How to show a map has the homomorphism property?
Consider the map $f$ defined in the answer from here. I'm trying to understand why $f$ has the homomorphism property. The hint given in the comments is to include the point $1/2in I$ in the partition. But the points of partition had the property that $gamma(I_j)$ lies entirely within $U_1$ or $U_2$ (see the question referred to above for the notation). Why should the subintervals corresponding to $1/2$ have this property? Further, even if they have this property, $G$ is still not abelian, which looks like an obstruction to proving the homomorphism property (or maybe I'm overthinking...).
abstract-algebra general-topology algebraic-topology
asked Nov 27 '18 at 18:27
user531587
25413
add a comment |
Consider the map $f$ defined in the answer from here. I'm trying to understand why $f$ has the homomorphism property. The hint given in the comments is to include the point $1/2in I$ in the partition. But the points of partition had the property that $gamma(I_j)$ lies entirely within $U_1$ or $U_2$ (see the question referred to above for the notation). Why should the subintervals corresponding to $1/2$ have this property? Further, even if they have this property, $G$ is still not abelian, which looks like an obstruction to proving the homomorphism property (or maybe I'm overthinking...).
abstract-algebra general-topology algebraic-topology
asked Nov 27 '18 at 18:27
user531587
25413
add a comment |
Consider the map $f$ defined in the answer from here. I'm trying to understand why $f$ has the homomorphism property. The hint given in the comments is to include the point $1/2in I$ in the partition. But the points of partition had the property that $gamma(I_j)$ lies entirely within $U_1$ or $U_2$ (see the question referred to above for the notation). Why should the subintervals corresponding to $1/2$ have this property? Further, even if they have this property, $G$ is still not abelian, which looks like an obstruction to proving the homomorphism property (or maybe I'm overthinking...).
abstract-algebra general-topology algebraic-topology
asked Nov 27 '18 at 18:27
user531587
25413
Consider the map $f$ defined in the answer from here. I'm trying to understand why $f$ has the homomorphism property. The hint given in the comments is to include the point $1/2in I$ in the partition. But the points of partition had the property that $gamma(I_j)$ lies entirely within $U_1$ or $U_2$ (see the question referred to above for the notation). Why should the subintervals corresponding to $1/2$ have this property? Further, even if they have this property, $G$ is still not abelian, which looks like an obstruction to proving the homomorphism property (or maybe I'm overthinking...).
abstract-algebra general-topology algebraic-topology
abstract-algebra general-topology algebraic-topology
asked Nov 27 '18 at 18:27
user531587
25413
asked Nov 27 '18 at 18:27
user531587
25413
asked Nov 27 '18 at 18:27
user531587
25413
asked Nov 27 '18 at 18:27
user531587
25413
asked Nov 27 '18 at 18:27
user531587
25413
25413
add a comment |
add a comment |
1 Answer
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Given closed paths $gamma_1, gamma_2$, the path $gamma = gamma_1 cdot gamma_2$ is defined by $gamma(t) = gamma_1(2t)$ for $t le 1/2$ and $gamma(t) = gamma_2(2t-1)$ for $t ge 1/2$. Forv $k = 1,2$ choose partitions $I^k_j = [x^k_{j-1},x^k_j]$ of $I$, $j = 1,dots,n_k$, such that $gamma_k(I^k_j)$ is contained in $U_1$ or $U_2$. Define a partition $I_j = [x_{j-1},x_j]$ of $I$, $j = 1,dots,n_1+n_2$, by $x_j = frac{1}{2}x^1_j$ for $j = 0,dots,n_1$ and $x_j = frac{1}{2}(x^2_j+1)$ for $j = n_1,dots, n_1 + n_2$. Then $1/2$ is a partition point. By construction $gamma(I_j)$ is contained in $U_1$ or $U_2$. Choose paths $beta_j^k$ in the appropriate $U_{r(k,j)}$ which connect $x$ and $gamma_k(x^k_j)$, where $beta_0^k$ and $beta^k_{n_k}$ are constant. The paths $beta_j^k$ provide us with paths $beta_j$ in $U_{r(j)}$ connecting $x$ and $gamma(x_j)$: Take $beta_j = beta_j^1$ for $j = 0,dots,n_1$ and $beta_j = beta^2_{n_1-j}$ for $j = n_1,dots,n_1+n_2$. But now it is clear that
$$f(gamma) = f_{r(1)}([Gamma_1]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1)}([Gamma_1]) dots f_{r(n_1)}([Gamma_{n_1}]) f_{r(n_1+1)}([Gamma_{n_1+1}]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1,1)}([Gamma^1_1]) dots f_{r(1,n_1)}([Gamma^1_{n_1}]) f_{r(2,1)}([Gamma^2_1]) dots f_{r(2,n_2)}([Gamma^2_{n_2}]) = f(gamma_1) f(gamma_2) .$$
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
I would have never come up with such an argument.
