"prime" or "indecomposable" morphisms in small category Let me start by saying I have had to teach myself all the category theory I know, if I can even call it knowing category theory haha. I am now in a class using Emily Riehl's Category Theory in Context. I have control-F-ed looking for something like this but haven't found it, so I would welcome information and/or thoughts:
I'm wondering about the concept of "prime" or "indecomposable" morphisms (arrows) in my category. I'm motivated by dealing with k-graphs seen as categories, in which I get to talk about edges in this kind of "fully-broken-down" way. I am hoping there is some kind of similar concept of breaking down to the arrows in a small category that are most building-block-ish, as far away as I can get reasonably from being compositions of other nonidentity morphisms... I say reasonably bc I've just learned about products and so it seems that any morphism would be able to factor through a product and a projection back down, if my category happens to have products? I am very new to all of this. I have been told that the idea of "A morphism $h: X \rightarrow Y$ is indecomposable if $h = fg$ implies $g = 1_Y$ or $f = 1_X$" is not useful, I guess bc of the product thing I mentioned above. If anyone can shed some light on the subject I would be very appreciative!
 A: When you're dealing with category theory, you always want to phrase a property "nicely".
In particular, whenever a property involves equality, you want to make sure that you're only making use of the statement $f = g$ when $f$ and $g$ are morphisms with the same domain and codomain. It's almost always a bad idea to compare two objects in a category for equality, and it's almost always a bad idea to compare two arrows for equality unless you know they have a common domain and codomain.
In this case, your statement is "$h : X \to Y$ is indecomposable iff for all $W$ and for all $f : W \to Y$, $g : X \to W$ such that $h = f \circ g$, either ($W = X$ and $g = 1_X$) or ($W = Y$ and $f = 1_Y$)".
Notice that your definition requires comparing the object $W$ to the objects $X$ and $Y$ and asking whether there is an equality between these objects. This is a clear indication that this is the wrong thing to do.
Concretely, let $i : X \to X$ be an isomorphism, and let $j : X \to X$ be its inverse. Then $h = (h \circ j) \circ i$. Suppose $h$ is indecomposable. Then either $i = 1_X$ or $h \circ j = 1_Y$. Clearly, if $Y$ and $X$ are distinct then $h \circ j = 1_Y$ is impossible. So it must be the case that $i = 1_X$. So there can only be one isomorphism $X \to X$.
So we've artificially limited the number of permitted isomorphisms $X \to X$ in our definition. This seems rather silly.
There are some formal proofs which state that statements which are phrased "nicely" are invariant under isomorphism and equivalence. Stating this precisely takes some time, but the proof is pretty straightforward.
In this case, a better definition would be
$h: X \to Y$ is indecomposable if and only if for all $W$ and for all $f : W \to Y$, $g : X \to W$ such that $h = f \circ g$, either $f$ is an isomorphism or $g$ is an isomorphism.
