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Sep
1 |
comment |
Ring epimorphisms, and epimorphism in the category of small preadditive cats
@MartinBrandenburg, I do not understand your comment. The functor $\Phi$ is clearly defined as the restriction of $\phi_!$. Then $\Phi^*$ is defined sending a functor $F:\mathrm{mod}(S)^{op}\to \mathrm{Ab}$ to $F\circ \Phi$. Finally $\Phi_!$ is the additive left Kan extension along $\Phi$, you can construct it "explicitly" using tensor product of bifunctors. |
Jul
22 |
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Ring epimorphisms, and epimorphism in the category of small preadditive cats
@JeremyRickard, Thanks for your comments! Since there is a bounty ending tomorrow, if you want to post your partial answer with some details, I will accept it before tomorrow. Then it will be possible to edit. Also, I think that to have this partial answer could help others to give a complete answer. |
Jul
21 |
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Ring epimorphisms, and epimorphism in the category of small preadditive cats
So you are asking whether $\Phi$ in my question is surjective on objects, provided $\phi$ is an epimorphism. I do not know if this is true but I do not see any easy counterexample right now. |
Jul
14 |
asked | Ring epimorphisms, and epimorphism in the category of small preadditive cats |
Jul
10 |
answered | Spliting of short exact exact sequences of partially ordered groups |
Jul
8 |
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Spliting of short exact exact sequences of partially ordered groups
Also, are you considering just Abelian groups, or arbitrary group? |
Jul
8 |
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Spliting of short exact exact sequences of partially ordered groups
It depends what you actually want. In Abelian categories, the fact of having a $\gamma$ which is right inverse to $\beta$, tells you that $G=H\oplus G/H$. On the other hand, in general categories that condition just tells you that $G/H$ is a retract of $G$, but not a summand. Of course, in any case, you want $\gamma$ to be a morphism in your category, so in this case you want it to be an order homomorphism but, the mere existence of such $\gamma$ may not imply the splitting. |
Jul
4 |
awarded | Yearling |
Mar
16 |
awarded | Enlightened |
Mar
16 |
awarded | Nice Answer |
Mar
16 |
answered | Regarding the definition of S-flows over a category (given a monoid S) |
Mar
12 |
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Lattice of subobjects of a particular coproduct
oh, you are right, if you want to post it as an answer I'd be happy to accept it. |
Mar
12 |
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Lattice of subobjects of a particular coproduct
Sorry, I wanted to say by an automorphism of the ring. I mean, let $\phi:R\to R$ be an automorphism. Using $\phi$ you can see $R$ as an $R-R$-bimodule (differently than in the standard way) and you can tensor by that bimodule. |
Mar
12 |
asked | Lattice of subobjects of a particular coproduct |
Jan
29 |
awarded | Notable Question |
Dec
13 |
comment |
Exactness of an additive left Kan extension
Many thanks for your edit! The Noetherian case (when you have Kernels) was clear to me but I did not know how to proceed in the general case... |
Dec
13 |
accepted | Exactness of an additive left Kan extension |
Dec
12 |
comment |
Exactness of an additive left Kan extension
So do you claim that also the left adjoint between the categories of contravariant functors on $fp(R)$ and $fp(S)$ is exact provided $\phi_!$ is exact? I'm trying to prove this but I do not see how to prove that the slice category is directed... could you add some details for this other case? (If you prefer I can ask "officially" a new question) |
Nov
25 |
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Exactness of an additive left Kan extension
@FilippoAlbertoEdoardo I'm not defining $F_!$ as a restriction, I define $F$ as a restriction. $F_!$ is the additive left Kan extension to $F$, in particular, while $F$ is a functor $fp(R)\to fp(S)$, $F_!$ goes from the category of additive functors on $fp(R)$ to Abelian groups, to the same category of presheaves over $fp(S)$. |
Nov
24 |
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Exactness of an additive left Kan extension
You are right, $\mathbb Z\to \mathbb Q$ is flat (an Abelian group is flat if and only if it is torsion free). So $f^*$ cannot be restricted to a functor $fp(\mathbb Q)\to fp(\mathbb Z)$, but I am not asking that. I am just restricting $f_!$ and using its exactness (in your example it is clear that the image of a finitely presented module is a finite dimensional rational vector space). |