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| Tony |
 Posted: Sun Feb 13, 2005 8:36 am |
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Guest
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Hi all,
I am trying to prove the following which I would appreciate any input on :
Denote |_| by the coproduct. I want to show that if M_i is a family of
R-modules and N_i are submodules of M_i, then
|_| (M_i / N_i) = ( |_| M_i ) / (|_| N_i )
I am trying to do this using definition of coproduct...but I'm not sure if I
am getting it right. So in order to see if ( |_| M_i ) / (|_| N_i ) is the
coproduct of (M_i / N_i), I need to show that ( |_| M_i ) / (|_| N_i )
satisfies the universal property of coproducts. So, I make the following
diagram :
M_i / N_i --------> ( |_| M_i ) / (|_| N_i )
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\/
X
The morphisms I specify F_i from M_i / N_i to ( |_| M_i ) / (|_| N_i )
which I need to specify as part of the definition of coproduct will be that
m_i + N_i goes to (0,0,...,0,m_i,0,0,...) + |_| N_i
where m_i is an element of M_i.
Now if f_i are a family of maps from M_i / N_i to the R-module X,
Now I define G from ( |_| M_i ) / (|_| N_i ) to X by sending (m1,m2,m3,...)
+ |_| N_ i to summation f_i (m_i + N_i).
I'm going to need to state of course that all of these tuples that I am
writing down have all but finitely many terms zero, as I tacitly identify
the coproduct with the direct sum (am I allowed to do that? That is, am I
allowed to state the words direct sum and use the fact that all but finitely
many terms are zero when I am trying to prove something entirely in terms of
category theory stuff like coproduct?)
Then, I think these maps work...and if H is another morphism from ( |_|
M_i ) / (|_| N_i ) to X making this diagram commute, I think I can show that
H equals G (I did it last night and I'll have to recreate it again but I
think it's ok)...
Am I right with my maps and with everything?
Thanks for any help,
Tony |
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| Tony |
| Posted: Sun Feb 13, 2005 9:20 am |
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"Lukas Horosiewicz" <horosiewicz@gmail.com> wrote in message
news:1108307385.012948.249670@g14g2000cwa.googlegroups.com...
[quote:f1a6e09d33]
Tony wrote:
Hi all,
I am trying to prove the following which I would appreciate any input
on :
Denote |_| by the coproduct. I want to show that if M_i is a family
of
R-modules and N_i are submodules of M_i, then
|_| (M_i / N_i) = ( |_| M_i ) / (|_| N_i )
They are not equal, just equivalent in the category.
I am trying to do this using definition of coproduct...but I'm not
sure if I
am getting it right. So in order to see if ( |_| M_i ) / (|_| N_i )
is the
coproduct of (M_i / N_i), I need to show that ( |_| M_i ) / (|_| N_i
)
satisfies the universal property of coproducts. So, I make the
following
diagram :
Again here just equivalent (isomorphic).
M_i / N_i --------> ( |_| M_i ) / (|_| N_i )
|
|
|
|
\/
X
The morphisms I specify F_i from M_i / N_i to ( |_| M_i ) / (|_| N_i
)
which I need to specify as part of the definition of coproduct will
be that
m_i + N_i goes to (0,0,...,0,m_i,0,0,...) + |_| N_i
where m_i is an element of M_i.
Now if f_i are a family of maps from M_i / N_i to the R-module X,
Now I define G from ( |_| M_i ) / (|_| N_i ) to X by sending
(m1,m2,m3,...)
+ |_| N_ i to summation f_i (m_i + N_i).
I hope you're not summing the morphisms here?
[/quote:f1a6e09d33]
No, I'm summing f_i (m_i + N_i) . The target of f_i is X, an R-module, so
this
sum makes sense. Also, this sum makes sense since in the element
(m1,m2,m3,...) + |_| N_i , only
finitely many of the m_i are non-zero.
[quote:f1a6e09d33]I'm going to need to state of course that all of these tuples that I
am
writing down have all but finitely many terms zero, as I tacitly
identify
the coproduct with the direct sum (am I allowed to do that? That is,
am I
allowed to state the words direct sum and use the fact that all but
finitely
many terms are zero when I am trying to prove something entirely in
terms of
category theory stuff like coproduct?)
Yes you are since you're using the universal property of the coproduct;
ie that any two objects in R-Mod satisfying the property will be
equivalent -- isomorphic.
Then, I think these maps work...and if H is another morphism from (
|_|
M_i ) / (|_| N_i ) to X making this diagram commute, I think I can
show that
H equals G (I did it last night and I'll have to recreate it again
but I
think it's ok)...
Am I right with my maps and with everything?
I didn't have time to check, hope someone else does. But why don't you
just define the projections pi_i: M_i -> M_i/N_i and look at the map
\sum pi_i: M-i -> M_i/N_i (generally \sum f_i is mono/epi iff all the
components are) and look at the kernel?
