Mixing
Finding a homotopy directly to show that two induced homomorphisms on a fundamental group are the same
$begingroup$
Consider 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 $f_n(z)=z^n$.
I can solve it using the fact that $pi_1(S^1)=mathbb Z$. But without assuming this fact, how can one directly solve that problem only using the definition of homotopy? I tried to construct a homotopy writing paths in the polar coordinates but I couldn’t find any.
algebraic-topology homotopy-theory
asked Jan 14 at 18:25
User12239User12239
453216
$endgroup$
add a comment |
$begingroup$
Consider 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 $f_n(z)=z^n$.
I can solve it using the fact that $pi_1(S^1)=mathbb Z$. But without assuming this fact, how can one directly solve that problem only using the definition of homotopy? I tried to construct a homotopy writing paths in the polar coordinates but I couldn’t find any.
algebraic-topology homotopy-theory
asked Jan 14 at 18:25
User12239User12239
453216
$endgroup$
1
$begingroup$
Assuming this is an introductory problem, I think the best place to start is to exhibit a homotopy between $f_1 * f_1$ and $f_2$.
$endgroup$
– Matthew Leingang
Jan 14 at 18:33
$begingroup$
Exactly I started with this case but I faced difficulty finding a homotopy
$endgroup$
– User12239
Jan 14 at 18:40
add a comment |
$begingroup$
Consider 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 $f_n(z)=z^n$.
I can solve it using the fact that $pi_1(S^1)=mathbb Z$. But without assuming this fact, how can one directly solve that problem only using the definition of homotopy? I tried to construct a homotopy writing paths in the polar coordinates but I couldn’t find any.
algebraic-topology homotopy-theory
asked Jan 14 at 18:25
User12239User12239
453216
$endgroup$
Consider 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 $f_n(z)=z^n$.
I can solve it using the fact that $pi_1(S^1)=mathbb Z$. But without assuming this fact, how can one directly solve that problem only using the definition of homotopy? I tried to construct a homotopy writing paths in the polar coordinates but I couldn’t find any.
algebraic-topology homotopy-theory
algebraic-topology homotopy-theory
asked Jan 14 at 18:25
User12239User12239
453216
asked Jan 14 at 18:25
User12239User12239
453216
edited Jan 14 at 19:03
User12239
asked Jan 14 at 18:25
User12239User12239
453216
asked Jan 14 at 18:25
User12239User12239
453216
asked Jan 14 at 18:25
User12239User12239
453216
453216
1
$begingroup$
Assuming this is an introductory problem, I think the best place to start is to exhibit a homotopy between $f_1 * f_1$ and $f_2$.
$endgroup$
– Matthew Leingang
Jan 14 at 18:33
$begingroup$
Exactly I started with this case but I faced difficulty finding a homotopy
$endgroup$
– User12239
Jan 14 at 18:40
add a comment |
1
$begingroup$
Assuming this is an introductory problem, I think the best place to start is to exhibit a homotopy between $f_1 * f_1$ and $f_2$.
$endgroup$
– Matthew Leingang
Jan 14 at 18:33
$begingroup$
Exactly I started with this case but I faced difficulty finding a homotopy
$endgroup$
– User12239
Jan 14 at 18:40
1
1
$begingroup$
Assuming this is an introductory problem, I think the best place to start is to exhibit a homotopy between $f_1 * f_1$ and $f_2$.
$endgroup$
– Matthew Leingang
Jan 14 at 18:33
$begingroup$
Assuming this is an introductory problem, I think the best place to start is to exhibit a homotopy between $f_1 * f_1$ and $f_2$.
$endgroup$
– Matthew Leingang
Jan 14 at 18:33
$begingroup$
Exactly I started with this case but I faced difficulty finding a homotopy
$endgroup$
– User12239
Jan 14 at 18:40
$begingroup$
Exactly I started with this case but I faced difficulty finding a homotopy
$endgroup$
– User12239
Jan 14 at 18:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Let's define loops in $X$ based at $x_0$ to be continuous maps $gammacolon[0,1] to X$ with $gamma(0) = gamma(1) = x_0$. This is equivalent to maps $f colon S^1 to X$ with $f(1,0) = x_0$, but it's a bit easier for me to think in terms of a real variable.
