Talk:Morphism

Citation Needed!

My god, almost none of this is sourced save for notes in basic algebra which probably don't even use the same wording in the definition. Every proposition in here that isn't trivial needs to be justified by a source. If someone doesn't explain why there aren't some I'm going to start removing non-trivial bits of information (such as the following passage: For example, in the category of sets, where morphisms are functions, two functions may be identical as sets of ordered pairs (may have the same range), while having different codomains. The two functions are distinct from the viewpoint of category theory. Thus many authors require that the hom-classes hom(X, Y) be disjoint. In practice, this is not a problem because if this disjointness does not hold, it can be assured by appending the domain and codomain to the morphisms, (say, as the second and third components of an ordered triple).)198.151.130.135 (talk) 11:46, 3 November 2011 (UTC) DDiaz

Definition

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.— Preceding unsigned comment added by 203.175.98.124 (talk) 05:44, 30 September 2003

Be bold, if you know the information, contribute :) Dysprosia 05:50, 30 Sep 2003 (UTC)
That's incorrect. 'Morphism' is used in a more general sense than the one you describe. See the page on category theory. 86.1.196.219 (talk) 03:40, 22 February 2009 (UTC)

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)

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 — Preceding unsigned comment added by 138.192.77.54 (talk) 20:30, 20 June 2005

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
${\displaystyle f\colon [0,1)\to S^{1}}$
from the half-open unit interval to the unit circle given by
${\displaystyle x\mapsto e^{2\pi ix}}$
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? — Preceding unsigned comment added by 138.192.77.54 (talk) 20:57, 20 June 2005

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 — Preceding unsigned comment added by 138.192.77.54 (talk) 21:35, 20 June 2005

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)

Another example of a bimorphism that is not an isomorphism: Consider the free monoid on one generator, i.e. the natural numbers with 0 under addition, as a category, in the standard way by having one object and each number corresponding to an arrow. Then every arrow is epi and mono, essentially by the freeness assumption, however, only the identity arrow has an inverse.

More concretely, we all know from basic facts about that integers that, given any number n, it is true that ${\displaystyle n+x=n+y}$ implies ${\displaystyle x=y}$ , and also that ${\displaystyle x+n=y+n}$ implies ${\displaystyle x=y}$ ; these are precisely the statements that n is mono and epi, respectively.

Thus in this category, every arrow is a bimorphism, but only n = 0, the identity, has an inverse and thus only n = 0 is an isomorphism.

It is difficult to get closer to the notion of isomorphism while remaining strictly weaker than with the concept of a bimorphism -- it is not possible to have an arrow with left-inverse and right-inverse which are not the same. Indeed, any monomorphism with right inverse, and similarly any epimorphism with a left inverse, must be an isomorphism. -- Render787 11:41, 18 Sep 2009 (UTC)

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)

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: RS1. 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)

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)
I suggest to use the term image instead of range, this might prevent the confusion above. 169.237.6.179 20:34, 17 April 2007 (UTC)
image is the right word to use here. Range and codomain are usually used as synonyms for each other. --Matthew Pocock (talk) 13:39, 2 July 2012 (UTC)

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 morphisms on a category are different (distinct). I don't understand the mechanism that it's talking about, perhaps that section could be expanded on a little and possibly include an example?

Answer: Consider the assignment ${\displaystyle x\mapsto x^{2}}$, first as a function ${\displaystyle f_{1}\colon \mathbb {R} \to \mathbb {R} }$, and then as a function ${\displaystyle f_{2}\colon \mathbb {R} \to [0,\infty [}$. Both functions have the same graph, namely ${\displaystyle \{(x,y):x\in \mathbb {R} ,y\in \mathbb {R} ,y\geq 0\}}$. So if your definition of a function is actually just the graph, then ${\displaystyle f_{1}=f_{2}}$, and annotating the codomain is just syntactic sugar. This is a very common viewpoint. However, strictly speaking, it makes good sense to keep ${\displaystyle f_{1}}$ and ${\displaystyle f_{2}}$ separate -- after all, one is surjective and one isn't! Therefore, an alternative definition of a function ${\displaystyle f\colon A\to B}$ is a triple of three sets ${\displaystyle (A,B,\Gamma )}$ so that for every ${\displaystyle a\in A}$ there exists a unique ${\displaystyle b\in B}$ with ${\displaystyle (a,b)\in \Gamma }$.

