Mixing
Inverse limits of groups
1
$begingroup$
Let $varprojlim {A_i} = P$ with $mu_{ji} : A_j to A_i$ be surjective. Show $pi_i:P to A_i$ is surjective for all $iinmathbb{Z}^+$.
Let $a_i in A_i$. Then for all $j > i$, the set $mu_{ji}^{-1}(a_i)$ is non empty.
Take $a = (mu_{i1}(a_i),mu_{i2}(a_i), ..., a_i, a_{i+1}, ... )$ where $a_j in mu_{ji}^{-1}(a_i)$ for $j>i$
Clearly $ain P$ since $mu_{ji}(a_j) = a_i$ and $pi_i(a) = a_i$.
Is my attempt correct?
group-theory proof-verification
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
$endgroup$
add a comment |
1
$begingroup$
Let $varprojlim {A_i} = P$ with $mu_{ji} : A_j to A_i$ be surjective. Show $pi_i:P to A_i$ is surjective for all $iinmathbb{Z}^+$.
Let $a_i in A_i$. Then for all $j > i$, the set $mu_{ji}^{-1}(a_i)$ is non empty.
Take $a = (mu_{i1}(a_i),mu_{i2}(a_i), ..., a_i, a_{i+1}, ... )$ where $a_j in mu_{ji}^{-1}(a_i)$ for $j>i$
Clearly $ain P$ since $mu_{ji}(a_j) = a_i$ and $pi_i(a) = a_i$.
Is my attempt correct?
group-theory proof-verification
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
$endgroup$
$begingroup$
What is your index set?
$endgroup$
– Arturo Magidin
Jan 30 at 17:05
$begingroup$
@ArturoMagidin it's $mathbb{N}$
$endgroup$
– Good Morning Captain
Feb 1 at 22:43
1
$begingroup$
Your $a$ need not be consistent as written, because you have no guarantee that $mu_{j+1,j}(a_{j+1}) = a_{j}$ under your definition; you only required that $mu_{j+1,i}(a_{j+1}) = mu_{j,i}(a_j)$. So your $a$ may fail to lie in the inverse limit. If your index set is $mathbb{N}$, then construct $a$ inductively, by choosing $a_{i+1}$ to map to $a_i$, and then $a_{k+1}$ to map to $a_k$ assuming you alreaady know what $a_k$ is.
$endgroup$
– Arturo Magidin
Feb 1 at 22:52
$begingroup$
For example, consider the inverse limit of $mathbb{Z}_{3^n}$ with the obvious projections, and take the element $1$ in $mathbb{Z}_3$. Your definition would allow me to pick $4$ for the $mathbb{Z}_{3^2}$ coordinate, and then pick $1$ for the $mathbb{Z}_{3^3}$ coordinate, but this element is not in the inverse limit because $1notequiv 4 pmod{3^2}$, even though both map to $1$ in $mathbb{Z}_3$.
$endgroup$
– Arturo Magidin
Feb 1 at 22:54
$begingroup$
@ArturoMagidin I appreciate the response. I will review this and make another attempt later on
$endgroup$
– Good Morning Captain
Feb 1 at 22:57
add a comment |
1
1
1
$begingroup$
Let $varprojlim {A_i} = P$ with $mu_{ji} : A_j to A_i$ be surjective. Show $pi_i:P to A_i$ is surjective for all $iinmathbb{Z}^+$.
Let $a_i in A_i$. Then for all $j > i$, the set $mu_{ji}^{-1}(a_i)$ is non empty.
Take $a = (mu_{i1}(a_i),mu_{i2}(a_i), ..., a_i, a_{i+1}, ... )$ where $a_j in mu_{ji}^{-1}(a_i)$ for $j>i$
Clearly $ain P$ since $mu_{ji}(a_j) = a_i$ and $pi_i(a) = a_i$.
Is my attempt correct?
group-theory proof-verification
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
$endgroup$
Let $varprojlim {A_i} = P$ with $mu_{ji} : A_j to A_i$ be surjective. Show $pi_i:P to A_i$ is surjective for all $iinmathbb{Z}^+$.
Let $a_i in A_i$. Then for all $j > i$, the set $mu_{ji}^{-1}(a_i)$ is non empty.