If $gamma_1$ and $gamma_2$ are two loops, the usual definition of their concatenation $gamma_1 * gamma_2$ is
$$
(gamma_1 * gamma_2)(t) = begin{cases}
gamma_1(2t) & 0 leq t leq frac{1}{2} \
gamma_2(2t-1) & frac{1}{2} leq t leq 1
end{cases}
$$
The idea is that you split up the interval $[0,1]$ into two sub-intervals, you do $gamma_1$ on the first half (scaling $[0,frac12]$ to $[0,1]$), then $gamma_2$ on the second half (scaling and shifting $[tfrac12,1]$ to $[0,1]$).
But you can also break up the interval into unequal pieces $[0,z]$ and $[z,1]$. You would get a weighted-concatenation operation
$$
(gamma_1 *_z gamma_2)(t) = begin{cases}
gamma_1left(frac{t}{z}right) & 0 leq t leq z \
gamma_2left(frac{t-z}{1-z}right) & z leq t leq 1
end{cases}
$$
Notice that $gamma_1 *_{1/2} gamma_2 = gamma_1 * gamma_2$. Also, all of the maps $gamma_1 *_z gamma_2$ are homotopic. You just continuously move that breakpoint from one $z$-value to each other. To be explicit:
$$
h(t,s) = (gamma_1 *_{s z_1 + (1-s)z_2} gamma_2)(t)
$$
is a homotopy between $gamma_1 *_{z_1} gamma_2$ and $gamma_1 *_{z_2} gamma_2$.
Let $eta_n colon[0,1] to S^1$ be the map $eta_n(t) = e^{2pi int}$. Notice $eta_n(t) = f_n(e^{2pi i t})$, so they induce the same homotopy classes in $pi_1(S^1)$. Let $m$ and $n$ be positive integers. We want to show that $eta_m * eta_n$ is homotopic to $eta_{m+n}$.
Now $eta_{m+n}$ completes $m+n$ loops, equally spaced in the interval $[0,1]$. $eta_m * eta_n$ also completes $m+n$ loops, but the first $m$ are equally spaced in the interval $[0,frac12]$, and the last $n$ are equally spaced in the interval $[frac12,1]$. So let's form $eta_m *_{m/(m+n)} eta_n$ instead. It completes its first $m$ loops equally spaced within the interval $[0,frac{m}{m+n}]$, and its last $n$ loops equally spaced within the interval $[frac{m}{m+n},1]$. But this means that each loop is completed in the same width interval: $frac{1}{m}frac{m}{m+n} = frac{1}{m+n}$ on the first part, and $frac{1}{n}left(1 - frac{m}{m+n}right) = frac{1}{m+n}$ on the second part. So in fact, $eta_m *_{m/(n+m)} eta_n = eta_{n+m}$. Since $eta_m *_{m/(n+m)} eta_n$ is homotopic to $eta_m * eta_n$, we are done.
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
add a comment |
$begingroup$
$S^1$ is a topological group under multiplication.
In general if $G$ is a topological group with identity $e$, then
there is an alternative multiplication on $pi_1(G,e)$ given by
$[f]cdot[g]=[h]$ where $h(t)=f(t)g(t)$ (multiplication in the group $G$).
The Hilton-Eckmann argument proves that this new multiplication
is the same as the standard group operation on $pi_1(G,e)$ (and is
also commutative).
As a consequence, if $f$ is a loop from $e$ to $e$ in $G$, the
class of $n[f]$ in $pi_1(G,e)$ is given by that of the path $tmapsto
f(t)^n$. Your example is $G=S^1$, considered as a subgroup of $Bbb C^*$.
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Thanks. It is a polished solution but the problem is given right after introducing homotopy so the solution should use only the definition of homotopy
$endgroup$
– User12239
Jan 14 at 18:37
$begingroup$
What else does this use? Do you mean the definition of topological group?
$endgroup$
– Justin Young
Jan 23 at 12:52
$begingroup$
@JustinYoung no, I haven’t seen the Hilton-Eckmann argument yet
$endgroup$
– User12239
Jan 24 at 17:14
$begingroup$
It's a purely algebraic argument, and you don't have to cite it, you can prove it for yourself. Show that in a topological group, the two operations on $pi_1$ agree and are both abelian. Hint: $Gtimes G to G$ induces a homomorphism for the "standard" multiplication on $pi_1$.
