Talk:Morphism

message: 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)
thread_title: Citation Needed!

message: 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)
replies: text: 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)
thread_title: Definition

message: I found this comment very useful in distinguishing a morphism from a function:
* Every partially ordered set (poset) is a category
        x→y iff x≤y. 
* The morphism here is the ordered pair ⟨x,y⟩, which is not a function: it’s not a set of ordered pairs. 
* Composition in this case is just ‘cutting out the middleman’; that it’s available is a consequence of transitivity.  
to-what-extent-are-morphisms-required-to-be-functions
—  (talk • contribs) 09:02, 29 September 2015 (UTC)
thread_title:  Demonstrate a morphism that isn't a function =

message: 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)
thread_title:  The problem of terminology 

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

text: 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)
thread_title: Example of a bimorphism which is not a isomorphism

message: 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)
replies: text: Hhmmm very nice. Thanks! =) Cyb3r 13:34, 5 December 2005 (UTC)
thread_title:  Holomorphism? 

message: 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)
replies: text: 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)
replies: text: 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)
thread_title:  split epimorphism = surjection? 

message: 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)
replies: text: 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)
replies: text: 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)

text: 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)

text: 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)
thread_title: Something seems wrong here

message: 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 x \mapsto x^2, first as a function f_1 \colon \mathbb{R} \to \mathbb{R}, and then as a function f_2 \colon \mathbb{R} \to [0,\infty[. Both functions have the same graph, namely \{(x,y): x \in \mathbb{R}, y \in \mathbb{R}, y \ge 0\}. So if your definition of a function is actually just the graph, then 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 f_1 and f_2 separate -- after all, one is surjective and one isn't! Therefore, an alternative definition of a function f \colon A \to B is a triple of three sets (A,B,\Gamma) so that for every a \in A there exists a unique b \in B with (a,b) \in \Gamma. 
169.237.6.179 20:29, 17 April 2007 (UTC)
thread_title:  Example please? 

message: 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)
thread_title:  Request for arrow decorations 

message: The entire point of the notion of morphisms as "structure-preserving mappings" is to be as general as possible. While this may cause some confusion or ambiguity at first, the abundance of examples on this page should clarify the power of such a general notion. --Daviddwd (talk) 20:30, 11 November 2014 (UTC)
thread_title:  Disambiguation needed for structure-preserving mapping?  

message: I am not sure what to do with  by  that introduced the term G-structure into the section Morphism#Some specific morphisms. These additions feel somewhat out of place to me, and more specifically,  that appended the words
:called the G-structure
to the sentence
:Thus h ∘ g: Y → Y is idempotent; that is, (h ∘ g)2  h ∘ (g ∘ h) ∘ g  h ∘ g.
looks like a paste error to me. —Tea2min (talk) 09:56, 4 May 2016 (UTC)
replies: text: I agree, these edits must be reverted, and I'll do it. In fact the article defines automorphisms for any structure, and there is no reason to choose, as a prominent example, a very specialized structure, much less known than many much more common examples (such as topological spaces or vector spaces). However, it would be useful to quote that the automorphisms form always a group. D.Lazard (talk) 17:46, 4 May 2016 (UTC)
thread_title:  G-structures