Talk:Monomorphism

Where does this fact fit?

In context of polymorphism, monomorphic means the opposite (not genetically different from). Where should this fact go? --134.36.64.135 (talk) 14:07, 27 October 2009 (UTC)

Where it currently is: at the top of the page, where there is a link directing readers to the page on polymorphism for other uses. Magidin (talk) 14:38, 27 October 2009 (UTC)

More general setting of category theory

"...a left-cancellative morphism, that is, a map f : X → Y such that..." — but in this generality a morphism is not a map, is it? Boris Tsirelson (talk) 11:05, 2 September 2010 (UTC)

Probably should be "arrow" instead of "map". Magidin (talk) 14:16, 2 September 2010 (UTC)

Suggestions

This page is rather impenetrable in spite of the fact that the subject is quite simple. The purpose of the article is (ought to be) to give a rigorous definition of monomorphism (so it does), and to illustrate that a monomorphism isn't necessarily injective. The example illustrating the latter point fails to bring the point home to the less advanced reader.

Suggestions:

• Either delete the section "Terminology" or rename it to "History" and place it in the end.
• Mention early that injective is equivalent to left invertible (barring pathologies).
• The term "monic" is used before it is defined.
• Emphasize that an injective function is never many-to-one.
• Emphasize that an monomorphism may be many-to-one.

(Those last two bullets together are the whole second point of the article. I got that after having penetrated the example, which took some time. I felt rather silly when i finally understood this almost trivial point.)

• Define Q, Z and Q/Z before using them.
• Using the definitions, point out that q:Q->Q/Z is many-to-one.
• Don't use h before G is introduced.
• Avoid abuse of notation. The equation x = (h(x) + 1)y is pretty damned awful in this respect.
• Skip the NB; it's not needed anyway as is pointed out. Waste of space.
• Reverse the order of things; First show that in Div the the conditions for monomorphism are equivalent.

Then there is a contradiction. In the example it says that monomormhisms and invertible maps are the same things for groups, rings, and all abelian categories (and proceeds to show that this certainly not is the case.) YohanN7 (talk) 14:43, 18 September 2012 (UTC)

I've given the definition of Q and Z. I'm not sure that this is the place to describe a quotient, though. Your points 4 and 5 above, though seem to utterly miss that the meaning of "injective" is exactly "one-to-one". Saying that one should emphasize that "injective" functions are not many-to-one is like saying that one should emphasize that if an object is "red", then it can never be "not red". Your antepenultimate point also seems to have missed the point: there is absolutely no abuse of notation in that equation! h(x) is an integer, so h(x)+1 is an integer, and we can multiply the rational number y by the integer h(x)+1. What is abusive about it, exactly? In any case, I've replaced the expression h(x)+1 with n, so perhaps that will prevent your confusion. Your second point is actually incorrect: injective is not equivalent to "left invertible" in many cases! For example, in the category of groups, if a map H->G is left-invertible, then G must be a semidirect product of a normal subgroup N by H. This almost never happens. Even in abelian groups, you would need H to be a direct summand of G; again, that is rare. This is not a pathology; in fact, categories in which injective is equivalent to left invertible are extremely rare. Finally, your final comment again missed the meaning of the statement. It says "the category of groups", not "any category of groups". And the category of divisible abelian groups is not an abelian category (even though its objects are abelian groups): in order to be an abelian category, it must have binary biproducts, every arrow must have a kernel and a cokernel, and every monomorphism must be a kernel and every epimorphism a cokernel. The category of divisible abelian groups is not an abelian category, among other reasons, because it does not have kernels: the map Q->Q/Z does not have a kernel, since Z is not an object in the category. So what contradiction is it that you believe is in that statement? Magidin (talk) 16:13, 18 September 2012 (UTC)

Notable improvements! Thanks, but why so hostile?

You indicate that left invertible and "injective" have deeper meanings than what I'm aware of. (Or am I missing the point again?) I take your answer as being that inverses of some kind must be morphisms too. This is far from obvious from the article. I thought these concepts had the usual set theoretical meaning.