[/quote:f1a6e09d33] |
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| Lukas Horosiewicz |
| Posted: Sun Feb 13, 2005 10:09 am |
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Guest
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Tony wrote:
[quote:bec3d78f9c]Hi all,
I am trying to prove the following which I would appreciate any input
on :
Denote |_| by the coproduct. I want to show that if M_i is a family
of
R-modules and N_i are submodules of M_i, then
|_| (M_i / N_i) = ( |_| M_i ) / (|_| N_i )
[/quote:bec3d78f9c]
They are not equal, just equivalent in the category.
[quote:bec3d78f9c]I am trying to do this using definition of coproduct...but I'm not
sure if I
am getting it right. So in order to see if ( |_| M_i ) / (|_| N_i )
is the
coproduct of (M_i / N_i), I need to show that ( |_| M_i ) / (|_| N_i
)
satisfies the universal property of coproducts. So, I make the
following
diagram :
[/quote:bec3d78f9c]
Again here just equivalent (isomorphic).
[quote:bec3d78f9c]M_i / N_i --------> ( |_| M_i ) / (|_| N_i )
|
|
|
|
\/
X
The morphisms I specify F_i from M_i / N_i to ( |_| M_i ) / (|_| N_i
)
which I need to specify as part of the definition of coproduct will
be that
m_i + N_i goes to (0,0,...,0,m_i,0,0,...) + |_| N_i
where m_i is an element of M_i.
Now if f_i are a family of maps from M_i / N_i to the R-module X,
Now I define G from ( |_| M_i ) / (|_| N_i ) to X by sending
(m1,m2,m3,...)
+ |_| N_ i to summation f_i (m_i + N_i).
[/quote:bec3d78f9c]
I hope you're not summing the morphisms here?
[quote:bec3d78f9c]I'm going to need to state of course that all of these tuples that I
am
writing down have all but finitely many terms zero, as I tacitly
identify
the coproduct with the direct sum (am I allowed to do that? That is,
am I
allowed to state the words direct sum and use the fact that all but
finitely
many terms are zero when I am trying to prove something entirely in
terms of
category theory stuff like coproduct?)
[/quote:bec3d78f9c]
Yes you are since you're using the universal property of the coproduct;
ie that any two objects in R-Mod satisfying the property will be
equivalent -- isomorphic.
[quote:bec3d78f9c]Then, I think these maps work...and if H is another morphism from (
|_|
M_i ) / (|_| N_i ) to X making this diagram commute, I think I can
show that
H equals G (I did it last night and I'll have to recreate it again
but I
think it's ok)...
Am I right with my maps and with everything?
[/quote:bec3d78f9c]
I didn't have time to check, hope someone else does. But why don't you
just define the projections pi_i: M_i -> M_i/N_i and look at the map
\sum pi_i: M-i -> M_i/N_i (generally \sum f_i is mono/epi iff all the
components are) and look at the kernel? |
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| Jannick Asmus |
| Posted: Sun Feb 13, 2005 10:10 am |
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Guest
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On 13.02.2005 14:36, Tony wrote:
[quote:f0d565453d]I am trying to prove the following which I would appreciate any input on :
Denote |_| by the coproduct. I want to show that if M_i is a family of
R-modules and N_i are submodules of M_i, then
|_| (M_i / N_i) = ( |_| M_i ) / (|_| N_i )
[/quote:f0d565453d]
I presume that R is a commutative ring with unity.
[quote:f0d565453d]
I am trying to do this using definition of coproduct...but I'm not sure if I
am getting it right. So in order to see if ( |_| M_i ) / (|_| N_i ) is the
coproduct of (M_i / N_i), I need to show that ( |_| M_i ) / (|_| N_i )
satisfies the universal property of coproducts. So, I make the following
diagram :
M_i / N_i --------> ( |_| M_i ) / (|_| N_i )
|
|
|
|
\/
X
The morphisms I specify F_i from M_i / N_i to ( |_| M_i ) / (|_| N_i )
which I need to specify as part of the definition of coproduct will be that
m_i + N_i goes to (0,0,...,0,m_i,0,0,...) + |_| N_i
where m_i is an element of M_i.
Now if f_i are a family of maps from M_i / N_i to the R-module X,
Now I define G from ( |_| M_i ) / (|_| N_i ) to X by sending (m1,m2,m3,...)
+ |_| N_ i to summation f_i (m_i + N_i).
I'm going to need to state of course that all of these tuples that I am
writing down have all but finitely many terms zero, as I tacitly identify
the coproduct with the direct sum (am I allowed to do that? That is, am I
allowed to state the words direct sum and use the fact that all but finitely
many terms are zero when I am trying to prove something entirely in terms of
category theory stuff like coproduct?)
[/quote:f0d565453d]
Yes, it is allowed, since the co-product and the direct sum are both the
representation of the functor
prod_{i e I} Hom_R(M_i,.),
hence they are naturally isomorphic.
[quote:f0d565453d]
Then, I think these maps work...and if H is another morphism from ( |_|
M_i ) / (|_| N_i ) to X making this diagram commute, I think I can show that
H equals G (I did it last night and I'll have to recreate it again but I
think it's ok)...
[/quote:f0d565453d]
If you know the notion of exact sequences and functors, it is easier to
show stuff like this by the gathering the so-called universal property
in the equivalence of functors
prod_{i e I} Hom_R(M_i,.) -> Hom_R( (+)_i M_i,.).
Best,
J. |
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