Fundamental Group Contains Infinite Cyclic Subgroup












2












$begingroup$



Let $X subseteq Bbb{R}^2$ s.t. $Bbb{R}^2$ such that $S^1 subseteq X subseteq Bbb{R}^2 - {(0,0)}$. Prove that contains $pi_1(X,(0,1))$ an infinite cyclic subgroup.




Here's what I did. Let $j : S^1 to X$ be the embedding of $S^1$ into $X$. I then showed that the induced homomorphism $j_* : pi_1(S^1,(0,1)) to pi_1(X, (0,1))$ injective, so $pi_1(X,(0,1))$ contains the subgroup $j_*(pi_1(S^1,(0,1))) simeq pi_1(S^1,(0,1)) simeq Bbb{Z}$. However, I didn't really use the fact that $X subseteq Bbb{R}^2 - {(0,0)}$ in my proof, which makes me nervous. Does what I did sound right? Here's my proof of injectivity.




Let $alpha$ be a loop $(0,1)$ such that $j_*([alpha]) = [1_{(0,1)}]$ (note that $1_{(0,1)}$ is the constant loop about $(0,1)$, not the indicator function). Then $[j circ alpha] = [1_{(0,1)}]$ which implies there is some path homotopy $H : I times I to X$ such that $H(s,0) = j(alpha (s)) = alpha (s)$ and $H(s,1) = 1_{(0,1)}(s) = (0,1)$ for every $s in I$; and $H(0,t) = (0,1) = H(1,t)$ for every $t in I$. But these are points in $S^1$, so the codomain of $H$ can be restricted and this gives a path homotopy between $alpha$ and $1_{(0,1)}$ itself.




...As I'm typing this up, I now see the issue. The "boundaries" of the map lie in $S^1$, but there is no guarantee that what is happening "in between" stays in $S^1$...I think I could use a hint at this point. The fact that we are subtracting an interior point of $S^1$ from $Bbb{R}^2$ should play some role.






algebraic-topology cyclic-groups fundamental-groups







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$endgroup$












  • $begingroup$
    You found the mistake in your proof; I added a hint to help you solve the problem
    $endgroup$
    – Max
    Dec 24 '18 at 22:16
















2












$begingroup$



Let $X subseteq Bbb{R}^2$ s.t. $Bbb{R}^2$ such that $S^1 subseteq X subseteq Bbb{R}^2 - {(0,0)}$. Prove that contains $pi_1(X,(0,1))$ an infinite cyclic subgroup.




Here's what I did. Let $j : S^1 to X$ be the embedding of $S^1$ into $X$. I then showed that the induced homomorphism $j_* : pi_1(S^1,(0,1)) to pi_1(X, (0,1))$ injective, so $pi_1(X,(0,1))$ contains the subgroup $j_*(pi_1(S^1,(0,1))) simeq pi_1(S^1,(0,1)) simeq Bbb{Z}$. However, I didn't really use the fact that $X subseteq Bbb{R}^2 - {(0,0)}$ in my proof, which makes me nervous. Does what I did sound right? Here's my proof of injectivity.




Let $alpha$ be a loop $(0,1)$ such that $j_*([alpha]) = [1_{(0,1)}]$ (note that $1_{(0,1)}$ is the constant loop about $(0,1)$, not the indicator function). Then $[j circ alpha] = [1_{(0,1)}]$ which implies there is some path homotopy $H : I times I to X$ such that $H(s,0) = j(alpha (s)) = alpha (s)$ and $H(s,1) = 1_{(0,1)}(s) = (0,1)$ for every $s in I$; and $H(0,t) = (0,1) = H(1,t)$ for every $t in I$. But these are points in $S^1$, so the codomain of $H$ can be restricted and this gives a path homotopy between $alpha$ and $1_{(0,1)}$ itself.




