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.
Best Answer
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.