– user531587
Nov 28 '18 at 18:08
add a comment |
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1 Answer
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Given closed paths $gamma_1, gamma_2$, the path $gamma = gamma_1 cdot gamma_2$ is defined by $gamma(t) = gamma_1(2t)$ for $t le 1/2$ and $gamma(t) = gamma_2(2t-1)$ for $t ge 1/2$. Forv $k = 1,2$ choose partitions $I^k_j = [x^k_{j-1},x^k_j]$ of $I$, $j = 1,dots,n_k$, such that $gamma_k(I^k_j)$ is contained in $U_1$ or $U_2$. Define a partition $I_j = [x_{j-1},x_j]$ of $I$, $j = 1,dots,n_1+n_2$, by $x_j = frac{1}{2}x^1_j$ for $j = 0,dots,n_1$ and $x_j = frac{1}{2}(x^2_j+1)$ for $j = n_1,dots, n_1 + n_2$. Then $1/2$ is a partition point. By construction $gamma(I_j)$ is contained in $U_1$ or $U_2$. Choose paths $beta_j^k$ in the appropriate $U_{r(k,j)}$ which connect $x$ and $gamma_k(x^k_j)$, where $beta_0^k$ and $beta^k_{n_k}$ are constant. The paths $beta_j^k$ provide us with paths $beta_j$ in $U_{r(j)}$ connecting $x$ and $gamma(x_j)$: Take $beta_j = beta_j^1$ for $j = 0,dots,n_1$ and $beta_j = beta^2_{n_1-j}$ for $j = n_1,dots,n_1+n_2$. But now it is clear that
$$f(gamma) = f_{r(1)}([Gamma_1]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1)}([Gamma_1]) dots f_{r(n_1)}([Gamma_{n_1}]) f_{r(n_1+1)}([Gamma_{n_1+1}]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1,1)}([Gamma^1_1]) dots f_{r(1,n_1)}([Gamma^1_{n_1}]) f_{r(2,1)}([Gamma^2_1]) dots f_{r(2,n_2)}([Gamma^2_{n_2}]) = f(gamma_1) f(gamma_2) .$$
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
I would have never come up with such an argument.
– user531587
Nov 28 '18 at 18:08
add a comment |
Given closed paths $gamma_1, gamma_2$, the path $gamma = gamma_1 cdot gamma_2$ is defined by $gamma(t) = gamma_1(2t)$ for $t le 1/2$ and $gamma(t) = gamma_2(2t-1)$ for $t ge 1/2$. Forv $k = 1,2$ choose partitions $I^k_j = [x^k_{j-1},x^k_j]$ of $I$, $j = 1,dots,n_k$, such that $gamma_k(I^k_j)$ is contained in $U_1$ or $U_2$. Define a partition $I_j = [x_{j-1},x_j]$ of $I$, $j = 1,dots,n_1+n_2$, by $x_j = frac{1}{2}x^1_j$ for $j = 0,dots,n_1$ and $x_j = frac{1}{2}(x^2_j+1)$ for $j = n_1,dots, n_1 + n_2$. Then $1/2$ is a partition point. By construction $gamma(I_j)$ is contained in $U_1$ or $U_2$. Choose paths $beta_j^k$ in the appropriate $U_{r(k,j)}$ which connect $x$ and $gamma_k(x^k_j)$, where $beta_0^k$ and $beta^k_{n_k}$ are constant. The paths $beta_j^k$ provide us with paths $beta_j$ in $U_{r(j)}$ connecting $x$ and $gamma(x_j)$: Take $beta_j = beta_j^1$ for $j = 0,dots,n_1$ and $beta_j = beta^2_{n_1-j}$ for $j = n_1,dots,n_1+n_2$. But now it is clear that
$$f(gamma) = f_{r(1)}([Gamma_1]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1)}([Gamma_1]) dots f_{r(n_1)}([Gamma_{n_1}]) f_{r(n_1+1)}([Gamma_{n_1+1}]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1,1)}([Gamma^1_1]) dots f_{r(1,n_1)}([Gamma^1_{n_1}]) f_{r(2,1)}([Gamma^2_1]) dots f_{r(2,n_2)}([Gamma^2_{n_2}]) = f(gamma_1) f(gamma_2) .$$
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
I would have never come up with such an argument.
