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Googling for "atomic morphism" gives me only 70 results. Is this concept so fruitless or does it have another standard name? What I mean is a morphism $f: A \rightarrow B$ such that

$$(\forall g,h)\ f = g \circ h \rightarrow (g = f\ \wedge\ h = id_A)\ \vee\ (g = id_B\ \wedge\ h = f).$$

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  • $\begingroup$ When you say f = gh, do you mean f : A \to B, h : A \to C, g : C \to B for an arbitrary choice of C? $\endgroup$
    – Qiaochu Yuan
    Commented Jan 4, 2010 at 9:44
  • $\begingroup$ Because that doesn't make sense with what you've written, and neither does C = A or C = B. So you must mean A = B = C, and you should say that. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 4, 2010 at 9:45
  • $\begingroup$ I want to say that f cannot be non-trivially decomposed. What is wrong with my formulation? $\endgroup$
    – Hans-Peter Stricker
    Commented Jan 4, 2010 at 10:04
  • $\begingroup$ If you don't specify that f is an endomorphism, none of the statements g = f, h = id, g = id, h = f make sense. $\endgroup$
    – Qiaochu Yuan
    Commented Jan 4, 2010 at 10:21
  • $\begingroup$ @Qiaochu, why don't those statements make sense? With your names for sources and targets, clearly g=f additionally means that A=C, etc. $\endgroup$
    – Kim Morrison
    Commented Jan 4, 2010 at 10:28

3 Answers 3

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As mentioned in the comments, I would probably call such a morphism "irreducible" or "prime." A "less evil," and perhaps more useful, version would be to ask that if $f = g \circ h$, then either $g$ or $h$ is an isomorphism. In this form, if you regard the multiplicative monoid of a ring as a category with one object, the (noninvertible) irreducible morphisms are precisely the irreducible elements of the ring.

I agree that in "concrete categories" such morphisms are unlikely to be very common or useful, but one other situation in which they arise is free categories on directed graphs. In such a category, the nonidentity irreducible morphisms are precisely the generators (the images of the edges of the directed graph you started from).

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    $\begingroup$ I agree that Mike's definition is more meaningful, but I would also consider an even stricter version. The problem is that even this definition is pretty much vacuous if the category has products or coproducts... (e.g., if there are products, any $f$ factors through $C=B\times A$, or in fact $C=B\times X$ for any $X$). So how about restricting the notion of atomic morphisms to monomorphisms (or dually, epimorphisms)? E.g.: A monomorphism $f$ is atomic if for any decomposition $f=g\circ h$ into a product of monomorphisms, either $g$ or $h$ is an isomorphism. $\endgroup$
    – t3suji
    Commented Jan 4, 2010 at 16:06
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Like Qiaochu said, the notion you have defined is rarely useful: for instance it is not invariant under replacing a category with an equivalent one (e.g., the unique morphism in the terminal category • is "atomic" but there are no atomic morphisms in the category with two objects related by a unique isomorphism). Even when your category is rigid (has no isomorphisms at all besides identites) there may be fewer "atomic" morphisms than you expect: for instance the simplex category has none (we can factor the identity morphism of [n] nontrivially through [m] for m > n).

However, I noticed in your other question you use the term in the case of categories which are posets. In that special case, the notion is a good one and has a name: if there is an "atomic" morphism x → y, we say that "y covers x". I think the morphism itself is also called a cover, but I'm not sure about the exact terminology in that case.

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  • $\begingroup$ Right, and an atom in a Boolean algebra is an element which covers 0. Is that the origin of the terminology? $\endgroup$
    – Pete L. Clark
    Commented Jan 4, 2010 at 15:44
  • $\begingroup$ Not consciously. $\endgroup$
    – Hans-Peter Stricker
    Commented Jan 4, 2010 at 15:50
  • $\begingroup$ I have also seen the term "immediate successor" (but I'm not a poset person). $\endgroup$
    – S. Carnahan ♦
    Commented Jan 4, 2010 at 16:49
  • $\begingroup$ Most poset people don't use category theoretic language. However, they certainly say that the pair (x,y) is a cover, so I think it is fair game to say that x -> y is a cover. $\endgroup$
    – David E Speyer
    Commented Jan 4, 2010 at 22:31
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A very similar notion is that of an irreducible morphism.

In an abelian category, a morphism is called a split monomorphism if it is an inclusion of an direct summand, and a split epimorphism if it is the projection onto a direct summand. A morphism is called split if it has either of the above properties.

A morphism $f$ is called irreducible if it is not split but, whenever $f=st$, either $s$ is a split monomorphism or $t$ is a split epimorphism.

I have never seen this definition used outside abelian categories, but I think it makes sense in any category: We can define split monomorphisms as the maps which have right inverses and split epimorphisms as the maps which have left inverses.

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