Take $a = (mu_{i1}(a_i),mu_{i2}(a_i), ..., a_i, a_{i+1}, ... )$ where $a_j in mu_{ji}^{-1}(a_i)$ for $j>i$
Clearly $ain P$ since $mu_{ji}(a_j) = a_i$ and $pi_i(a) = a_i$.
Is my attempt correct?
group-theory proof-verification
group-theory proof-verification
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
edited Jan 27 at 5:27
YuiTo Cheng
2,1212837
2,1212837
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
asked Jan 27 at 1:09
Good Morning CaptainGood Morning Captain
551420
551420
$begingroup$
What is your index set?
$endgroup$
– Arturo Magidin
Jan 30 at 17:05
$begingroup$
@ArturoMagidin it's $mathbb{N}$
$endgroup$
– Good Morning Captain
Feb 1 at 22:43
1
$begingroup$
Your $a$ need not be consistent as written, because you have no guarantee that $mu_{j+1,j}(a_{j+1}) = a_{j}$ under your definition; you only required that $mu_{j+1,i}(a_{j+1}) = mu_{j,i}(a_j)$. So your $a$ may fail to lie in the inverse limit. If your index set is $mathbb{N}$, then construct $a$ inductively, by choosing $a_{i+1}$ to map to $a_i$, and then $a_{k+1}$ to map to $a_k$ assuming you alreaady know what $a_k$ is.
$endgroup$
– Arturo Magidin
Feb 1 at 22:52
$begingroup$
For example, consider the inverse limit of $mathbb{Z}_{3^n}$ with the obvious projections, and take the element $1$ in $mathbb{Z}_3$. Your definition would allow me to pick $4$ for the $mathbb{Z}_{3^2}$ coordinate, and then pick $1$ for the $mathbb{Z}_{3^3}$ coordinate, but this element is not in the inverse limit because $1notequiv 4 pmod{3^2}$, even though both map to $1$ in $mathbb{Z}_3$.
$endgroup$
– Arturo Magidin
Feb 1 at 22:54
$begingroup$
@ArturoMagidin I appreciate the response. I will review this and make another attempt later on
$endgroup$
– Good Morning Captain
Feb 1 at 22:57
add a comment |
$begingroup$
What is your index set?
$endgroup$
– Arturo Magidin
Jan 30 at 17:05
$begingroup$
@ArturoMagidin it's $mathbb{N}$
$endgroup$
– Good Morning Captain
Feb 1 at 22:43
1
$begingroup$
Your $a$ need not be consistent as written, because you have no guarantee that $mu_{j+1,j}(a_{j+1}) = a_{j}$ under your definition; you only required that $mu_{j+1,i}(a_{j+1}) = mu_{j,i}(a_j)$. So your $a$ may fail to lie in the inverse limit. If your index set is $mathbb{N}$, then construct $a$ inductively, by choosing $a_{i+1}$ to map to $a_i$, and then $a_{k+1}$ to map to $a_k$ assuming you alreaady know what $a_k$ is.
$endgroup$
– Arturo Magidin
Feb 1 at 22:52
$begingroup$
For example, consider the inverse limit of $mathbb{Z}_{3^n}$ with the obvious projections, and take the element $1$ in $mathbb{Z}_3$. Your definition would allow me to pick $4$ for the $mathbb{Z}_{3^2}$ coordinate, and then pick $1$ for the $mathbb{Z}_{3^3}$ coordinate, but this element is not in the inverse limit because $1notequiv 4 pmod{3^2}$, even though both map to $1$ in $mathbb{Z}_3$.
$endgroup$
– Arturo Magidin
Feb 1 at 22:54
$begingroup$
@ArturoMagidin I appreciate the response. I will review this and make another attempt later on
$endgroup$
– Good Morning Captain
Feb 1 at 22:57
$begingroup$
What is your index set?
$endgroup$
– Arturo Magidin
Jan 30 at 17:05
$begingroup$
What is your index set?
