# Simple group-like objects in other categories

I was thinking about simple groups (as one does) and noticed that the first isomorphism theorem implies that every group homomorphism $$\varphi: G \to H$$, where $$G$$ is simple, must either be injective or degenerate; that is $$\varphi(g) = e_H$$ for all $$g \in G$$. This is seen easily since $$\ker \varphi$$ must be normal in $$G$$, which leaves the options of $$\{e\}$$, making $$\varphi$$ injective, or $$G$$, making $$\varphi$$ degenerate. After some more thinking, the same is true of fields, for the same reason; if $$\psi: F \to R$$ is a ring homomorphism and $$F$$ is a field, then $$\psi$$ is again either injective or degenerate, since the kernel must be an ideal of $$F$$, of which there are none both nontrivial and proper.

Now, I don't know much category theory, but in that language I think we can say that simple groups in Grp and fields in Ring have the following property:

If $$B$$ is such an object, then all morphisms with codomain $$B$$ are either monomorphisms or "degenerate" morphisms, where a degenerate morphism is defined as follows: $$\varphi$$ is denegerate if, in the following diagram

There exists a unique $$d: A \to C$$ such that for all $$f: A \to B$$, $$f \circ \varphi = d$$. In other words, $$\varphi$$ collapses all composition chains that contain it into a single morphism determined by the final domain and codomain.

My question to you is: does this type of object, one where all morphisms with it as a codomain are monomorphisms or degenerate, exist in other categories? In Set it's just the one-element sets, and I doubt it exists (nontrivially; again one-element spaces will suffice) in Top though I couldn't prove that one way or the other. I'm not sure about more exotic categories. It's reminiscent of a coproduct, but I think a bit stronger since it must work for all pairs of objects it connects to rather than just two.

• Irreducible representations come to mind Commented Sep 2, 2022 at 20:51
• More generally simple modules Commented Sep 2, 2022 at 20:52

There are several different possible definitions of simple objects depending on what you're trying to do. Recall that a category has zero morphisms if for every pair of objects $$a, b$$ there exist morphisms $$0_{a, b} : a \to b$$ which are absorbing with respect to composition in the sense that every composition containing a zero morphism is another zero morphism (formally, if $$g : b \to c$$ is a morphism then $$g \circ 0_{a, b} = 0_{a, c}$$, and if $$f : c \to a$$ is a morphism then $$0_{a, b} \circ f = 0_{c, b}$$). The simplest way for a category to have zero morphisms is to have a zero object, namely an object $$0$$ which is both initial and terminal; then the zero morphism is the unique morphism factoring through $$0$$.
In a category with zero morphisms, we can define an object $$s$$ to be simple if every morphism $$f : s \to t$$ out of $$s$$ is either a monomorphism or zero; this is slightly stronger than the condition you ask for. In $$\text{Grp}$$ this recovers the simple groups, and more generally in the category of $$R$$-modules for $$R$$ a ring this recovers the simple modules. Even more generally, in an abelian category this condition is equivalent to the more typical condition that a simple object have no subobjects other than itself and zero (edit: except when applied to the zero object itself), but note that this isn't equivalent to simplicity in $$\text{Grp}$$. This condition also recovers, for example, the simple Lie algebras (together with the $$1$$-dimensional abelian Lie algebra).
The category of rings doesn't have a zero object or zero morphisms so this definition doesn't apply to it. However, fields do have the closely related property that every morphism $$f : F \to R$$ out of a field is either a monomorphism or constant, where a constant morphism is a morphism that factors through the terminal object (which for rings is the zero ring), and this property characterizes simple rings among rings.