Is a categorical coproduct of epimorphisms (monomorphisms) always an epimorphism (a monomorphism)?
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Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.
I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.
Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
reference-request ct.category-theory abelian-categories products
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up vote
7
down vote
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Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.
I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.
Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
reference-request ct.category-theory abelian-categories products
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.
I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.
Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
reference-request ct.category-theory abelian-categories products
Let $mathbf{C}$ be a category (that does not necessary have a coproduct for every collection of objects). Suppose that we have two families of objects $(A_i)_{iin I}$ and $(B_i)_{iin I}$ in $mathbf{C}$ indexed by the same index set $I$. Assume further that there exist coproducts $A$ and $B$ of $(A_i)_{iin I}$ and of $(B_i)_{iin I}$, respectively. If $f_i:A_ito B_i$ are morphisms for all $iin I$, then there exists a unique morphism $f:Ato B$, which is the coproduct of $(f_i)_{iin I}$.
I know the following results. The morphism $f$ need not be monic, even if all $f_i$ are monic. If $mathbf{C}$ is an abelian category and if all $f_i$ are epic, then $f$ is epic. Therefore, I have two related questions. All references are very welcome.
Question I. In general (where $mathbf{C}$ need not be an abelian category), does $f$ have to be epic, provided that all $f_i$ are epic? If so, could you please provide a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
Question II. If $mathbf{C}$ is an abelian category and if all $f_i$ are monic, then does it follow that $f$ is monic? If so, could you please provide
a proof or a reference? If not, could you please provide a counterexample (for a finite $I$ and for an infinite $I$, if possible)?
reference-request ct.category-theory abelian-categories products
reference-request ct.category-theory abelian-categories products
edited Dec 10 at 0:55
asked Dec 9 at 23:44
Batominovski
22815
22815
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Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.
Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).
Regarding the changes in your question after edits:
In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
$$
(gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
$$
where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.
As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
11
down vote
accepted
Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.
Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).
Regarding the changes in your question after edits:
In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
$$
(gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
$$
where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.
As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.
add a comment |
up vote
11
down vote
accepted
Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.
Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).
Regarding the changes in your question after edits:
In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
$$
(gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
$$
where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.
As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.
add a comment |
up vote
11
down vote
accepted
up vote
11
down vote
accepted
Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.
Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).
Regarding the changes in your question after edits:
In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
$$
(gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
$$
where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.
As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.
Question 1: Yes. The $I$-coproduct-functor $bigsqcup_Icolonprod_{iin I}mathbf{C}tomathbf{C}$ is left-adjoint (its right adjoint is the diagonal functor $Delta_{mathbf{C}}^Icolon mathbf{C}toprod_{iin I}mathbf{C}$), hence always preserves epimorphisms.
Question 2: No, in general (even if $mathbf{C}$ is $I$-coproduct-complete). Cocomplete abelian categories with such property are called satisfying axiom AB4. See nlab article on Grothendieck axioms. Some counterexamples were discussed on SE; see, for example Zhen Lin's answer on MSE. The original source for this axiom is the Tôhoku paper (A.Grothendieck, "Sur quelques points d'algèbre homologique", 1957).
Regarding the changes in your question after edits:
In the situation when $mathbf{C}$ is not $I$-coproduct-complete, the previous proof becomes not entirely correct (probably it may be improved by regarding relative adjoint functors, but I don't think it's relevant to this question). However, there is a straightforward proof of this fact:
$$
(gcirc f=hcirc f)Rightarrow(gcirc fcirc s_i=hcirc fcirc s_i)Rightarrow(gcirc s_icirc f_i=hcirc s_icirc f_i)Rightarrow(gcirc s_i=hcirc s_i)Rightarrow(gcirc h),
$$
where $g$ and $h$ are arbitrary morphisms of $mathbf{C}$ with domain $B$ and $s_i$, $iin I$, are canonical injections of coproducts.
As for the finite $I$ in your second question: there is no such counterexamples. An abelian category is always finitely complete, finitely cocomplete and finite coproducts coincide with finite products, hence if $I$ is finite, then the $I$-coproduct functor of an abelian (or even additive) category is both left- and right-adjoint, hence exact. So the axiom AB4 requires only that the infinite coproducts of monomorphisms should be monomorphisms.
edited Dec 10 at 3:48
answered Dec 10 at 0:28
Oskar
481159
481159
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