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

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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.

  • $\begingroup$ Irreducible representations come to mind $\endgroup$
    – Vincent
    Commented Sep 2, 2022 at 20:51
  • $\begingroup$ More generally simple modules $\endgroup$
    – Vincent
    Commented Sep 2, 2022 at 20:52

1 Answer 1


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.

Finally, it's worth noting that the nLab uses a different definition of simple object phrased in terms of quotient objects, which should be equivalent to the above under some mild hypotheses but not always. I didn't want to use this one because it takes a little effort to define what quotient objects are at this level of generality, and (I did not realize this until writing this answer) it still doesn't apply to rings.

  • 1
    $\begingroup$ Also I've written these definitions in such a way that they apply to the zero object but they really shouldn't. I'm not sure what a clean way is to fix this. $\endgroup$ Commented Sep 2, 2022 at 21:53

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