James
Posts:
17
Registered:
3/4/05
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Re: Category question....
Posted:
Mar 9, 2005 8:21 AM
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"Ron Sperber" <ronsperber@optonline.net> wrote in message
news:EW6Xd.10954$WM2.10693@fe11.lga...
> James wrote:
>> "Ron Sperber" <ronsperber@optonline.net> wrote in message
>> news:fWPWd.1610$Kl3.23@fe10.lga...
>>
>>>James wrote:
>>>
>>>>"James Dolan" <jdolan@math-cl-n03.math.ucr.edu> wrote in message
>>>>news:d0fekj$6jh$1@glue.ucr.edu...
>>>>
>>>>
>>>>>in article <d0f89q$cduj$1@netnews.upenn.edu>,
>>>>>james <jpacket1@hotmail.com> wrote:
>>>>>
>>>>>|Hi all, I have a question about categories,
>>>>>|
>>>>>|I have to show that every left exact covariant functor T : R-Mod --->
>>>>>|Ab preserves pullbacks, where R-Mod is the category of left
>>>>>|R-modules. Conclude that if B and C are submodules of a module A,
>>>>>|then for every module M, we have
>>>>>|
>>>>>|Hom_R (M, B /\ C) = Hom_R (M,B) /\ Hom_R (M,C)
>>>>>|
>>>>>|What exactly does it mean to "preserve pullbacks"?
>>>>>|
>>>>>|If I have C --> A and B ---> A, then we can pullback to D, in R-Mod.
>>>>>|They always exist in R-Mod. Now, does this question mean that if I
>>>>>|apply T to A,B,C, and D, then all the arrows stay the same direction,
>>>>>|and I need to try to show that D still exists in Ab? How do I use
>>>>>|the fact that T would be left exact? I don't see how exactness
>>>>>|affects anything because I am only dealing with a "square" of A,B,C,
>>>>>|and D.
>>>>>
>>>>>it's obvious that if you have a commutative square in r-mod with
>>>>>objects a,b,c,d then applying t to it results in a corresponding
>>>>>commutative square in ab with the objects t(a),t(b),t(c),t(d). what's
>>>>>not so obvious is that if the original commutative square in r-mod is
>>>>>a pullback square then the corresponding commutative square in ab is
>>>>>also a pullback square. you won't be able to prove it without the
>>>>>extra assumption that t is left-exact, because it's pretty easy to
>>>>>find an example of a pullback square in r-mod and a (non-left-exact)
>>>>>functor t: r-mod -> ab such that the corresponding commutative square
>>>>>in ab is not a pullback square.
>>>>>
>>>>
>>>>
>>>>Ok, I have done some more research on pullbacks, and I found (without
>>>>proof) that the
>>>>pullback of two maps f : B ---> A and g : C ---> A in R-Mod is
>>>>
>>>>D = {(b,c) in B (+) C : f(b) = g(c)}
>>>>
>>>>I have now managed to prove this. However, I can't seem to figure out
>>>>why
>>>>left exactness will give that the commutative square in Ab is now a
>>>>pullback
>>>>square. Why is this?
>>>>
>>>>Thank you so much for your help,
>>>>
>>>>James
>>>
>>>Perhaps the following could help. One can write D as the kernel of the
>>>homomorphism h:B (+} C -> A given by h(b,c)=f(b)-g(c). What you need to
>>>do is show that TD is the pullback of Tf: TB--->TA and Tg: TC ---> TA.
>>>So first you might show that the pullback of this square is
>>>{(b',c') in (TB (+) TC) | Tf(b')=Tg(c') }. Which is again the kernel of
>>>Th:TB (+) TC -> TA with Th(b',c')=Tf(b')-Tf(c').
>>
>>
>> I'm sorry to ask such a question, but how do you know that
>> Th(b',c')=Tf(b')-Tf(c')?
>> h is a morphism, and Th is a different morphism, in Ab. But who knows
>> what that
>> morphism could be? There's something about functors I don't understand I
>> guess.
>>
>> Are you saying that Th (b',c') = T(h(b',c')) and since h(b,c) = f(b) -
>> g(c), then h(b',c') = f(b') - g(c')?
>> I doubt that's what you're saying since b' and c' live in T(B) and T(C)
>> and h was not defined on those two objects.
> No, what I'm saying is this. h is a morphism from B (+) C -> A.
> Thus Th is a morphism from T(B) (+) T(C) -> TA. I guess your question is
> why is Th=Tf-Tg when all you know is that Tf=T(f-g). I guess part of the
> question is how are we thinking of the categories R-Mod and Ab. I was
> making the assumption that T preserves the additive structure of Hom sets.
> i.e. given f,g in Hom(A,B) that T(f+g)=Tf+Tg in Hom (TA,TB).
You also assumed that T(B (+) C) = T(B) (+) T(C). Is this always true?
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