Talk:Epimorphism

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I've never encountered the use of "monomorphism" and "epimorphism" as synonymous with "1-1" and "onto" respectively, as apparently Bourbaki intended. I have always been thought to consider what category one was working in when encountering these terms. In other words, it makes no sense to say "f is an epimorphism from N to Z". If you mean in Set, then just say surjective. If you mean in monoid category, say so. I wasn't aware there was a confusion over the terminology. In other words, it makes perfect sense to me to say, "This epimorphism isn't surjective". Revolver 22:11, 25 January 2006 (UTC)

I see, it's universal algebra that has different definitions. Revolver 22:15, 25 January 2006 (UTC)

Contents

[edit] Recent changes

I'm not sure of the tone of recent changes. "Beginners often make the mistake..." seems needlessly judgemental, and there are plenty of textbooks that use "epimorphism", rightly or wrongly, as a synonym for surjective (such as Bourbaki, to go no further). If the person who added the stuff on the further concepts of epimorphisms could give their precise definition, I do not see why new pages should be added, except perhaps as redirects to this page. Magidin 05:15, 6 February 2006 (UTC)

Well, I've added back the algebraic definitions. It's an unfortunately fact of life that the two concepts don't agree, but we can't ignore the fact the both meanings are in use. -- Fropuff 05:22, 6 February 2006 (UTC)
I fully agree. I do see that the comment on "beginners being often mistaken" is still there. While the comments following do seem appropriate, I still think the first sentence is needlessly judgemental. Magidin 05:24, 6 February 2006 (UTC)

[edit] "Algebraic" epimorphisms

I have never seen "epimorphism" used for a surjection except in Bourbaki. In my professional experience, epimorphism refers exclusively to the categorical notion. I am familiar with the book of Bergman cited in the references and it does not include "algebraic" epimorphisms.

Please provide a citation for this if you insist on keeping it. As it is, I believe the article to be incorrect and misleading. 141.211.62.20 22:23, 6 February 2006 (UTC)

I cited Bergman's book with reference to the history and to Mac Lane's attempt at clearing up the ambiguities. As for references, I'm pretty sure Lang's Algebra uses epimorphism as a synonym for surjective, and will be happy to withdraw the claim or confirm it tomorrow morning when I can check in my office, as well as provide (or disprove) other references. Magidin 01:03, 7 February 2006 (UTC)
A quick perusal through my bookshelf revealed the following books in which epimorphism is defined simply as a surjective map: Lattice Theory by Birkhoff, Revised Edition; Homological Algebra by Cartan and Eilenberg; Rings and Categories of Modules by Anderson and Fuller (they give the categorical definition in an exercise); Algebra by Lang; A Basic Course in Algebraic Topology by Massey; and Basic Algebra I by Jacobson. All I can say is that my professional experience does not agree with yours: both notions appear to be widespread, and there are few books that even remark on the conflicting use. Magidin 15:37, 7 February 2006 (UTC)
I will add Advanced Linear Algebra (2nd edition) by Steven Roman; the book includes rings, modules, and algebras, and defines epimorphism as a surjective morphism. Magidin 15:49, 7 February 2006 (UTC)
Well, in that case, I suppose it's worth keeping. It seems odd to me, though--maybe it's because I've only ever heard it used in the categorical sense. 141.211.62.20 20:03, 7 February 2006 (UTC)

[edit] Epimorphisms in the category of groups

Why are group epimorphisms surjective? AxelBoldt 02:07, 8 March 2006 (UTC)

Adámek, Herrlich, and Strecker (Abstract and Concrete Categories) claim the proof is far from obvious and assign it as an exercise. They do give some hints. I haven't worked it out for myself. -- Fropuff 06:26, 8 March 2006 (UTC)
The nicest proof I know is from Carl E. Linderholm, who published it in a two page paper in the American Mathematical Monthly, vol. 77 (1970), pp. 176-177. The proof has the virtue of also showing that epimorphisms in the category of finite groups are surjective (the standard proof for groups uses the amalgamated coproduct of Schreier, but that leads to an infinite group). You can see the proof in a post I made on sci.math,
http://groups.google.com/group/sci.math/msg/6d4023d93a2b4300
Magidin 14:33, 8 March 2006 (UTC)
Great, thanks, I'll add the reference to the article. AxelBoldt 17:26, 8 March 2006 (UTC)
I removed your attribution of the two results to Linderholm. He is merely a reference. The proof is essentially due to Schreier, via the amalgamated free product; their existence, plus the properties he established, yield both results as trivial corollaries, and were (and still are) the standard reference. In fact, it proves more: it proves that any subgroup is an equalizer in either category, which is much stronger. The virtue of Linderholm's argument is that it does not use the amalgamated free product, and the same proof establishes the fact in both Grp and Fin Grp. Magidin 20:55, 8 March 2006 (UTC)

[edit] Surjective implies epi

Paul writes

I'm not sure why Axel removed the bit about every "surjecton is an epi"?

It's mentioned in the paragraph above, although not quite as explicitly. It probably can't hurt to mention it twice though. AxelBoldt 05:50, 9 March 2006 (UTC)

Well not quite ;-) The previous paragraph says that the surjective homomorphisms studied in univeral algebra are epis. But in fact all surjections are epis. Paul August ☎ 15:04, 9 March 2006 (UTC)
Good point. So it's all good right now. AxelBoldt 20:38, 9 March 2006 (UTC)

[edit] Bimorphisms in Top

Magidin writes:

bimorphisms are homeos in Top; changed to Haus.

Not so. The map [0,1) → S1 which sends x to exp(2πix) is a bimorphism but not a homeomorphism in Top. The present example works as well. -- Fropuff 18:37, 10 March 2006 (UTC)

Ah, I see my mistake. Of course, "bijective continuous map" is not the same thing as homeomorphism. Magidin 14:35, 13 March 2006 (UTC)
Correct, the inverse has to be continuous as well. Paul August ☎ 16:15, 13 March 2006 (UTC)

[edit] Localizations

any map from a commutative ring R to one of its localizations is an epimorphism.

Surely this is only true for the natural map from R to a localization? Otherwise every endomorphism of a commutative ring would be an epi in Ring, but Z[X] → Z[X] which maps X to 0 isn't. AxelBoldt 17:38, 13 March 2006 (UTC)

[edit] Dazed And Confused

I found your article to be useless and uninformative. It was too confusing to be useful. The catagory definition was too obscure. Rigor does not produce understanding for the user.71.251.189.131 11:55, 25 October 2007 (UTC)