...As I'm typing this up, I now see the issue. The "boundaries" of the map lie in $S^1$, but there is no guarantee that what is happening "in between" stays in $S^1$...I think I could use a hint at this point. The fact that we are subtracting an interior point of $S^1$ from $Bbb{R}^2$ should play some role.






algebraic-topology cyclic-groups fundamental-groups







share|cite|improve this question









$endgroup$












  • $begingroup$
    You found the mistake in your proof; I added a hint to help you solve the problem
    $endgroup$
    – Max
    Dec 24 '18 at 22:16














2












2








2





$begingroup$



Let $X subseteq Bbb{R}^2$ s.t. $Bbb{R}^2$ such that $S^1 subseteq X subseteq Bbb{R}^2 - {(0,0)}$. Prove that contains $pi_1(X,(0,1))$ an infinite cyclic subgroup.




Here's what I did. Let $j : S^1 to X$ be the embedding of $S^1$ into $X$. I then showed that the induced homomorphism $j_* : pi_1(S^1,(0,1)) to pi_1(X, (0,1))$ injective, so $pi_1(X,(0,1))$ contains the subgroup $j_*(pi_1(S^1,(0,1))) simeq pi_1(S^1,(0,1)) simeq Bbb{Z}$. However, I didn't really use the fact that $X subseteq Bbb{R}^2 - {(0,0)}$ in my proof, which makes me nervous. Does what I did sound right? Here's my proof of injectivity.




Let $alpha$ be a loop $(0,1)$ such that $j_*([alpha]) = [1_{(0,1)}]$ (note that $1_{(0,1)}$ is the constant loop about $(0,1)$, not the indicator function). Then $[j circ alpha] = [1_{(0,1)}]$ which implies there is some path homotopy $H : I times I to X$ such that $H(s,0) = j(alpha (s)) = alpha (s)$ and $H(s,1) = 1_{(0,1)}(s) = (0,1)$ for every $s in I$; and $H(0,t) = (0,1) = H(1,t)$ for every $t in I$. But these are points in $S^1$, so the codomain of $H$ can be restricted and this gives a path homotopy between $alpha$ and $1_{(0,1)}$ itself.




...As I'm typing this up, I now see the issue. The "boundaries" of the map lie in $S^1$, but there is no guarantee that what is happening "in between" stays in $S^1$...I think I could use a hint at this point. The fact that we are subtracting an interior point of $S^1$ from $Bbb{R}^2$ should play some role.






algebraic-topology cyclic-groups fundamental-groups







share|cite|improve this question









$endgroup$





Let $X subseteq Bbb{R}^2$ s.t. $Bbb{R}^2$ such that $S^1 subseteq X subseteq Bbb{R}^2 - {(0,0)}$. Prove that contains $pi_1(X,(0,1))$ an infinite cyclic subgroup.




Here's what I did. Let $j : S^1 to X$ be the embedding of $S^1$ into $X$. I then showed that the induced homomorphism $j_* : pi_1(S^1,(0,1)) to pi_1(X, (0,1))$ injective, so $pi_1(X,(0,1))$ contains the subgroup $j_*(pi_1(S^1,(0,1))) simeq pi_1(S^1,(0,1)) simeq Bbb{Z}$. However, I didn't really use the fact that $X subseteq Bbb{R}^2 - {(0,0)}$ in my proof, which makes me nervous. Does what I did sound right? Here's my proof of injectivity.




Let $alpha$ be a loop $(0,1)$ such that $j_*([alpha]) = [1_{(0,1)}]$ (note that $1_{(0,1)}$ is the constant loop about $(0,1)$, not the indicator function). Then $[j circ alpha] = [1_{(0,1)}]$ which implies there is some path homotopy $H : I times I to X$ such that $H(s,0) = j(alpha (s)) = alpha (s)$ and $H(s,1) = 1_{(0,1)}(s) = (0,1)$ for every $s in I$; and $H(0,t) = (0,1) = H(1,t)$ for every $t in I$. But these are points in $S^1$, so the codomain of $H$ can be restricted and this gives a path homotopy between $alpha$ and $1_{(0,1)}$ itself.




