From: "Charles H. Giffen" Subject: Re: Surjectivity of epimorphisms of groups Date: Tue, 07 Mar 2000 12:25:33 -0500 Newsgroups: sci.math Summary: [missing] Arturo Magidin wrote: > [snip] > > In the case of objects with underlying sets, it is easy to show that > every surjective morphism is an epimorphism, adn every injective > morphism is an epimorphism. Is this a typo? (I don't mean "adn" for "and") If there is also a well-defined notion of > "free object on one generator", then it is easy to show that any > monomorphism is necessarily injective. Usually, showing that > epimorphisms are surjective is the hardest (and often it is not even > correct). > > As another quick example, in the category of Haussdorf Hausdorff! topological > spaces, a continuous map f:T->V is an epimorphism if and only if the > image f(T) is dense in V. > Or, in the category of all topological spaces and continuous maps, a map f: X -> Y is a monomorphism iff it is an injection, and it is an epimorphism iff it is a surjection. Hence, for any space X other than a singleton or the empty set, we have the maps X_discrete -> X -> X_indiscrete where X_discrete is X with every subset open, and X_indiscrete is X with only the empty set and X open. Each of the maps above is both epi and mono, but at least one of them is not a homeomorphism (isomorphism in the category). For algebraists, another example: Consider associative rings with unit. The inclusion of the integers in the rationals is both an epimorphism and a monomorphism, but clearly not an isomorphism. --Chuck Giffen ============================================================================== From: magidin@bosco.berkeley.edu (Arturo Magidin) Subject: Re: Surjectivity of epimorphisms of groups Date: 7 Mar 2000 17:47:22 GMT Newsgroups: sci.math In article <38C53B8D.D503BF47@virginia.edu>, Charles H. Giffen wrote: >Arturo Magidin wrote: >> >[snip] >> >> In the case of objects with underlying sets, it is easy to show that >> every surjective morphism is an epimorphism, adn every injective >> morphism is an epimorphism. > >Is this a typo? (I don't mean "adn" for "and") Yes, I meant "and every injective morphism is a monomorphism." Sorry. That's what I get for not proof-reading >> As another quick example, in the category of Haussdorf > >Hausdorff! Argh. I can never keep straight which is the letter that is supposed to be doubled. Thanks. I'll write it in my blackboard a couple hundred times. [.snip.] >Or, in the category of all topological spaces and continuous >maps, a map f: X -> Y is a monomorphism iff it is an injection, >and it is an epimorphism iff it is a surjection. Hence, for any >space X other than a singleton or the empty set, we have the >maps > > X_discrete -> X -> X_indiscrete > >where X_discrete is X with every subset open, and X_indiscrete >is X with only the empty set and X open. Each of the maps >above is both epi and mono, but at least one of them is not >a homeomorphism (isomorphism in the category). Yes. There is a nice (oldish) article with a bunch of examples, namely "The meaning of mono and epi in some familiar categories", W. Burgess, Canad. Math. Bull. 8 (1965) 759-769. >For algebraists, another example: Consider associative rings with >unit. The inclusion of the integers in the rationals is both an >epimorphism and a monomorphism, but clearly not an isomorphism. In general, for algebras (in the sense of UA), if A->B is any map we can factor it into a quotient map followed by an immersion. If the original map is a nonsurjective epimorphism, then the immersion from this factorization will be an epimorphism and an injection, hence a monomorphism, but not an isomorphism. I believe the name for a category in which epi+mono -> iso is "balanced." ====================================================================== "It's not denial. I'm just very selective about what I accept as reality." --- Calvin ("Calvin and Hobbes") ====================================================================== Arturo Magidin magidin@math.berkeley.edu