169.237.6.179 20:29, 17 April 2007 (UTC)

Request for arrow decorations

I don't know much category theory, but having been given a few diagrams to chase, I've found out that monomorphisms are written as arrows with a hook (↪), epimorphisms are written with double-headed arrows (↠), and isomorphisms are written with double arrows (⇒). Furthermore, the algebraists who use these things consider them so standard that the usage isn't glossed in a typical article. It would be really nice if this usage (and any other ways that convey the same information) were described on the morphism page, and were used in Category theory and Commutative diagram articles.

I'm thinking of scattering some pointers to this request around, unless there is a better way of accomplishing it. I also don't know how this is best accomplished in text (as above), where the use of various browsers that don't support Unicode must be taken into account. I didn't see anything in WP:MSM or Wikipedia:Mathematical_symbols that suggests ways of accomplishing this. Still, I'm mostly concerned with the commutative diagrams, where I've had one of those "working mathematician disappointed by wikipedia" moments. Perhaps if I were less averse to m*thw*rld I would have found it easier. –Dan Hoeytalk 13:51, 22 May 2007 (UTC)

This is either incorrect, or technical yet unexplained

"There are two operations defined on every morphism, the domain (or source) and the codomain (or target)."

In what sense of the word "operation" are domain and codomain "operations" on a morphism? In lay terms they are properties or characteristics or just parts of a morphism, but "operations"?

"If a morphism f has domain X and codomain Y, we write f : X → Y. Thus a morphism is an arrow from its domain to its codomain."

In what sense can a morphism actually BE an arrow? Surely this is confusing the concept "morphism" with the symbol used to represent it? Or is arrow written in italics some special technical term? Gwideman (talk) 18:22, 12 December 2009 (UTC)

I have slightly reworded this in the hope of clarifying. I think the meaning of "two operations defined on every morphism" is that there are two operations, and that each of these two can be applied to any morphism. However, if anyone knows better please correct it. JamesBWatson (talk) 14:46, 14 December 2009 (UTC)
For any category C, there are functions (you can also call them operations) dom, and cod on the class of all morphisms of C (hom(C)) which associates each morphism f:XY with its domain, dom(f) = X, and codomain, cod(f) = Y respectively. The term "arrow" is simply a synonym for "morphism". It is used to emphasize the abstract nature of morphisms, as distinct from the more concrete and prototypical example of functions. Paul August 19:20, 14 December 2009 (UTC)
Thanks James and Paul. I guess the issue here was that order of sentences gets in the way of exposition for us readers who don't already know the material. Within the Definition section, introducing the domain and codomain operations before introducing the domain and codomain features of a morphism creates a mental hiccup. I think the section would be more helpful if presented the same info in a slightly different order:
Definition
Given a morphism f which has domain X and codomain Y, we write f : X → Y. Thus a morphism is represented by an arrow from its domain to its codomain. We can also write formal statements about the domain and codomain of f, using dom(f) = X, and codomain, cod(f) = Y respectively.
Hope that helps. Gwideman (talk) 22:19, 14 December 2009 (UTC)

Just how abstract is it?

In the first sentence of the intro we read: "In mathematics, a morphism is an abstraction derived from structure-preserving mappings between two mathematical structures."

Does the intro sentence need to be so hedgey and ethereal-sounding? Isn't the following more straightforward statement true, or at least close enough for a good start? "In mathematics, a morphism is a structure-preserving mapping between two mathematical structures."

