The fundamental group of $(S^1times S^1)/(S^1times {x})$
$begingroup$
What is the fundamental group of $(S^1times S^1)/((S^1times{x})$ where $x$ is a point in $S^1$? My guess is that $S^1$ is a deformation retract of $(S^1times S^1)/(S^1times {x})$. Thus $pi_1(S^1times S^1/S^1times {x})congpi_1(S^1)congmathbb{Z}$. At the same time, I know that the torus $S^1times S^1$ does not deformation retract to $S^1$ (since their fundamental groups are different). If the above claim is true, how identifying all points of $S^1times {x}$ with a single point in $(S^1times S^1)$ makes it possible to deformation retracts $(S^1times S^1)(/(S^1times {x})$ onto $S^1$?
algebraic-topology fundamental-groups
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
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$begingroup$
What is the fundamental group of $(S^1times S^1)/((S^1times{x})$ where $x$ is a point in $S^1$? My guess is that $S^1$ is a deformation retract of $(S^1times S^1)/(S^1times {x})$. Thus $pi_1(S^1times S^1/S^1times {x})congpi_1(S^1)congmathbb{Z}$. At the same time, I know that the torus $S^1times S^1$ does not deformation retract to $S^1$ (since their fundamental groups are different). If the above claim is true, how identifying all points of $S^1times {x}$ with a single point in $(S^1times S^1)$ makes it possible to deformation retracts $(S^1times S^1)(/(S^1times {x})$ onto $S^1$?
algebraic-topology fundamental-groups
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
$endgroup$
add a comment |
$begingroup$
What is the fundamental group of $(S^1times S^1)/((S^1times{x})$ where $x$ is a point in $S^1$? My guess is that $S^1$ is a deformation retract of $(S^1times S^1)/(S^1times {x})$. Thus $pi_1(S^1times S^1/S^1times {x})congpi_1(S^1)congmathbb{Z}$. At the same time, I know that the torus $S^1times S^1$ does not deformation retract to $S^1$ (since their fundamental groups are different). If the above claim is true, how identifying all points of $S^1times {x}$ with a single point in $(S^1times S^1)$ makes it possible to deformation retracts $(S^1times S^1)(/(S^1times {x})$ onto $S^1$?
algebraic-topology fundamental-groups
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
$endgroup$
What is the fundamental group of $(S^1times S^1)/((S^1times{x})$ where $x$ is a point in $S^1$? My guess is that $S^1$ is a deformation retract of $(S^1times S^1)/(S^1times {x})$. Thus $pi_1(S^1times S^1/S^1times {x})congpi_1(S^1)congmathbb{Z}$. At the same time, I know that the torus $S^1times S^1$ does not deformation retract to $S^1$ (since their fundamental groups are different). If the above claim is true, how identifying all points of $S^1times {x}$ with a single point in $(S^1times S^1)$ makes it possible to deformation retracts $(S^1times S^1)(/(S^1times {x})$ onto $S^1$?
algebraic-topology fundamental-groups
algebraic-topology fundamental-groups
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
asked Jan 8 at 2:56
gladimetcampbellsgladimetcampbells
387111
387111
add a comment |
add a comment |
1 Answer
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$begingroup$
The space $X = (S^1 times S^1)/ (S^1 times {x})$ does not deformation retract to $S^1$. It is more useful to think of it as the quotient of $S^2$ where two points are identified, as can be seen from the surface diagram below:
Then for $p,q in S^2$, we have $S^2 / (p sim q)$ is homotopy equivalent to a 2-sphere with a line segment glued between $p$ and $q$, which is in turn homotopy equivalent to $S^2 vee S^1$. Hence, the fundamental group of $X$ is indeed $mathbb{Z}$.
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
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$begingroup$
The space $X = (S^1 times S^1)/ (S^1 times {x})$ does not deformation retract to $S^1$. It is more useful to think of it as the quotient of $S^2$ where two points are identified, as can be seen from the surface diagram below:
Then for $p,q in S^2$, we have $S^2 / (p sim q)$ is homotopy equivalent to a 2-sphere with a line segment glued between $p$ and $q$, which is in turn homotopy equivalent to $S^2 vee S^1$. Hence, the fundamental group of $X$ is indeed $mathbb{Z}$.
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
$endgroup$
add a comment |
$begingroup$
The space $X = (S^1 times S^1)/ (S^1 times {x})$ does not deformation retract to $S^1$. It is more useful to think of it as the quotient of $S^2$ where two points are identified, as can be seen from the surface diagram below:
Then for $p,q in S^2$, we have $S^2 / (p sim q)$ is homotopy equivalent to a 2-sphere with a line segment glued between $p$ and $q$, which is in turn homotopy equivalent to $S^2 vee S^1$. Hence, the fundamental group of $X$ is indeed $mathbb{Z}$.
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
$endgroup$
add a comment |
$begingroup$
The space $X = (S^1 times S^1)/ (S^1 times {x})$ does not deformation retract to $S^1$. It is more useful to think of it as the quotient of $S^2$ where two points are identified, as can be seen from the surface diagram below:
Then for $p,q in S^2$, we have $S^2 / (p sim q)$ is homotopy equivalent to a 2-sphere with a line segment glued between $p$ and $q$, which is in turn homotopy equivalent to $S^2 vee S^1$. Hence, the fundamental group of $X$ is indeed $mathbb{Z}$.
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
$endgroup$
The space $X = (S^1 times S^1)/ (S^1 times {x})$ does not deformation retract to $S^1$. It is more useful to think of it as the quotient of $S^2$ where two points are identified, as can be seen from the surface diagram below:
Then for $p,q in S^2$, we have $S^2 / (p sim q)$ is homotopy equivalent to a 2-sphere with a line segment glued between $p$ and $q$, which is in turn homotopy equivalent to $S^2 vee S^1$. Hence, the fundamental group of $X$ is indeed $mathbb{Z}$.
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
answered Jan 8 at 3:18
Joshua MundingerJoshua Mundinger
2,9521028
2,9521028
add a comment |
add a comment |
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