$endgroup$
– Justin Young
Jan 24 at 17:22
add a comment |
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2 Answers
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active
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2 Answers
2
active
oldest
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$begingroup$
Let's define loops in $X$ based at $x_0$ to be continuous maps $gammacolon[0,1] to X$ with $gamma(0) = gamma(1) = x_0$. This is equivalent to maps $f colon S^1 to X$ with $f(1,0) = x_0$, but it's a bit easier for me to think in terms of a real variable.
If $gamma_1$ and $gamma_2$ are two loops, the usual definition of their concatenation $gamma_1 * gamma_2$ is
$$
(gamma_1 * gamma_2)(t) = begin{cases}
gamma_1(2t) & 0 leq t leq frac{1}{2} \
gamma_2(2t-1) & frac{1}{2} leq t leq 1
end{cases}
$$
The idea is that you split up the interval $[0,1]$ into two sub-intervals, you do $gamma_1$ on the first half (scaling $[0,frac12]$ to $[0,1]$), then $gamma_2$ on the second half (scaling and shifting $[tfrac12,1]$ to $[0,1]$).
But you can also break up the interval into unequal pieces $[0,z]$ and $[z,1]$. You would get a weighted-concatenation operation
$$
(gamma_1 *_z gamma_2)(t) = begin{cases}
gamma_1left(frac{t}{z}right) & 0 leq t leq z \
gamma_2left(frac{t-z}{1-z}right) & z leq t leq 1
end{cases}
$$
Notice that $gamma_1 *_{1/2} gamma_2 = gamma_1 * gamma_2$. Also, all of the maps $gamma_1 *_z gamma_2$ are homotopic. You just continuously move that breakpoint from one $z$-value to each other. To be explicit:
$$
h(t,s) = (gamma_1 *_{s z_1 + (1-s)z_2} gamma_2)(t)
$$
is a homotopy between $gamma_1 *_{z_1} gamma_2$ and $gamma_1 *_{z_2} gamma_2$.
Let $eta_n colon[0,1] to S^1$ be the map $eta_n(t) = e^{2pi int}$. Notice $eta_n(t) = f_n(e^{2pi i t})$, so they induce the same homotopy classes in $pi_1(S^1)$. Let $m$ and $n$ be positive integers. We want to show that $eta_m * eta_n$ is homotopic to $eta_{m+n}$.
Now $eta_{m+n}$ completes $m+n$ loops, equally spaced in the interval $[0,1]$. $eta_m * eta_n$ also completes $m+n$ loops, but the first $m$ are equally spaced in the interval $[0,frac12]$, and the last $n$ are equally spaced in the interval $[frac12,1]$. So let's form $eta_m *_{m/(m+n)} eta_n$ instead. It completes its first $m$ loops equally spaced within the interval $[0,frac{m}{m+n}]$, and its last $n$ loops equally spaced within the interval $[frac{m}{m+n},1]$. But this means that each loop is completed in the same width interval: $frac{1}{m}frac{m}{m+n} = frac{1}{m+n}$ on the first part, and $frac{1}{n}left(1 - frac{m}{m+n}right) = frac{1}{m+n}$ on the second part. So in fact, $eta_m *_{m/(n+m)} eta_n = eta_{n+m}$. Since $eta_m *_{m/(n+m)} eta_n$ is homotopic to $eta_m * eta_n$, we are done.
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
add a comment |
$begingroup$
Let's define loops in $X$ based at $x_0$ to be continuous maps $gammacolon[0,1] to X$ with $gamma(0) = gamma(1) = x_0$. This is equivalent to maps $f colon S^1 to X$ with $f(1,0) = x_0$, but it's a bit easier for me to think in terms of a real variable.
If $gamma_1$ and $gamma_2$ are two loops, the usual definition of their concatenation $gamma_1 * gamma_2$ is
$$
(gamma_1 * gamma_2)(t) = begin{cases}
gamma_1(2t) & 0 leq t leq frac{1}{2} \
gamma_2(2t-1) & frac{1}{2} leq t leq 1
end{cases}
$$
The idea is that you split up the interval $[0,1]$ into two sub-intervals, you do $gamma_1$ on the first half (scaling $[0,frac12]$ to $[0,1]$), then $gamma_2$ on the second half (scaling and shifting $[tfrac12,1]$ to $[0,1]$).