– user531587
Nov 28 '18 at 18:08
add a comment |
Given closed paths $gamma_1, gamma_2$, the path $gamma = gamma_1 cdot gamma_2$ is defined by $gamma(t) = gamma_1(2t)$ for $t le 1/2$ and $gamma(t) = gamma_2(2t-1)$ for $t ge 1/2$. Forv $k = 1,2$ choose partitions $I^k_j = [x^k_{j-1},x^k_j]$ of $I$, $j = 1,dots,n_k$, such that $gamma_k(I^k_j)$ is contained in $U_1$ or $U_2$. Define a partition $I_j = [x_{j-1},x_j]$ of $I$, $j = 1,dots,n_1+n_2$, by $x_j = frac{1}{2}x^1_j$ for $j = 0,dots,n_1$ and $x_j = frac{1}{2}(x^2_j+1)$ for $j = n_1,dots, n_1 + n_2$. Then $1/2$ is a partition point. By construction $gamma(I_j)$ is contained in $U_1$ or $U_2$. Choose paths $beta_j^k$ in the appropriate $U_{r(k,j)}$ which connect $x$ and $gamma_k(x^k_j)$, where $beta_0^k$ and $beta^k_{n_k}$ are constant. The paths $beta_j^k$ provide us with paths $beta_j$ in $U_{r(j)}$ connecting $x$ and $gamma(x_j)$: Take $beta_j = beta_j^1$ for $j = 0,dots,n_1$ and $beta_j = beta^2_{n_1-j}$ for $j = n_1,dots,n_1+n_2$. But now it is clear that
$$f(gamma) = f_{r(1)}([Gamma_1]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1)}([Gamma_1]) dots f_{r(n_1)}([Gamma_{n_1}]) f_{r(n_1+1)}([Gamma_{n_1+1}]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1,1)}([Gamma^1_1]) dots f_{r(1,n_1)}([Gamma^1_{n_1}]) f_{r(2,1)}([Gamma^2_1]) dots f_{r(2,n_2)}([Gamma^2_{n_2}]) = f(gamma_1) f(gamma_2) .$$
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
Given closed paths $gamma_1, gamma_2$, the path $gamma = gamma_1 cdot gamma_2$ is defined by $gamma(t) = gamma_1(2t)$ for $t le 1/2$ and $gamma(t) = gamma_2(2t-1)$ for $t ge 1/2$. Forv $k = 1,2$ choose partitions $I^k_j = [x^k_{j-1},x^k_j]$ of $I$, $j = 1,dots,n_k$, such that $gamma_k(I^k_j)$ is contained in $U_1$ or $U_2$. Define a partition $I_j = [x_{j-1},x_j]$ of $I$, $j = 1,dots,n_1+n_2$, by $x_j = frac{1}{2}x^1_j$ for $j = 0,dots,n_1$ and $x_j = frac{1}{2}(x^2_j+1)$ for $j = n_1,dots, n_1 + n_2$. Then $1/2$ is a partition point. By construction $gamma(I_j)$ is contained in $U_1$ or $U_2$. Choose paths $beta_j^k$ in the appropriate $U_{r(k,j)}$ which connect $x$ and $gamma_k(x^k_j)$, where $beta_0^k$ and $beta^k_{n_k}$ are constant. The paths $beta_j^k$ provide us with paths $beta_j$ in $U_{r(j)}$ connecting $x$ and $gamma(x_j)$: Take $beta_j = beta_j^1$ for $j = 0,dots,n_1$ and $beta_j = beta^2_{n_1-j}$ for $j = n_1,dots,n_1+n_2$. But now it is clear that
$$f(gamma) = f_{r(1)}([Gamma_1]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1)}([Gamma_1]) dots f_{r(n_1)}([Gamma_{n_1}]) f_{r(n_1+1)}([Gamma_{n_1+1}]) dots f_{r(n_1+n_2)}([Gamma_{n_1+n_2}])$$
$$= f_{r(1,1)}([Gamma^1_1]) dots f_{r(1,n_1)}([Gamma^1_{n_1}]) f_{r(2,1)}([Gamma^2_1]) dots f_{r(2,n_2)}([Gamma^2_{n_2}]) = f(gamma_1) f(gamma_2) .$$
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
answered Nov 27 '18 at 23:17
Paul Frost
9,1511631
9,1511631
I would have never come up with such an argument.
– user531587
Nov 28 '18 at 18:08
add a comment |
I would have never come up with such an argument.
– user531587
Nov 28 '18 at 18:08
I would have never come up with such an argument.
– user531587
Nov 28 '18 at 18:08
I would have never come up with such an argument.
– user531587
Nov 28 '18 at 18:08
add a comment |
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