begingroup$
No, they are not. The fundamental group of "O" is isomorphic to the group of integers. However, using Van Kampen's theorem, one can see that the fundamental group of the figure eight is isomorphic to the free product on two sets of integers: Z*Z. The first group is abelian, while the second is not. Therefore they do not have isomorphic fundamental groups and the spaces cannot be homeomorphic.
Another proof - using only point-set topology: Assume they are homeomorphic. Then the figure eight minus the center point p must be homeomorphic to the circle O minus some point q. However, removing the point from the figure eight results in a non-connected space, while the circle minus a point is still connected. Contradiction, as homeomorphism preserves connectedness.
answered Jan 21 at 1:29
MH1992MH1992
212
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
You have correctly observed that your feeling is wrong: since removing a point from a figure-eight may result in a disconnected space, but removing a point from a circle never will, the two spaces are not homeomorphic.
But this isn't quite the end of the story. Why did you feel that they should be homeomorphic? And, what do the facts leading to that feeling actually imply?
I can think of two ways one might be led to believe that they are "essentially the same shape:
Artistic license. The process of drawing a figure-eight seems the same as the process of drawing a circle (you just "slightly change" the details of how the pencil moves around). What's going on here is that this process of drawing is really a continuous function (your pencil doesn't "jump") from TIME - which we can think of as $[0,1]$, in the sense that there's a start time $0$, a stop time $1$, and time is "line-like" - to the plane $mathbb{R}^2$, and the resulting curve is the image of this function. That is: both the circle and the figure-eight are the continuous image of the same space. Moreover, the continuous functions "drawing" each shape - taking the codomains as the curves being drawn, not $mathbb{R}^2$ - are "almost bijective:" each is trivially surjective, and going from $[0,1]$ to the circle we only have one non-injectivity (we wind up where we start, so $0$ and $1$ get "glued together") and going from $[0,1]$ to the figure eight we only have two non-injectivities (we wind up where we start and along the way we cross our path). But this kind of "almost homeomorphism" is quite deceptive: e.g. there is no almost-injective continuous surjection from the figure-eight to the circle, or from the circle to the line (but there is one from the circle to the figure-eight)!
- Incidentally, drawing pictures in spaces is really important, but that's a more advanced topic.
Accidentally using scissors We often think of topology as "ignoring small distortions," and so one might feel that we should be able to "unpinch" the x-point in the figure-eight to get a circle(-ish shape). However, this is not in fact a "topologically innocent" transformation - there is a sense in which it is "small," but it does change the homeomorphism type of the space.
None of the ideas above are silly - "small deviations" from nice behavior are often interesting, and later on you may see e.g. the notion of an immersion - but they do not correspond to genuine homeomorphicitude.
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
You have correctly observed that your feeling is wrong: since removing a point from a figure-eight may result in a disconnected space, but removing a point from a circle never will, the two spaces are not homeomorphic.
But this isn't quite the end of the story. Why did you feel that they should be homeomorphic? And, what do the facts leading to that feeling actually imply?
I can think of two ways one might be led to believe that they are "essentially the same shape:
Artistic license. The process of drawing a figure-eight seems the same as the process of drawing a circle (you just "slightly change" the details of how the pencil moves around). What's going on here is that this process of drawing is really a continuous function (your pencil doesn't "jump") from TIME - which we can think of as $[0,1]$, in the sense that there's a start time $0$, a stop time $1$, and time is "line-like" - to the plane $mathbb{R}^2$, and the resulting curve is the image of this function. That is: both the circle and the figure-eight are the continuous image of the same space. Moreover, the continuous functions "drawing" each shape - taking the codomains as the curves being drawn, not $mathbb{R}^2$ - are "almost bijective:" each is trivially surjective, and going from $[0,1]$ to the circle we only have one non-injectivity (we wind up where we start, so $0$ and $1$ get "glued together") and going from $[0,1]$ to the figure eight we only have two non-injectivities (we wind up where we start and along the way we cross our path). But this kind of "almost homeomorphism" is quite deceptive: e.g. there is no almost-injective continuous surjection from the figure-eight to the circle, or from the circle to the line (but there is one from the circle to the figure-eight)!