But you can also break up the interval into unequal pieces $[0,z]$ and $[z,1]$. You would get a weighted-concatenation operation
$$
(gamma_1 *_z gamma_2)(t) = begin{cases}
gamma_1left(frac{t}{z}right) & 0 leq t leq z \
gamma_2left(frac{t-z}{1-z}right) & z leq t leq 1
end{cases}
$$
Notice that $gamma_1 *_{1/2} gamma_2 = gamma_1 * gamma_2$. Also, all of the maps $gamma_1 *_z gamma_2$ are homotopic. You just continuously move that breakpoint from one $z$-value to each other. To be explicit:
$$
h(t,s) = (gamma_1 *_{s z_1 + (1-s)z_2} gamma_2)(t)
$$
is a homotopy between $gamma_1 *_{z_1} gamma_2$ and $gamma_1 *_{z_2} gamma_2$.
Let $eta_n colon[0,1] to S^1$ be the map $eta_n(t) = e^{2pi int}$. Notice $eta_n(t) = f_n(e^{2pi i t})$, so they induce the same homotopy classes in $pi_1(S^1)$. Let $m$ and $n$ be positive integers. We want to show that $eta_m * eta_n$ is homotopic to $eta_{m+n}$.
Now $eta_{m+n}$ completes $m+n$ loops, equally spaced in the interval $[0,1]$. $eta_m * eta_n$ also completes $m+n$ loops, but the first $m$ are equally spaced in the interval $[0,frac12]$, and the last $n$ are equally spaced in the interval $[frac12,1]$. So let's form $eta_m *_{m/(m+n)} eta_n$ instead. It completes its first $m$ loops equally spaced within the interval $[0,frac{m}{m+n}]$, and its last $n$ loops equally spaced within the interval $[frac{m}{m+n},1]$. But this means that each loop is completed in the same width interval: $frac{1}{m}frac{m}{m+n} = frac{1}{m+n}$ on the first part, and $frac{1}{n}left(1 - frac{m}{m+n}right) = frac{1}{m+n}$ on the second part. So in fact, $eta_m *_{m/(n+m)} eta_n = eta_{n+m}$. Since $eta_m *_{m/(n+m)} eta_n$ is homotopic to $eta_m * eta_n$, we are done.
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
add a comment |
$begingroup$
Let's define loops in $X$ based at $x_0$ to be continuous maps $gammacolon[0,1] to X$ with $gamma(0) = gamma(1) = x_0$. This is equivalent to maps $f colon S^1 to X$ with $f(1,0) = x_0$, but it's a bit easier for me to think in terms of a real variable.
If $gamma_1$ and $gamma_2$ are two loops, the usual definition of their concatenation $gamma_1 * gamma_2$ is
$$
(gamma_1 * gamma_2)(t) = begin{cases}
gamma_1(2t) & 0 leq t leq frac{1}{2} \
gamma_2(2t-1) & frac{1}{2} leq t leq 1
end{cases}
$$
The idea is that you split up the interval $[0,1]$ into two sub-intervals, you do $gamma_1$ on the first half (scaling $[0,frac12]$ to $[0,1]$), then $gamma_2$ on the second half (scaling and shifting $[tfrac12,1]$ to $[0,1]$).
But you can also break up the interval into unequal pieces $[0,z]$ and $[z,1]$. You would get a weighted-concatenation operation
$$
(gamma_1 *_z gamma_2)(t) = begin{cases}
gamma_1left(frac{t}{z}right) & 0 leq t leq z \
gamma_2left(frac{t-z}{1-z}right) & z leq t leq 1
end{cases}
$$
Notice that $gamma_1 *_{1/2} gamma_2 = gamma_1 * gamma_2$. Also, all of the maps $gamma_1 *_z gamma_2$ are homotopic. You just continuously move that breakpoint from one $z$-value to each other. To be explicit:
$$
h(t,s) = (gamma_1 *_{s z_1 + (1-s)z_2} gamma_2)(t)
$$
is a homotopy between $gamma_1 *_{z_1} gamma_2$ and $gamma_1 *_{z_2} gamma_2$.
