Talk:Morphism

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Morphism is not just a map f:X->Y, where X and Y are some mathematical objects but it is a map that brings us from (X,*) into (X',*') i.e. f:(X,*)->(X',*') is a morphism in sense that

                      f(x*y)=(fx) *' f(y)

where * and *' are two binary operations.

For example the logarithm to any base is a morphism

                     log(x*y)=log(x)+log(y).

There should be more examples on this topic that can elaborate how to find a specific function or map that is actually a morphism because every function on some objects is not a morphism e.g.

                    sin(x*y) != sin(x) + sin(y)

i.e sin is not a morphism of multiplication over addition.

Be bold, if you know the information, contribute :) Dysprosia 05:50, 30 Sep 2003 (UTC)

Contents

[edit] The problem of terminology

I think the article is currently incorrect in stating that the set of morphisms in a category is denoted HomC(A,B). As far as I know, the majority of recent works in the subject use Hom to denote the abelian group of morphisms, and Mor for the underlying set. Maybe I'm wrong on this point, but certainly Osborne's "Basic Homological Algebra" uses this convention, Lang's "Algebra" certainly uses the Mor notation. Another common notation which is not even mentioned in the article is C(A,B). I might find time to hunt this question down in the next few days & see if I should change the article. Anyone with strong views on this subject, please chip in. Ben 11:05, 11 August 2006 (UTC)

[edit] Example of a bimorphism which is not a isomorphism

If someone could give me examples, and functions or equivalent objects therein, of a category in which there exists a so-called 'bimorphism' which is NOT an isomorphism, by all means discuss it here. As it stands, with my undergraduate background, I see no reason why a bimorphism does not always imply an isomorphism, both being precisely bijective homomorphisms. The more general information on morphisms, on which I am unaware, would be helpful. -Cory

In most algebraic categories, like groups, rings, vector spaces, bimorphisms are always isomorphisms. One generally needs to go to a nonalgebraic category for a counterexample. Take the category of topological spaces. Here the morphisms are continuous functions and the isomorphisms are homeomorphisms. Any bijective continuous map is a bimorphism in this category, however such a map is not necessarily a homeomorphism; the inverse map need not be continuous. Take for example the map
f\colon [0,1) \to S^1
from the half-open unit interval to the unit circle given by
x \mapsto e^{2\pi i x}
This map is bijective and continuous, but the inverse is not continuous (it cuts the circle apart). The map f is then a bimorphism but not an isomorphism. -- Fropuff 20:52, 20 Jun 2005 (UTC)

Damn, that was fast, thanks. One more thing: how does the addition of an implication table (limited to certain categories) at the top of the page grab folks?

I'm not sure what you mean by an implication table. -- Fropuff 21:18, 20 Jun 2005 (UTC)

merely a graphic showing which morphisms imply which others in a 'normal' context. Iso points to epi and mono, which point to surjection and injection, respectively, endo points to homomorphism (and nothing else), etc. I believe the table would be accurate, and give a feel for the interdependence of all these various named morphisms, if we could qualify which categories it holds in. -Cory

Better to state only those implications which are true in all categories. It is too problematic to qualify what constitutes a "normal" category. A quick glance at such at qualified implication table is likely to confuse readers. -- Fropuff 22:14, 20 Jun 2005 (UTC)

[edit] Holomorphism?

Holomorphism exits? And I'm not talking about "holomorphic functions"! —The preceding unsigned comment was added by Cyb3r (talk • contribs) .

A holomorphism, as opposed to a holomorphic function, generally means a holomorphic map between complex manifolds, in other words a morphism in the category of complex manifolds. Of course, a holomorphic function is just a holomorphic map to C. -lethe talk 01:22, 5 December 2005 (UTC)
Hhmmm very nice. Thanks! =) Cyb3r 13:34, 5 December 2005 (UTC)

[edit] split epimorphism = surjection?

In a concrete category, every surjective morphism has a set-theoretic right-inverse function. But does it always have a right-inverse morphism? I know for example that this is the case in any category of algebraic objects (in the sense of universal algebra). Is it true more generally? I know examples of epics which are not surjective, but none of them are split epic. So for example, is there a continuous surjective map who has no continuous right-inverse? -lethe talk + 11:05, 25 March 2006 (UTC)

Ah, probably the map exp: R → S1. It has set-theoretic right-inverse function (the logarithm), but this cannot be continuous on the whole circle, I think. If it were, then the right-inverse would be a homeomorphism onto a subset of the line? -lethe talk + 11:08, 25 March 2006 (UTC)
Therefore surjectivity is weaker than split epimorphism, as the above example shows. And epimorphism is weaker than surjectivity, as the dense image example shows. Does this same heirarchy hold for injections? -lethe talk + 11:21, 25 March 2006 (UTC)

[edit] Something seems wrong here

The definition section states:

For example, in the category of sets, where morphisms are functions, two functions may be identical as set of ordered pairs, but have different codomains.

I can't make sense of that statement (and the sentance that follows it). In the category of sets, the codomain is going to be the set of all of the second elements of the pairs. So if two sets of ordered pairs are identical, then thier codomains are identical, (as are thier domains). So this sentance sure seems wrong to me. linas 23:24, 28 June 2006 (UTC)

JA: No, you are confusing codomain and range. But some of these articles have been worked over in a way that encourages that confusion, so it might help to rewrite things more clearly. Jon Awbrey 02:34, 29 June 2006 (UTC)
Dohh. Of course. Thanks. I'll try to tweak that sentance to say "..ordered pairs, having the same range (mathematics) in different codomains." thus eminding sloppy readers such as I that range and codomain are not the same. I'll make the change if you don't like it, revert, I'm not picky. linas 23:04, 29 June 2006 (UTC)

[edit] Example please?

In the definition section the last paragraph give a quick blurb on how two functions on sets when viewed as functions on sets are the same, but when viewed as morhpisims on a category are diffrent (distinct). I don't understand the mechanisim that it's talking about, perhaps that section could be expanded on a little and possibily include an example?

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