- Incidentally, drawing pictures in spaces is really important, but that's a more advanced topic.
Accidentally using scissors We often think of topology as "ignoring small distortions," and so one might feel that we should be able to "unpinch" the x-point in the figure-eight to get a circle(-ish shape). However, this is not in fact a "topologically innocent" transformation - there is a sense in which it is "small," but it does change the homeomorphism type of the space.
None of the ideas above are silly - "small deviations" from nice behavior are often interesting, and later on you may see e.g. the notion of an immersion - but they do not correspond to genuine homeomorphicitude.
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
$endgroup$
add a comment |
$begingroup$
You have correctly observed that your feeling is wrong: since removing a point from a figure-eight may result in a disconnected space, but removing a point from a circle never will, the two spaces are not homeomorphic.
But this isn't quite the end of the story. Why did you feel that they should be homeomorphic? And, what do the facts leading to that feeling actually imply?
I can think of two ways one might be led to believe that they are "essentially the same shape:
Artistic license. The process of drawing a figure-eight seems the same as the process of drawing a circle (you just "slightly change" the details of how the pencil moves around). What's going on here is that this process of drawing is really a continuous function (your pencil doesn't "jump") from TIME - which we can think of as $[0,1]$, in the sense that there's a start time $0$, a stop time $1$, and time is "line-like" - to the plane $mathbb{R}^2$, and the resulting curve is the image of this function. That is: both the circle and the figure-eight are the continuous image of the same space. Moreover, the continuous functions "drawing" each shape - taking the codomains as the curves being drawn, not $mathbb{R}^2$ - are "almost bijective:" each is trivially surjective, and going from $[0,1]$ to the circle we only have one non-injectivity (we wind up where we start, so $0$ and $1$ get "glued together") and going from $[0,1]$ to the figure eight we only have two non-injectivities (we wind up where we start and along the way we cross our path). But this kind of "almost homeomorphism" is quite deceptive: e.g. there is no almost-injective continuous surjection from the figure-eight to the circle, or from the circle to the line (but there is one from the circle to the figure-eight)!
- Incidentally, drawing pictures in spaces is really important, but that's a more advanced topic.
Accidentally using scissors We often think of topology as "ignoring small distortions," and so one might feel that we should be able to "unpinch" the x-point in the figure-eight to get a circle(-ish shape). However, this is not in fact a "topologically innocent" transformation - there is a sense in which it is "small," but it does change the homeomorphism type of the space.
None of the ideas above are silly - "small deviations" from nice behavior are often interesting, and later on you may see e.g. the notion of an immersion - but they do not correspond to genuine homeomorphicitude.
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
$endgroup$
You have correctly observed that your feeling is wrong: since removing a point from a figure-eight may result in a disconnected space, but removing a point from a circle never will, the two spaces are not homeomorphic.
But this isn't quite the end of the story. Why did you feel that they should be homeomorphic? And, what do the facts leading to that feeling actually imply?
I can think of two ways one might be led to believe that they are "essentially the same shape:
Artistic license. The process of drawing a figure-eight seems the same as the process of drawing a circle (you just "slightly change" the details of how the pencil moves around). What's going on here is that this process of drawing is really a continuous function (your pencil doesn't "jump") from TIME - which we can think of as $[0,1]$, in the sense that there's a start time $0$, a stop time $1$, and time is "line-like" - to the plane $mathbb{R}^2$, and the resulting curve is the image of this function. That is: both the circle and the figure-eight are the continuous image of the same space. Moreover, the continuous functions "drawing" each shape - taking the codomains as the curves being drawn, not $mathbb{R}^2$ - are "almost bijective:" each is trivially surjective, and going from $[0,1]$ to the circle we only have one non-injectivity (we wind up where we start, so $0$ and $1$ get "glued together") and going from $[0,1]$ to the figure eight we only have two non-injectivities (we wind up where we start and along the way we cross our path). But this kind of "almost homeomorphism" is quite deceptive: e.g. there is no almost-injective continuous surjection from the figure-eight to the circle, or from the circle to the line (but there is one from the circle to the figure-eight)!