Let $eta_n colon[0,1] to S^1$ be the map $eta_n(t) = e^{2pi int}$. Notice $eta_n(t) = f_n(e^{2pi i t})$, so they induce the same homotopy classes in $pi_1(S^1)$. Let $m$ and $n$ be positive integers. We want to show that $eta_m * eta_n$ is homotopic to $eta_{m+n}$.
Now $eta_{m+n}$ completes $m+n$ loops, equally spaced in the interval $[0,1]$. $eta_m * eta_n$ also completes $m+n$ loops, but the first $m$ are equally spaced in the interval $[0,frac12]$, and the last $n$ are equally spaced in the interval $[frac12,1]$. So let's form $eta_m *_{m/(m+n)} eta_n$ instead. It completes its first $m$ loops equally spaced within the interval $[0,frac{m}{m+n}]$, and its last $n$ loops equally spaced within the interval $[frac{m}{m+n},1]$. But this means that each loop is completed in the same width interval: $frac{1}{m}frac{m}{m+n} = frac{1}{m+n}$ on the first part, and $frac{1}{n}left(1 - frac{m}{m+n}right) = frac{1}{m+n}$ on the second part. So in fact, $eta_m *_{m/(n+m)} eta_n = eta_{n+m}$. Since $eta_m *_{m/(n+m)} eta_n$ is homotopic to $eta_m * eta_n$, we are done.
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
$endgroup$
Let's define loops in $X$ based at $x_0$ to be continuous maps $gammacolon[0,1] to X$ with $gamma(0) = gamma(1) = x_0$. This is equivalent to maps $f colon S^1 to X$ with $f(1,0) = x_0$, but it's a bit easier for me to think in terms of a real variable.
If $gamma_1$ and $gamma_2$ are two loops, the usual definition of their concatenation $gamma_1 * gamma_2$ is
$$
(gamma_1 * gamma_2)(t) = begin{cases}
gamma_1(2t) & 0 leq t leq frac{1}{2} \
gamma_2(2t-1) & frac{1}{2} leq t leq 1
end{cases}
$$
The idea is that you split up the interval $[0,1]$ into two sub-intervals, you do $gamma_1$ on the first half (scaling $[0,frac12]$ to $[0,1]$), then $gamma_2$ on the second half (scaling and shifting $[tfrac12,1]$ to $[0,1]$).
But you can also break up the interval into unequal pieces $[0,z]$ and $[z,1]$. You would get a weighted-concatenation operation
$$
(gamma_1 *_z gamma_2)(t) = begin{cases}
gamma_1left(frac{t}{z}right) & 0 leq t leq z \
gamma_2left(frac{t-z}{1-z}right) & z leq t leq 1
end{cases}
$$
Notice that $gamma_1 *_{1/2} gamma_2 = gamma_1 * gamma_2$. Also, all of the maps $gamma_1 *_z gamma_2$ are homotopic. You just continuously move that breakpoint from one $z$-value to each other. To be explicit:
$$
h(t,s) = (gamma_1 *_{s z_1 + (1-s)z_2} gamma_2)(t)
$$
is a homotopy between $gamma_1 *_{z_1} gamma_2$ and $gamma_1 *_{z_2} gamma_2$.
Let $eta_n colon[0,1] to S^1$ be the map $eta_n(t) = e^{2pi int}$. Notice $eta_n(t) = f_n(e^{2pi i t})$, so they induce the same homotopy classes in $pi_1(S^1)$. Let $m$ and $n$ be positive integers. We want to show that $eta_m * eta_n$ is homotopic to $eta_{m+n}$.
Now $eta_{m+n}$ completes $m+n$ loops, equally spaced in the interval $[0,1]$. $eta_m * eta_n$ also completes $m+n$ loops, but the first $m$ are equally spaced in the interval $[0,frac12]$, and the last $n$ are equally spaced in the interval $[frac12,1]$. So let's form $eta_m *_{m/(m+n)} eta_n$ instead. It completes its first $m$ loops equally spaced within the interval $[0,frac{m}{m+n}]$, and its last $n$ loops equally spaced within the interval $[frac{m}{m+n},1]$. But this means that each loop is completed in the same width interval: $frac{1}{m}frac{m}{m+n} = frac{1}{m+n}$ on the first part, and $frac{1}{n}left(1 - frac{m}{m+n}right) = frac{1}{m+n}$ on the second part. So in fact, $eta_m *_{m/(n+m)} eta_n = eta_{n+m}$. Since $eta_m *_{m/(n+m)} eta_n$ is homotopic to $eta_m * eta_n$, we are done.