I note that page Homomorphism says : "In abstract algebra, a homomorphism is a structure-preserving map between two algebraic structures...". No suggestion that a homomorphism is "an abstraction derived from...". Gwideman (talk) 09:53, 27 February 2013 (UTC)

It really is abstract. Many categories are concrete, i.e. it is "possible to think of the objects of the category as sets with additional structure, and of its morphisms as structure-preserving functions", but some are not. E.g. every (pre-)ordered set can be seen as a category, with the elements of the set being the objects of the category and a morphism/arrow between to objects x and y iff xy, i.e. the morphisms are not functions, and there is no structure to preserve. The nomenclature is certainly derived from morphisms as "structure-preserving mappings between two mathematical structures" but it is more abstract and general than that. — Tobias Bergemann (talk) 10:51, 27 February 2013 (UTC)
I am not following your line of reasoning. Your concluding sentence appears to say "The nomenclature ["morphism"] is certainly derived from morphisms, but is more abstract and general than [morphisms]". Ie: morphisms are more abstract than morphisms.
To me, for morphisms to be "an abstraction derived from structure-preserving mappings between two mathematical structures", as in the intro,
• a morphisms has to be more abstract than a map, or
• the map has to apply to something more abstract than mathematical structures, or
• whatever the map preserves has to be more abstract than structure.
That would seem a tall order to fill. Your discussion of morphisms as functions, and the example where they are not, doesn't seem to be more abstract than "map". Gwideman (talk) 22:49, 27 February 2013 (UTC)
I think the challenge here is that there are at least two uses of the word morphism in mathematics: (1) an arrow in category theory and (2) the general concept of a structure preserving mapping. The lead has to discuss both these uses without confusing the reader. How about

In category theory, a field of mathematics, a morphism or arrow is a map from one object to another. In other fields of mathematics, morphism is also used as a synonym for a structure-preserving mapping from one mathematical structure to another. The notion of morphism recurs in much of contemporary mathematics. In set theory, morphisms are functions; in linear algebra, linear transformations; in group theory, group homomorphisms; in topology, continuous functions, and so on. Each structure-preserving mapping corresponds to a morphism in the associated category; for instance in the category of groups, an object is a group and a morphism is a group homomorphism mapping from one group to another. In this sense, the morphism in category theory is a useful abstraction of the universal properties that all these structure-preserving mappings share.

--Mark viking (talk) 23:31, 27 February 2013 (UTC)
Re: "In category theory, a field of mathematics, a morphism or arrow is a map from one object to another." Isn't that the point? That the objects of a category need not be sets and the morphisms need not be maps/mappings/functions? Or perhaps I am missing some subtlety in your use of the word "map". — Tobias Bergemann (talk) 14:37, 28 February 2013 (UTC)
There is no subtlety here; the idea is to give a context for the term morphism and to explain to the non-specialist reader what morphisms are. From MOS:MATH#Article introduction we have an obligation to make the lead as accessible as possible. I used "map" as term somewhat less concrete than "function", but still familiar enough to someone with a little math training to give them an intuitive idea of what role a morphism plays in category theory --Mark viking (talk) 18:05, 28 February 2013 (UTC)
Thanks Mark for the comment that there might be two distinct meanings of "morphism" in play here, very useful. However, your comment about "synonym" doesn't seem to be describing a synonym, but rather an alternative definition. And having decided to use "map" as the relationship in your first (category theory) definition, the two definitions are barely distinguished, returning to my original comment/question.
How about: In many fields of mathematics, morphism refers to a structure-preserving mapping from one mathematical structure to another...[examples]. In category theory, morphism is a broadly similar idea, but somewhat more abstract: the mathematical objects involved need not be sets, and the relationship between them may be something more general than a map. Gwideman (talk) 23:14, 28 February 2013 (UTC)
As long as the lead section makes it clear from the start that morphisms need not be functions at all I should not protest too loudly if something like Gwideman's suggestion for the lead sentences is used. (We could probably work in a link to the article homomorphism somewhere.) I still personally prefer the intro as it is currently worded. But I agree that the lead section should be as accessible as possible ("as simple as possible, but not simpler"). — Tobias Bergemann (talk) 08:03, 1 March 2013 (UTC)
Reading what I wrote again I am now afraid I may have stifled the discussion (it's a knack, sorry). That was of course not my intention. There is always room for improvement, and if you believe that the lead section may be incomprehensible to the lay reader, by all means, go on and improve it. — Tobias Bergemann (talk) 13:16, 4 March 2013 (UTC)
Thanks for the discussion, and no apology called for -- my fault for not checking back. I have gone boldly ahead to revise the intro so that it combines my suggested first sentence with the original para. I've not added your link to homomorphism, as I'm not sure where to work it in, so please do so. Gwideman (talk) 22:12, 12 March 2013 (UTC)