- Incidentally, drawing pictures in spaces is really important, but that's a more advanced topic.
Accidentally using scissors We often think of topology as "ignoring small distortions," and so one might feel that we should be able to "unpinch" the x-point in the figure-eight to get a circle(-ish shape). However, this is not in fact a "topologically innocent" transformation - there is a sense in which it is "small," but it does change the homeomorphism type of the space.
None of the ideas above are silly - "small deviations" from nice behavior are often interesting, and later on you may see e.g. the notion of an immersion - but they do not correspond to genuine homeomorphicitude.
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
edited Jan 20 at 18:34
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
answered Jan 20 at 18:26
Noah SchweberNoah Schweber
126k10151290
126k10151290
add a comment |
add a comment |
$begingroup$
As you have realized, a useful topological invariant is the set of path components $pi_0$. A continuous map $fcolon Xrightarrow Y$ must induce a map on path components $fcolon pi_0(X)rightarrow pi_0(Y)$. This is because it respects the equivalence relation which defines path components; if $betacolon Irightarrow X$ is a path from $x$ to $x'$ (both points in $X$), then $fcirc beta colon Irightarrow Y$ is a path from $f(x)$ to $f(x')$. Importantly, homeomorphisms induce bijections on $pi_0$ (this is an easy exercise).
This is often useful in "local" considerations like the one you make in your question. Indeed, if $xin X$ then $f$ must induce a homeomorphism from $Xsetminus x$ to $Ysetminus f(x)$. But this will then induce a bijection between $pi_0 (Xsetminus x)$ and $pi_0 (Ysetminus f(x))$.
answered Jan 20 at 18:13
o.h.o.h.
4717
$endgroup$
add a comment |
$begingroup$
As you have realized, a useful topological invariant is the set of path components $pi_0$. A continuous map $fcolon Xrightarrow Y$ must induce a map on path components $fcolon pi_0(X)rightarrow pi_0(Y)$. This is because it respects the equivalence relation which defines path components; if $betacolon Irightarrow X$ is a path from $x$ to $x'$ (both points in $X$), then $fcirc beta colon Irightarrow Y$ is a path from $f(x)$ to $f(x')$. Importantly, homeomorphisms induce bijections on $pi_0$ (this is an easy exercise).
This is often useful in "local" considerations like the one you make in your question. Indeed, if $xin X$ then $f$ must induce a homeomorphism from $Xsetminus x$ to $Ysetminus f(x)$. But this will then induce a bijection between $pi_0 (Xsetminus x)$ and $pi_0 (Ysetminus f(x))$.
answered Jan 20 at 18:13
o.h.o.h.
4717
$endgroup$
add a comment |
$begingroup$
As you have realized, a useful topological invariant is the set of path components $pi_0$. A continuous map $fcolon Xrightarrow Y$ must induce a map on path components $fcolon pi_0(X)rightarrow pi_0(Y)$. This is because it respects the equivalence relation which defines path components; if $betacolon Irightarrow X$ is a path from $x$ to $x'$ (both points in $X$), then $fcirc beta colon Irightarrow Y$ is a path from $f(x)$ to $f(x')$. Importantly, homeomorphisms induce bijections on $pi_0$ (this is an easy exercise).
This is often useful in "local" considerations like the one you make in your question. Indeed, if $xin X$ then $f$ must induce a homeomorphism from $Xsetminus x$ to $Ysetminus f(x)$. But this will then induce a bijection between $pi_0 (Xsetminus x)$ and $pi_0 (Ysetminus f(x))$.
answered Jan 20 at 18:13
o.h.o.h.
4717
$endgroup$
As you have realized, a useful topological invariant is the set of path components $pi_0$. A continuous map $fcolon Xrightarrow Y$ must induce a map on path components $fcolon pi_0(X)rightarrow pi_0(Y)$. This is because it respects the equivalence relation which defines path components; if $betacolon Irightarrow X$ is a path from $x$ to $x'$ (both points in $X$), then $fcirc beta colon Irightarrow Y$ is a path from $f(x)$ to $f(x')$. Importantly, homeomorphisms induce bijections on $pi_0$ (this is an easy exercise).