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
edited Jan 14 at 21:30
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
answered Jan 14 at 20:00
Matthew LeingangMatthew Leingang
16.4k12244
16.4k12244
add a comment |
add a comment |
$begingroup$
$S^1$ is a topological group under multiplication.
In general if $G$ is a topological group with identity $e$, then
there is an alternative multiplication on $pi_1(G,e)$ given by
$[f]cdot[g]=[h]$ where $h(t)=f(t)g(t)$ (multiplication in the group $G$).
The Hilton-Eckmann argument proves that this new multiplication
is the same as the standard group operation on $pi_1(G,e)$ (and is
also commutative).
As a consequence, if $f$ is a loop from $e$ to $e$ in $G$, the
class of $n[f]$ in $pi_1(G,e)$ is given by that of the path $tmapsto
f(t)^n$. Your example is $G=S^1$, considered as a subgroup of $Bbb C^*$.
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Thanks. It is a polished solution but the problem is given right after introducing homotopy so the solution should use only the definition of homotopy
$endgroup$
– User12239
Jan 14 at 18:37
$begingroup$
What else does this use? Do you mean the definition of topological group?
$endgroup$
– Justin Young
Jan 23 at 12:52
$begingroup$
@JustinYoung no, I haven’t seen the Hilton-Eckmann argument yet
$endgroup$
– User12239
Jan 24 at 17:14
$begingroup$
It's a purely algebraic argument, and you don't have to cite it, you can prove it for yourself. Show that in a topological group, the two operations on $pi_1$ agree and are both abelian. Hint: $Gtimes G to G$ induces a homomorphism for the "standard" multiplication on $pi_1$.
$endgroup$
– Justin Young
Jan 24 at 17:22
add a comment |
$begingroup$
$S^1$ is a topological group under multiplication.
In general if $G$ is a topological group with identity $e$, then
there is an alternative multiplication on $pi_1(G,e)$ given by
$[f]cdot[g]=[h]$ where $h(t)=f(t)g(t)$ (multiplication in the group $G$).
The Hilton-Eckmann argument proves that this new multiplication
is the same as the standard group operation on $pi_1(G,e)$ (and is
also commutative).
As a consequence, if $f$ is a loop from $e$ to $e$ in $G$, the
class of $n[f]$ in $pi_1(G,e)$ is given by that of the path $tmapsto
f(t)^n$. Your example is $G=S^1$, considered as a subgroup of $Bbb C^*$.
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$begingroup$
Thanks. It is a polished solution but the problem is given right after introducing homotopy so the solution should use only the definition of homotopy
$endgroup$
– User12239
Jan 14 at 18:37
$begingroup$
What else does this use? Do you mean the definition of topological group?
$endgroup$
– Justin Young
Jan 23 at 12:52
$begingroup$
@JustinYoung no, I haven’t seen the Hilton-Eckmann argument yet
$endgroup$
– User12239
Jan 24 at 17:14
$begingroup$
It's a purely algebraic argument, and you don't have to cite it, you can prove it for yourself. Show that in a topological group, the two operations on $pi_1$ agree and are both abelian. Hint: $Gtimes G to G$ induces a homomorphism for the "standard" multiplication on $pi_1$.
$endgroup$
– Justin Young
Jan 24 at 17:22
add a comment |
$begingroup$
$S^1$ is a topological group under multiplication.
In general if $G$ is a topological group with identity $e$, then
there is an alternative multiplication on $pi_1(G,e)$ given by
$[f]cdot[g]=[h]$ where $h(t)=f(t)g(t)$ (multiplication in the group $G$).
The Hilton-Eckmann argument proves that this new multiplication
is the same as the standard group operation on $pi_1(G,e)$ (and is
also commutative).
As a consequence, if $f$ is a loop from $e$ to $e$ in $G$, the
class of $n[f]$ in $pi_1(G,e)$ is given by that of the path $tmapsto
f(t)^n$. Your example is $G=S^1$, considered as a subgroup of $Bbb C^*$.
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
$S^1$ is a topological group under multiplication.