$endgroup$
– Arturo Magidin
Jan 30 at 17:05
$begingroup$
@ArturoMagidin it's $mathbb{N}$
$endgroup$
– Good Morning Captain
Feb 1 at 22:43
$begingroup$
@ArturoMagidin it's $mathbb{N}$
$endgroup$
– Good Morning Captain
Feb 1 at 22:43
1
1
$begingroup$
Your $a$ need not be consistent as written, because you have no guarantee that $mu_{j+1,j}(a_{j+1}) = a_{j}$ under your definition; you only required that $mu_{j+1,i}(a_{j+1}) = mu_{j,i}(a_j)$. So your $a$ may fail to lie in the inverse limit. If your index set is $mathbb{N}$, then construct $a$ inductively, by choosing $a_{i+1}$ to map to $a_i$, and then $a_{k+1}$ to map to $a_k$ assuming you alreaady know what $a_k$ is.
$endgroup$
– Arturo Magidin
Feb 1 at 22:52
$begingroup$
Your $a$ need not be consistent as written, because you have no guarantee that $mu_{j+1,j}(a_{j+1}) = a_{j}$ under your definition; you only required that $mu_{j+1,i}(a_{j+1}) = mu_{j,i}(a_j)$. So your $a$ may fail to lie in the inverse limit. If your index set is $mathbb{N}$, then construct $a$ inductively, by choosing $a_{i+1}$ to map to $a_i$, and then $a_{k+1}$ to map to $a_k$ assuming you alreaady know what $a_k$ is.
$endgroup$
– Arturo Magidin
Feb 1 at 22:52
$begingroup$
For example, consider the inverse limit of $mathbb{Z}_{3^n}$ with the obvious projections, and take the element $1$ in $mathbb{Z}_3$. Your definition would allow me to pick $4$ for the $mathbb{Z}_{3^2}$ coordinate, and then pick $1$ for the $mathbb{Z}_{3^3}$ coordinate, but this element is not in the inverse limit because $1notequiv 4 pmod{3^2}$, even though both map to $1$ in $mathbb{Z}_3$.
$endgroup$
– Arturo Magidin
Feb 1 at 22:54
$begingroup$
For example, consider the inverse limit of $mathbb{Z}_{3^n}$ with the obvious projections, and take the element $1$ in $mathbb{Z}_3$. Your definition would allow me to pick $4$ for the $mathbb{Z}_{3^2}$ coordinate, and then pick $1$ for the $mathbb{Z}_{3^3}$ coordinate, but this element is not in the inverse limit because $1notequiv 4 pmod{3^2}$, even though both map to $1$ in $mathbb{Z}_3$.
$endgroup$
– Arturo Magidin
Feb 1 at 22:54
$begingroup$
@ArturoMagidin I appreciate the response. I will review this and make another attempt later on
$endgroup$
– Good Morning Captain
Feb 1 at 22:57
$begingroup$
@ArturoMagidin I appreciate the response. I will review this and make another attempt later on
$endgroup$
– Good Morning Captain
Feb 1 at 22:57
add a comment |
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$begingroup$
What is your index set?
$endgroup$
– Arturo Magidin
Jan 30 at 17:05
$begingroup$
@ArturoMagidin it's $mathbb{N}$
$endgroup$
– Good Morning Captain
Feb 1 at 22:43
1
$begingroup$
Your $a$ need not be consistent as written, because you have no guarantee that $mu_{j+1,j}(a_{j+1}) = a_{j}$ under your definition; you only required that $mu_{j+1,i}(a_{j+1}) = mu_{j,i}(a_j)$. So your $a$ may fail to lie in the inverse limit. If your index set is $mathbb{N}$, then construct $a$ inductively, by choosing $a_{i+1}$ to map to $a_i$, and then $a_{k+1}$ to map to $a_k$ assuming you alreaady know what $a_k$ is.
$endgroup$
– Arturo Magidin
Feb 1 at 22:52
$begingroup$
For example, consider the inverse limit of $mathbb{Z}_{3^n}$ with the obvious projections, and take the element $1$ in $mathbb{Z}_3$. Your definition would allow me to pick $4$ for the $mathbb{Z}_{3^2}$ coordinate, and then pick $1$ for the $mathbb{Z}_{3^3}$ coordinate, but this element is not in the inverse limit because $1notequiv 4 pmod{3^2}$, even though both map to $1$ in $mathbb{Z}_3$.
$endgroup$
– Arturo Magidin
Feb 1 at 22:54
$begingroup$
@ArturoMagidin I appreciate the response. I will review this and make another attempt later on
$endgroup$
– Good Morning Captain
Feb 1 at 22:57