Fundamental Group of Polygon
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I have a question. I am trying to calculate $pi _{1}(X)$ where X is a polygon as you see in the picture such that $p_{1}$ and $p_{2}$ two punctures by identifying edges.
enter image description here
Here is my idea. I think, if I identify edges I will have 2-Sphere with two punctured points. One point punctured sphere homotopy equivalent to $mathbb{R}^{2}$, then by using the second one, it is homotopy equivalent to $S^{1}$. And finally, I will have $pi _{1}(X)=mathbb{Z}$. Am I right?
Thank you for your help.
algebraic-topology
asked Nov 25 '18 at 17:27
user619499user619499
214
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add a comment |
$begingroup$
I have a question. I am trying to calculate $pi _{1}(X)$ where X is a polygon as you see in the picture such that $p_{1}$ and $p_{2}$ two punctures by identifying edges.
enter image description here
Here is my idea. I think, if I identify edges I will have 2-Sphere with two punctured points. One point punctured sphere homotopy equivalent to $mathbb{R}^{2}$, then by using the second one, it is homotopy equivalent to $S^{1}$. And finally, I will have $pi _{1}(X)=mathbb{Z}$. Am I right?
Thank you for your help.
algebraic-topology
asked Nov 25 '18 at 17:27
user619499user619499
214
$endgroup$
$begingroup$
Yes, the space is homeomorphic to a sphere minus two points, which has fundamental group $mathbb Z$ by your argument.
$endgroup$
– Cheerful Parsnip
Nov 25 '18 at 17:30
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Yes, you are right. Except that for spaces I think it is more common to use the phrase "homotopy equivalent" rather than the word "homotopic" (which is reserved for two mappings).
$endgroup$
– Mees de Vries
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Thank you for the correction.
$endgroup$
– user619499
Nov 25 '18 at 17:32
add a comment |
$begingroup$
I have a question. I am trying to calculate $pi _{1}(X)$ where X is a polygon as you see in the picture such that $p_{1}$ and $p_{2}$ two punctures by identifying edges.
enter image description here
Here is my idea. I think, if I identify edges I will have 2-Sphere with two punctured points. One point punctured sphere homotopy equivalent to $mathbb{R}^{2}$, then by using the second one, it is homotopy equivalent to $S^{1}$. And finally, I will have $pi _{1}(X)=mathbb{Z}$. Am I right?
Thank you for your help.
algebraic-topology
asked Nov 25 '18 at 17:27
user619499user619499
214
$endgroup$
I have a question. I am trying to calculate $pi _{1}(X)$ where X is a polygon as you see in the picture such that $p_{1}$ and $p_{2}$ two punctures by identifying edges.
enter image description here
Here is my idea. I think, if I identify edges I will have 2-Sphere with two punctured points. One point punctured sphere homotopy equivalent to $mathbb{R}^{2}$, then by using the second one, it is homotopy equivalent to $S^{1}$. And finally, I will have $pi _{1}(X)=mathbb{Z}$. Am I right?
Thank you for your help.
algebraic-topology
algebraic-topology
asked Nov 25 '18 at 17:27
user619499user619499
214
asked Nov 25 '18 at 17:27
user619499user619499
214
edited Nov 25 '18 at 17:33
user619499
asked Nov 25 '18 at 17:27
user619499user619499
214
asked Nov 25 '18 at 17:27
user619499user619499
214
asked Nov 25 '18 at 17:27
user619499user619499
214
214
$begingroup$
Yes, the space is homeomorphic to a sphere minus two points, which has fundamental group $mathbb Z$ by your argument.
$endgroup$
– Cheerful Parsnip
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Except that for spaces I think it is more common to use the phrase "homotopy equivalent" rather than the word "homotopic" (which is reserved for two mappings).
$endgroup$
– Mees de Vries
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Thank you for the correction.
$endgroup$
– user619499
Nov 25 '18 at 17:32
add a comment |
$begingroup$
Yes, the space is homeomorphic to a sphere minus two points, which has fundamental group $mathbb Z$ by your argument.
$endgroup$
– Cheerful Parsnip
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Except that for spaces I think it is more common to use the phrase "homotopy equivalent" rather than the word "homotopic" (which is reserved for two mappings).
$endgroup$
– Mees de Vries
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Thank you for the correction.
$endgroup$
– user619499
Nov 25 '18 at 17:32
$begingroup$
Yes, the space is homeomorphic to a sphere minus two points, which has fundamental group $mathbb Z$ by your argument.
$endgroup$
– Cheerful Parsnip
Nov 25 '18 at 17:30
$begingroup$
Yes, the space is homeomorphic to a sphere minus two points, which has fundamental group $mathbb Z$ by your argument.
$endgroup$
– Cheerful Parsnip
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Except that for spaces I think it is more common to use the phrase "homotopy equivalent" rather than the word "homotopic" (which is reserved for two mappings).
$endgroup$
– Mees de Vries
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Except that for spaces I think it is more common to use the phrase "homotopy equivalent" rather than the word "homotopic" (which is reserved for two mappings).
$endgroup$
– Mees de Vries
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Thank you for the correction.
$endgroup$
– user619499
Nov 25 '18 at 17:32
$begingroup$
Yes, you are right. Thank you for the correction.
$endgroup$
– user619499
Nov 25 '18 at 17:32
add a comment |
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$begingroup$
Yes, the space is homeomorphic to a sphere minus two points, which has fundamental group $mathbb Z$ by your argument.
$endgroup$
– Cheerful Parsnip
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Except that for spaces I think it is more common to use the phrase "homotopy equivalent" rather than the word "homotopic" (which is reserved for two mappings).
$endgroup$
– Mees de Vries
Nov 25 '18 at 17:30
$begingroup$
Yes, you are right. Thank you for the correction.
$endgroup$
– user619499
Nov 25 '18 at 17:32