So, my last point was clearly wrong then. There is no contradiction. Thanks for a thorough explanation though. Perhaps part of the explanation could go in the article?

About the notation: y and h(x) reside in different spaces for one thing. And, in particular, implied n times repeated group multiplication (which happens to be +) indicated by juxtaposition is certainly abuse of notation in a context where it hasn't been explained. This is fine, even recommendable, for a textbook after the abuse is explained. Here it's potentially confusing. Please note that I don't say that anything is wrong. It's perfectly correct. Let's not have an argument about this. YohanN7 (talk) 17:27, 18 September 2012 (UTC)

Let H be a subgroup of G such that there is no normal subgroup N such that G=NH (semidirect product). Let f:H->G denote the injection map. Let l, at first, denote the identity map on G. Restrict the domain of l to H. Finally extend l to all of G by setting l(g) = anything in H for g in G\H. Then l ° f = Id_H. This qualifies l as a left inverse according to the article (first section after leade). Obviously, l isn't necessarily a homomorphism.

Suggestion: (I hope I'm not completely off as usual;) Define "Left inverse, Right inverse" and "Inverse" somewhere in the context of Category Theory and link it. Or, if there are other established terms for this (inversomorphisms?;), then use those terms. YohanN7 (talk) 19:42, 18 September 2012 (UTC)

A morphism f:X->Y is "left invertible" if and only if there exists a morphism g:Y->X such that gf is the identity morphism of X; f is "right invertible" if and only if there exists a morphism h:Y->X such that fh is the identity morphism of Y. Note that we start with a morphism. Those are the basic, standard definitions. In the category of Set, of course, injective = monomorphism, and injective+(empty codomain <->empty domain) = left invertible; while surjective=epimorphism and (surjective=right invertible) is equivalent to the Axiom of Choice. But that's in the category Set.

As to your l: in fact, l is never a morphism under the listed conditions. You are defining a set-theoretic inverse, not an inverse (that is, you are first mapping to the category Set using the forgetful functor, and then you are considering your functions there; you've left the building already). In the context of category theory, any arrow must be a morphism. If you are talking about the category of groups, you are talking about groups, and group morphisms between them, not arbitrary functions between them. So your l does not "live" in the category in question, it lives in a different category and so is not even invited to the party. Left inverse, right inverse, and inverse are morphisms of the category that have the relevant properties. That paragraph clearly needs some tightening, though that will only make it more intimidating. Magidin (talk) 20:43, 18 September 2012 (UTC)

Thanks for your reply. Now I'm with you 99%. The remaining 1% is that I don't agree that an edit clarifying these points will make the article more intimidating. The first occurrences of the terms involved could well be linked and live in their own (small) article, like "Inverse (Category Theory)". I stand by what I said at the outset. The actual subject of the article is quite simple. But the slightest deviation from "rigor" in terminology opens up for confusion.

I am slightly embarrassed that I didn't understand immediately from the context. But still, inside every category, I bet that even some mathematicians could use left inverse in the set theoretic sense. Surely, a topologist might say that an injective continuous function has a left inverse. He might or might not add that then we aren't talking about a homeomorphism depending on his mood. YohanN7 (talk) 21:37, 18 September 2012 (UTC)

I was afraid that clarifying would require an extensive discussion about how even though a function "exists" it may not "count" because it is outside the category, etc. Turns out that simply changing "function"/"map" to "morphism" did the trick, in my opinion. As to your second paragraph, my actual experience is that people (including myself) are very clear when they deal with such set-theoretic inverses that they are set-theoretic. For example, a set-theoretic right inverse of the group quotient map G->G/N has a special name (a "transversal of N") precisely to make the distinction between morphisms and set-theoretic maps between groups that are not morphisms. When no special terminology exists, "set-theoretic" is often added to make it clear. Now, sure, there are sloppy writers everywhere, and people may not feel like being precise when there is no danger of confusion. Magidin (talk) 15:16, 19 September 2012 (UTC)