In general if $G$ is a topological group with identity $e$, then
there is an alternative multiplication on $pi_1(G,e)$ given by
$[f]cdot[g]=[h]$ where $h(t)=f(t)g(t)$ (multiplication in the group $G$).
The Hilton-Eckmann argument proves that this new multiplication
is the same as the standard group operation on $pi_1(G,e)$ (and is
also commutative).
As a consequence, if $f$ is a loop from $e$ to $e$ in $G$, the
class of $n[f]$ in $pi_1(G,e)$ is given by that of the path $tmapsto
f(t)^n$. Your example is $G=S^1$, considered as a subgroup of $Bbb C^*$.
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Jan 14 at 18:32
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
$begingroup$
Thanks. It is a polished solution but the problem is given right after introducing homotopy so the solution should use only the definition of homotopy
$endgroup$
– User12239
Jan 14 at 18:37
$begingroup$
What else does this use? Do you mean the definition of topological group?
$endgroup$
– Justin Young
Jan 23 at 12:52
$begingroup$
@JustinYoung no, I haven’t seen the Hilton-Eckmann argument yet
$endgroup$
– User12239
Jan 24 at 17:14
$begingroup$
It's a purely algebraic argument, and you don't have to cite it, you can prove it for yourself. Show that in a topological group, the two operations on $pi_1$ agree and are both abelian. Hint: $Gtimes G to G$ induces a homomorphism for the "standard" multiplication on $pi_1$.
$endgroup$
– Justin Young
Jan 24 at 17:22
add a comment |
$begingroup$
Thanks. It is a polished solution but the problem is given right after introducing homotopy so the solution should use only the definition of homotopy
$endgroup$
– User12239
Jan 14 at 18:37
$begingroup$
What else does this use? Do you mean the definition of topological group?
$endgroup$
– Justin Young
Jan 23 at 12:52
$begingroup$
@JustinYoung no, I haven’t seen the Hilton-Eckmann argument yet
$endgroup$
– User12239
Jan 24 at 17:14
$begingroup$
It's a purely algebraic argument, and you don't have to cite it, you can prove it for yourself. Show that in a topological group, the two operations on $pi_1$ agree and are both abelian. Hint: $Gtimes G to G$ induces a homomorphism for the "standard" multiplication on $pi_1$.
$endgroup$
– Justin Young
Jan 24 at 17:22
$begingroup$
Thanks. It is a polished solution but the problem is given right after introducing homotopy so the solution should use only the definition of homotopy
$endgroup$
– User12239
Jan 14 at 18:37
$begingroup$
Thanks. It is a polished solution but the problem is given right after introducing homotopy so the solution should use only the definition of homotopy
$endgroup$
– User12239
Jan 14 at 18:37
$begingroup$
What else does this use? Do you mean the definition of topological group?
$endgroup$
– Justin Young
Jan 23 at 12:52
$begingroup$
What else does this use? Do you mean the definition of topological group?
$endgroup$
– Justin Young
Jan 23 at 12:52
$begingroup$
@JustinYoung no, I haven’t seen the Hilton-Eckmann argument yet
$endgroup$
– User12239
Jan 24 at 17:14
$begingroup$
@JustinYoung no, I haven’t seen the Hilton-Eckmann argument yet
$endgroup$
– User12239
Jan 24 at 17:14
$begingroup$
It's a purely algebraic argument, and you don't have to cite it, you can prove it for yourself. Show that in a topological group, the two operations on $pi_1$ agree and are both abelian. Hint: $Gtimes G to G$ induces a homomorphism for the "standard" multiplication on $pi_1$.
$endgroup$
– Justin Young
Jan 24 at 17:22
$begingroup$
It's a purely algebraic argument, and you don't have to cite it, you can prove it for yourself. Show that in a topological group, the two operations on $pi_1$ agree and are both abelian. Hint: $Gtimes G to G$ induces a homomorphism for the "standard" multiplication on $pi_1$.
$endgroup$
– Justin Young
Jan 24 at 17:22
add a comment |
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$begingroup$
Assuming this is an introductory problem, I think the best place to start is to exhibit a homotopy between $f_1 * f_1$ and $f_2$.
$endgroup$
– Matthew Leingang
Jan 14 at 18:33
$begingroup$
Exactly I started with this case but I faced difficulty finding a homotopy
$endgroup$
– User12239
Jan 14 at 18:40