C* algebra exact sequences and ideals
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if you have C* algebras $A,B$ and $C$ and $exists$ a short exact sequence as follows $0rightarrow Arightarrow B rightarrow C rightarrow 0 $ where the functions are $phi$ and $psi$ respectively, can you assure that $phi(A)$ is ideal in B and if so why?
How would multiplying $phi(a)$ with an arbitrary $bin B$ assure you still land in $phi(A)$ ?
ideals c-star-algebras exact-sequence
asked Jan 12 at 16:29
sirjoesirjoe
726
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add a comment |
$begingroup$
if you have C* algebras $A,B$ and $C$ and $exists$ a short exact sequence as follows $0rightarrow Arightarrow B rightarrow C rightarrow 0 $ where the functions are $phi$ and $psi$ respectively, can you assure that $phi(A)$ is ideal in B and if so why?
How would multiplying $phi(a)$ with an arbitrary $bin B$ assure you still land in $phi(A)$ ?
ideals c-star-algebras exact-sequence
asked Jan 12 at 16:29
sirjoesirjoe
726
$endgroup$
$begingroup$
kernels of morphisms are ideals and by exactness the kernel of $B to C$ is the image of $A to B$.
$endgroup$
– Paul K
Jan 12 at 16:53
add a comment |
$begingroup$
if you have C* algebras $A,B$ and $C$ and $exists$ a short exact sequence as follows $0rightarrow Arightarrow B rightarrow C rightarrow 0 $ where the functions are $phi$ and $psi$ respectively, can you assure that $phi(A)$ is ideal in B and if so why?
How would multiplying $phi(a)$ with an arbitrary $bin B$ assure you still land in $phi(A)$ ?
ideals c-star-algebras exact-sequence
asked Jan 12 at 16:29
sirjoesirjoe
726
$endgroup$
if you have C* algebras $A,B$ and $C$ and $exists$ a short exact sequence as follows $0rightarrow Arightarrow B rightarrow C rightarrow 0 $ where the functions are $phi$ and $psi$ respectively, can you assure that $phi(A)$ is ideal in B and if so why?
How would multiplying $phi(a)$ with an arbitrary $bin B$ assure you still land in $phi(A)$ ?
ideals c-star-algebras exact-sequence
ideals c-star-algebras exact-sequence
asked Jan 12 at 16:29
sirjoesirjoe
726
asked Jan 12 at 16:29
sirjoesirjoe
726
asked Jan 12 at 16:29
sirjoesirjoe
726
asked Jan 12 at 16:29
sirjoesirjoe
726
asked Jan 12 at 16:29
sirjoesirjoe
726
726
$begingroup$
kernels of morphisms are ideals and by exactness the kernel of $B to C$ is the image of $A to B$.
$endgroup$
– Paul K
Jan 12 at 16:53
add a comment |
$begingroup$
kernels of morphisms are ideals and by exactness the kernel of $B to C$ is the image of $A to B$.
$endgroup$
– Paul K
Jan 12 at 16:53
$begingroup$
kernels of morphisms are ideals and by exactness the kernel of $B to C$ is the image of $A to B$.
$endgroup$
– Paul K
Jan 12 at 16:53
$begingroup$
kernels of morphisms are ideals and by exactness the kernel of $B to C$ is the image of $A to B$.
$endgroup$
– Paul K
Jan 12 at 16:53
add a comment |
1 Answer
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By exactness, you know $psi(phi(a)b) = 0$, hence $phi(a)b in A$ for all $a in A$ and $b in B$.
$endgroup$
$begingroup$
Ah okay that was obvious thank you.
$endgroup$
– sirjoe
Jan 12 at 16:50
add a comment |
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$begingroup$
By exactness, you know $psi(phi(a)b) = 0$, hence $phi(a)b in A$ for all $a in A$ and $b in B$.
$endgroup$
$begingroup$
Ah okay that was obvious thank you.
$endgroup$
– sirjoe
Jan 12 at 16:50
add a comment |
$begingroup$
By exactness, you know $psi(phi(a)b) = 0$, hence $phi(a)b in A$ for all $a in A$ and $b in B$.
$endgroup$
$begingroup$
Ah okay that was obvious thank you.
$endgroup$
– sirjoe
Jan 12 at 16:50
add a comment |
$begingroup$
By exactness, you know $psi(phi(a)b) = 0$, hence $phi(a)b in A$ for all $a in A$ and $b in B$.
$endgroup$
By exactness, you know $psi(phi(a)b) = 0$, hence $phi(a)b in A$ for all $a in A$ and $b in B$.
answered Jan 12 at 16:42
user42761
$begingroup$
Ah okay that was obvious thank you.
$endgroup$
– sirjoe
Jan 12 at 16:50
add a comment |
$begingroup$
Ah okay that was obvious thank you.
$endgroup$
– sirjoe
Jan 12 at 16:50
$begingroup$
Ah okay that was obvious thank you.
$endgroup$
– sirjoe
Jan 12 at 16:50
$begingroup$
Ah okay that was obvious thank you.
$endgroup$
– sirjoe
Jan 12 at 16:50
add a comment |
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$begingroup$
kernels of morphisms are ideals and by exactness the kernel of $B to C$ is the image of $A to B$.
$endgroup$
– Paul K
Jan 12 at 16:53