Yes, that should do the trick. I appreciate the effort, and especially the detailed explanations you have given here. While we are at it... One last thing (promise). How about adding a sentence after "Left invertible morphisms are necessarily monic" (or rather after the equation) mentioning that the converse isn't necessarily true? If so, then the parenthetical (also called "one-to-one functions") in the lead could probably removed. [It's kind of my fault that it is there.] YohanN7 (talk) 21:37, 19 September 2012 (UTC)

Done on the invertibility section. Magidin (talk) 14:38, 20 September 2012 (UTC)

Ambiguous example

First of all, my most sincere thanks to Magidin and everyone who has helped write this article :)

As a student of undergraduate abstract algebra, reading this article took slightly more time than I expected. If someone more knowledgeable about algebra could insert a clarification of the notation used in the section named "Example", it would have helped me read the section the way it was intended to be.

1. A group has one operation only. In (Q, +), juxtaposition could ambiguosly refer to the only operation, addition, or, if either of the juxtaposed elements is an integer, multiplication by that integer. On Groupprops, juxtaposition would have referred to addition, but here it refers to multiplication by an integer. I got quite confused there for a moment.
2. The division and integrality test at the end of the section can be performed in the field of rational numbers, where division is defined, as the integrality of a rational number is independent of the containing algebraic entity.

In hope for a forgiveful example section for students previously exposed to the notation used in The Group Properties Wiki, SvartMan (talk) 17:43, 22 October 2015 (UTC)

I am not clear on what your question/objection is. On the example that you deleted calling it "incorrect", we are talking about additive groups, so the operation of the group is addition. It is standard, when you have an additive group, to use multiplicative notation by an integer for repeated addition (just like it is standard in a multiplicative group to use exponentiation by an integer to denote repeated multiplication). So, technically, in (Q,+), when you write nq with n in Z and q in Q, you are refering to the result of adding q to itself n times in Q (or the additive inverse of adding q to itself -n times if n is negative); however, this just happens to agree with the result of multiplication in the ring structure of Q because that is precisely how that structure is defined. So the potential ambiguity that you mention in 1 is actually immaterial, since regardless of how you interpret the notation you will get the exact same answer in the end. Note that the example explicitly states that Q and Z are both "groups under addition" (that is, the operation is addition, written as addition). One should not use juxtaposition to stand for the operation in this setting because that would be ambiguous: would "7(1/3)" refer to 7+(1/3), or would it refer to (1/3) operated with itself (added to itself) 7 times? Or to something else? And will $(1/3)^{2}$ mean (1/9), or mean (2/3)?

I don't know what you mean by "integrality test" or "integrality of a rational number is independent of the containing algebraic entity" (I just don't understand what that final sentence means at all). Since Q is uniquely divisible (for every x in Q and for every positive integer n there exists a unique y in Q such that ny =x), the notation (1/n)x to refer to this y is also pretty standard; again, it turns out that it does not matter whether you think of it as notation for "the unique solution to the equation ny=x", or if you think of it as a multiplication by a rational number: you get the same answer.

What the example is doing is showing that if G is divisible, and you have an group homomorphism from G to Q, then either everything is mapped to 0 or not everything is mapped to an integer. Because, if g gets mapped to the positive integer n, then divisibility of G guarantees that you can find an element y in G such that (n+1)y=x (if G is written additively; which makes sense to do here since we are dealing with abelian groups in any case), and then what does y get mapped to? It needs to mapped to some rational q with the property that (n+1)q=n. But this means that either n=q=0 (which contradicts our assumption that n is positive), or else that q is not an integer. There is no "integrality test", just the observation that if n is a positive integer, then n/(n+1) is not an integer. This is all happening inside of Q. Magidin (talk) 20:01, 22 October 2015 (UTC)

Thank you so much for this lucid explanation! I apologise for my completely unwarranted edit. — Preceding unsigned comment added by SvartMan (talk • contribs) 10:01, 11 November 2015 (UTC)

Properties

The 'Properties' section could do with an expansion, if anybody gets the time. Joel Brennan (talk) 20:05, 15 July 2019 (UTC)