Hi, How can I show that, in a connected category, projections (of the product) are epimorphisms? Thank you, Flavio Leonardo.
Of course it depends rather heavily on what you mean by connected: (1) if you mean that there is a way to get between any two objects via arrows -- and one is allowed to go backwards along arrows -- then this is not true. Any category with products is necessarily connected in this manner and the category of Sets provides a counter-example. Any projection p_0: A x 0 -> A where 0 is the empty set and A is non-empty is non-epic. (2) if you mean that given any objects A and B there is always an arrow f: A -> B (differs from (1) in that you are not allowed to go backwards along arrows) -- that is homsets are non-empty -- then this IS true. This is because every projection in such a category has a section as the composite <1_A,f> p_0 A --------> A x B --------> A is the identity. This makes the projection a retraction and thus epic. (2) if you mean (stretching a bit) that every object has a (regular) epic onto the final object (all objects have global support) then all you need in addition is that the product functors _ x A preserves these epics. This will be the case, for example, if the category is cartesian closed ... however, such a category better not have an initial object! -robin On 16 May, Flavio Leonardo Cavalcanti de Moura wrote:
Hi,
How can I show that, in a connected category, projections (of the product) are epimorphisms?
Thank you,
Flavio Leonardo.
The quickest natural example I know of a connected category in which projections from products needn't be epi is the category of commutative rings. Well, actually, the opposite category. The coproduct of Z_2 and Z_3 is the terminal ring. The two co-projections fail to be monic (the coproduct of a pair of objects in this category is their tensor product).
Peter Freyd writes:
The quickest natural example I know of a connected category in which projections from products needn't be epi is the category of commutative rings. Well, actually, the opposite category. The coproduct of Z_2 and Z_3 is the terminal ring. The two co-projections fail to be monic (the coproduct of a pair of objects in this category is their tensor product).
It is interesting that construction of the coproduct in the category of "just" rings (associative, unitary, but not necessary commutative) is not easy to find in the literature. It seems, however, that it is relatively easy to construct: for given rings R1 and R2, let M1 and M2 be their multiplicative monoids and M be coproduct of M1 and M2 in the category of monoids. Form monoid ring of M over Z. That is, form free abelian group generated by underlying set of M and extend multiplication from generators by distributivity. Let's call resulting ring G. Let K be ideal in G generated by all elements of the form (A + B) - A - B, where both A and B belong to the same ring: either R1 or R2. That is, in ((A + B) - A - B) "+" is addition in R1 or R2, and "-" is subtraction in G. Now, quotient G/K is a coproduct of R1 and R2 in the category of rings, which is easy to prove using universal properties of coproduct M, monoid ring G, and epicness of the projection G ->> G/K. Is there more "direct" description of coproduct in the category of rings?
Nikita.
participants (4)
-
Flavio Leonardo Cavalcanti de Moura -
Nikita Danilov -
Peter Freyd -
Robin Cockett