...As I'm typing this up, I now see the issue. The "boundaries" of the map lie in $S^1$, but there is no guarantee that what is happening "in between" stays in $S^1$...I think I could use a hint at this point. The fact that we are subtracting an interior point of $S^1$ from $Bbb{R}^2$ should play some role.






algebraic-topology cyclic-groups fundamental-groups




algebraic-topology cyclic-groups fundamental-groups






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asked Dec 24 '18 at 22:04









user193319user193319

2,4102925




2,4102925












  • $begingroup$
    You found the mistake in your proof; I added a hint to help you solve the problem
    $endgroup$
    – Max
    Dec 24 '18 at 22:16


















  • $begingroup$
    You found the mistake in your proof; I added a hint to help you solve the problem
    $endgroup$
    – Max
    Dec 24 '18 at 22:16
















$begingroup$
You found the mistake in your proof; I added a hint to help you solve the problem
$endgroup$
– Max
Dec 24 '18 at 22:16




$begingroup$
You found the mistake in your proof; I added a hint to help you solve the problem
$endgroup$
– Max
Dec 24 '18 at 22:16










1 Answer
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$begingroup$

Hint: there is $r: mathbb{R}^2setminus{0} to S^1$ such that $S^1to mathbb{R}^2setminus{0}to S^1$ is the identity on $S^1$; and $S^1to Xto mathbb{R}^2setminus {0} = S^1to Xto mathbb{R}^2setminus{0}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In other words, $S^1$ is a retract of $X$.
    $endgroup$
    – Paul Frost
    Jan 13 at 15:46











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1 Answer
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1












$begingroup$

Hint: there is $r: mathbb{R}^2setminus{0} to S^1$ such that $S^1to mathbb{R}^2setminus{0}to S^1$ is the identity on $S^1$; and $S^1to Xto mathbb{R}^2setminus {0} = S^1to Xto mathbb{R}^2setminus{0}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In other words, $S^1$ is a retract of $X$.
    $endgroup$
    – Paul Frost
    Jan 13 at 15:46
















1












$begingroup$

Hint: there is $r: mathbb{R}^2setminus{0} to S^1$ such that $S^1to mathbb{R}^2setminus{0}to S^1$ is the identity on $S^1$; and $S^1to Xto mathbb{R}^2setminus {0} = S^1to Xto mathbb{R}^2setminus{0}$






share|cite|improve this answer









$endgroup$









  • 1




    $begingroup$
    In other words, $S^1$ is a retract of $X$.
    $endgroup$
    – Paul Frost
    Jan 13 at 15:46














1












1








1





$begingroup$

Hint: there is $r: mathbb{R}^2setminus{0} to S^1$ such that $S^1to mathbb{R}^2setminus{0}to S^1$ is the identity on $S^1$; and $S^1to Xto mathbb{R}^2setminus {0} = S^1to Xto mathbb{R}^2setminus{0}$






share|cite|improve this answer









$endgroup$



Hint: there is $r: mathbb{R}^2setminus{0} to S^1$ such that $S^1to mathbb{R}^2setminus{0}to S^1$ is the identity on $S^1$; and $S^1to Xto mathbb{R}^2setminus {0} = S^1to Xto mathbb{R}^2setminus{0}$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 22:16









MaxMax

14.7k11143




14.7k11143








  • 1




    $begingroup$
    In other words, $S^1$ is a retract of $X$.
    $endgroup$
    – Paul Frost
    Jan 13 at 15:46














  • 1




    $begingroup$
    In other words, $S^1$ is a retract of $X$.
    $endgroup$
    – Paul Frost
    Jan 13 at 15:46








1




1




$begingroup$
In other words, $S^1$ is a retract of $X$.
$endgroup$
– Paul Frost
Jan 13 at 15:46




$begingroup$
In other words, $S^1$ is a retract of $X$.
$endgroup$
– Paul Frost
Jan 13 at 15:46


















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