The meaning of $pi_1(S^1,(1,0))$












0












$begingroup$


I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:



“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?



Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.










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  • 3




    $begingroup$
    $(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
    $endgroup$
    – Max
    Jan 14 at 15:41










  • $begingroup$
    Oh right, I’ve become so absent-minded. Thanks @Max
    $endgroup$
    – User12239
    Jan 14 at 15:41






  • 1




    $begingroup$
    It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
    $endgroup$
    – Magdiragdag
    Jan 14 at 22:49










  • $begingroup$
    @Magdiragdag yes exactly, maybe that’s why I got confused
    $endgroup$
    – User12239
    Jan 14 at 22:52
















0












$begingroup$


I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:



“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?



Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.










share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    $(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
    $endgroup$
    – Max
    Jan 14 at 15:41










  • $begingroup$
    Oh right, I’ve become so absent-minded. Thanks @Max
    $endgroup$
    – User12239
    Jan 14 at 15:41






  • 1




    $begingroup$
    It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
    $endgroup$
    – Magdiragdag
    Jan 14 at 22:49










  • $begingroup$
    @Magdiragdag yes exactly, maybe that’s why I got confused
    $endgroup$
    – User12239
    Jan 14 at 22:52














0












0








0





$begingroup$


I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:



“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?



Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.










share|cite|improve this question











$endgroup$




I know what $pi_1(X,star)$ means. But what does $pi_1(S^1,(1,0))$ mean? I came across it when wanting to solve the following problem:



“Let $Iin pi_1(S^1,(1,0))$ be the class of the identity map. Show that $nI$ is the class of the map $f_n:S^1rightarrow S^1$ given by $f_n(z)=z^n$.” What does this problem want?



Is $(1,0)$ the two element set consisting of $1$ and $0$, or is it an interval? But in both cases it again does not make sense to me as $0$ is not in the circle.







algebraic-topology homotopy-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 14 at 15:39







User12239

















asked Jan 14 at 15:33









User12239User12239

341216




341216








  • 3




    $begingroup$
    $(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
    $endgroup$
    – Max
    Jan 14 at 15:41










  • $begingroup$
    Oh right, I’ve become so absent-minded. Thanks @Max
    $endgroup$
    – User12239
    Jan 14 at 15:41






  • 1




    $begingroup$
    It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
    $endgroup$
    – Magdiragdag
    Jan 14 at 22:49










  • $begingroup$
    @Magdiragdag yes exactly, maybe that’s why I got confused
    $endgroup$
    – User12239
    Jan 14 at 22:52














  • 3




    $begingroup$
    $(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
    $endgroup$
    – Max
    Jan 14 at 15:41










  • $begingroup$
    Oh right, I’ve become so absent-minded. Thanks @Max
    $endgroup$
    – User12239
    Jan 14 at 15:41






  • 1




    $begingroup$
    It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
    $endgroup$
    – Magdiragdag
    Jan 14 at 22:49










  • $begingroup$
    @Magdiragdag yes exactly, maybe that’s why I got confused
    $endgroup$
    – User12239
    Jan 14 at 22:52








3




3




$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41




$begingroup$
$(1,0)$ is the couple that has $1$ as its first coordinate, $0$ its second coordinate
$endgroup$
– Max
Jan 14 at 15:41












$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41




$begingroup$
Oh right, I’ve become so absent-minded. Thanks @Max
$endgroup$
– User12239
Jan 14 at 15:41




1




1




$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49




$begingroup$
It's a bit weird that when defining the maps $f_n$, $S^1$ is treated as a subset of ${mathbb C}$, but when specifying the base point $(1,0)$ as a subset of ${mathbb R}^2$. It seems to make more sense, in this context, to write $pi_1(S^1,1)$.
$endgroup$
– Magdiragdag
Jan 14 at 22:49












$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52




$begingroup$
@Magdiragdag yes exactly, maybe that’s why I got confused
$endgroup$
– User12239
Jan 14 at 22:52










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