This is often useful in "local" considerations like the one you make in your question. Indeed, if $xin X$ then $f$ must induce a homeomorphism from $Xsetminus x$ to $Ysetminus f(x)$. But this will then induce a bijection between $pi_0 (Xsetminus x)$ and $pi_0 (Ysetminus f(x))$.
answered Jan 20 at 18:13
o.h.o.h.
4717
answered Jan 20 at 18:13
o.h.o.h.
4717
answered Jan 20 at 18:13
o.h.o.h.
4717
answered Jan 20 at 18:13
o.h.o.h.
4717
4717
add a comment |
add a comment |
$begingroup$
No, they are not. The fundamental group of "O" is isomorphic to the group of integers. However, using Van Kampen's theorem, one can see that the fundamental group of the figure eight is isomorphic to the free product on two sets of integers: Z*Z. The first group is abelian, while the second is not. Therefore they do not have isomorphic fundamental groups and the spaces cannot be homeomorphic.
Another proof - using only point-set topology: Assume they are homeomorphic. Then the figure eight minus the center point p must be homeomorphic to the circle O minus some point q. However, removing the point from the figure eight results in a non-connected space, while the circle minus a point is still connected. Contradiction, as homeomorphism preserves connectedness.
answered Jan 21 at 1:29
MH1992MH1992
212
$endgroup$
add a comment |
$begingroup$
No, they are not. The fundamental group of "O" is isomorphic to the group of integers. However, using Van Kampen's theorem, one can see that the fundamental group of the figure eight is isomorphic to the free product on two sets of integers: Z*Z. The first group is abelian, while the second is not. Therefore they do not have isomorphic fundamental groups and the spaces cannot be homeomorphic.
Another proof - using only point-set topology: Assume they are homeomorphic. Then the figure eight minus the center point p must be homeomorphic to the circle O minus some point q. However, removing the point from the figure eight results in a non-connected space, while the circle minus a point is still connected. Contradiction, as homeomorphism preserves connectedness.
answered Jan 21 at 1:29
MH1992MH1992
212
$endgroup$
add a comment |
$begingroup$
No, they are not. The fundamental group of "O" is isomorphic to the group of integers. However, using Van Kampen's theorem, one can see that the fundamental group of the figure eight is isomorphic to the free product on two sets of integers: Z*Z. The first group is abelian, while the second is not. Therefore they do not have isomorphic fundamental groups and the spaces cannot be homeomorphic.
Another proof - using only point-set topology: Assume they are homeomorphic. Then the figure eight minus the center point p must be homeomorphic to the circle O minus some point q. However, removing the point from the figure eight results in a non-connected space, while the circle minus a point is still connected. Contradiction, as homeomorphism preserves connectedness.
answered Jan 21 at 1:29
MH1992MH1992
212
$endgroup$
No, they are not. The fundamental group of "O" is isomorphic to the group of integers. However, using Van Kampen's theorem, one can see that the fundamental group of the figure eight is isomorphic to the free product on two sets of integers: Z*Z. The first group is abelian, while the second is not. Therefore they do not have isomorphic fundamental groups and the spaces cannot be homeomorphic.
Another proof - using only point-set topology: Assume they are homeomorphic. Then the figure eight minus the center point p must be homeomorphic to the circle O minus some point q. However, removing the point from the figure eight results in a non-connected space, while the circle minus a point is still connected. Contradiction, as homeomorphism preserves connectedness.
answered Jan 21 at 1:29
MH1992MH1992
212
answered Jan 21 at 1:29
MH1992MH1992
212
answered Jan 21 at 1:29
MH1992MH1992
212
answered Jan 21 at 1:29
MH1992MH1992
212
212
add a comment |
add a comment |
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7
$begingroup$
This shows that feelings may be inaccurate
$endgroup$
– Hagen von Eitzen
Jan 20 at 18:06