fundamental group of quotient space
$begingroup$
I found this exercise but I can't do it.
the text says:
Consider the quotient space $X = T^2/sim$, where $T^2 = S^1times S^1$ is the $2$-dimensional torus and this $sim$ is equivalence relation which identifies two distinct points $p$, $q$ of $T^2$. Prove that fundamental group of $X$ is $(mathbb Ztimesmathbb Z)*mathbb Z$.
algebraic-topology
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
asked Dec 12 '18 at 22:44
MAXMAX
94
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add a comment |
$begingroup$
I found this exercise but I can't do it.
the text says:
Consider the quotient space $X = T^2/sim$, where $T^2 = S^1times S^1$ is the $2$-dimensional torus and this $sim$ is equivalence relation which identifies two distinct points $p$, $q$ of $T^2$. Prove that fundamental group of $X$ is $(mathbb Ztimesmathbb Z)*mathbb Z$.
algebraic-topology
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
asked Dec 12 '18 at 22:44
MAXMAX
94
$endgroup$
add a comment |
$begingroup$
I found this exercise but I can't do it.
the text says:
Consider the quotient space $X = T^2/sim$, where $T^2 = S^1times S^1$ is the $2$-dimensional torus and this $sim$ is equivalence relation which identifies two distinct points $p$, $q$ of $T^2$. Prove that fundamental group of $X$ is $(mathbb Ztimesmathbb Z)*mathbb Z$.
algebraic-topology
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
asked Dec 12 '18 at 22:44
MAXMAX
94
$endgroup$
I found this exercise but I can't do it.
the text says:
Consider the quotient space $X = T^2/sim$, where $T^2 = S^1times S^1$ is the $2$-dimensional torus and this $sim$ is equivalence relation which identifies two distinct points $p$, $q$ of $T^2$. Prove that fundamental group of $X$ is $(mathbb Ztimesmathbb Z)*mathbb Z$.
algebraic-topology
algebraic-topology
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
asked Dec 12 '18 at 22:44
MAXMAX
94
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
asked Dec 12 '18 at 22:44
MAXMAX
94
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
edited Dec 13 '18 at 0:43
Aweygan
14.7k21442
14.7k21442
asked Dec 12 '18 at 22:44
MAXMAX
94
asked Dec 12 '18 at 22:44
MAXMAX
94
asked Dec 12 '18 at 22:44
MAXMAX
94
94
add a comment |
add a comment |
1 Answer
1
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$begingroup$
HINT
Note that $X$ looks like $T^2$ with two points pinched together. This is homotopy equivalent to the space $Y$ which is $T^2$ with a path connecting the points $p$ and $q$ added, which is then homotopy equivalent to $T^2vee S^1$. Then apply the Seifert-van Kampen theorem, or use a well known corollary if you're familiar with it.
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
$endgroup$
$begingroup$
Ok, thank you :)
$endgroup$
– MAX
Dec 13 '18 at 8:22
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
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votes
$begingroup$
HINT
Note that $X$ looks like $T^2$ with two points pinched together. This is homotopy equivalent to the space $Y$ which is $T^2$ with a path connecting the points $p$ and $q$ added, which is then homotopy equivalent to $T^2vee S^1$. Then apply the Seifert-van Kampen theorem, or use a well known corollary if you're familiar with it.
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
$endgroup$
$begingroup$
Ok, thank you :)
$endgroup$
– MAX
Dec 13 '18 at 8:22
add a comment |
$begingroup$
HINT
Note that $X$ looks like $T^2$ with two points pinched together. This is homotopy equivalent to the space $Y$ which is $T^2$ with a path connecting the points $p$ and $q$ added, which is then homotopy equivalent to $T^2vee S^1$. Then apply the Seifert-van Kampen theorem, or use a well known corollary if you're familiar with it.
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
$endgroup$
$begingroup$
Ok, thank you :)
$endgroup$
– MAX
Dec 13 '18 at 8:22
add a comment |
$begingroup$
HINT
Note that $X$ looks like $T^2$ with two points pinched together. This is homotopy equivalent to the space $Y$ which is $T^2$ with a path connecting the points $p$ and $q$ added, which is then homotopy equivalent to $T^2vee S^1$. Then apply the Seifert-van Kampen theorem, or use a well known corollary if you're familiar with it.
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
$endgroup$
HINT
Note that $X$ looks like $T^2$ with two points pinched together. This is homotopy equivalent to the space $Y$ which is $T^2$ with a path connecting the points $p$ and $q$ added, which is then homotopy equivalent to $T^2vee S^1$. Then apply the Seifert-van Kampen theorem, or use a well known corollary if you're familiar with it.
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
answered Dec 13 '18 at 0:50
AweyganAweygan
14.7k21442
14.7k21442
$begingroup$
Ok, thank you :)
$endgroup$
– MAX
Dec 13 '18 at 8:22
add a comment |
$begingroup$
Ok, thank you :)
$endgroup$
– MAX
Dec 13 '18 at 8:22
$begingroup$
Ok, thank you :)
$endgroup$
– MAX
Dec 13 '18 at 8:22
$begingroup$
Ok, thank you :)
$endgroup$
– MAX
Dec 13 '18 at 8:22
add a comment |
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