# What are some interesting cases where the two obvious definitions of “discrete object” diverge?

The nLab page defines "discrete object" as follows:

Definition. [nLab] Let $\mathbf{C}$ denote a concrete category whose forgetful functor $U$ has a left adjoint $F$. Call the counit of this adjunction $\varepsilon$. Then $X \in \mathbf{C}$ is discrete iff the morphism $\varepsilon_X : X \leftarrow FUX$ is an isomorphism.

Another possible definition, that seems to give the correct answer in most cases, makes sense for objects of any category with a terminal object:

Definition. [Alternative] Let $\mathbf{C}$ denote a category with a terminal object. Then $X \in \mathbf{C}$ is discrete iff it is a small coproduct of terminal objects.

If $\mathbf{C}$ furthermore has all small coproducts, this can be rephrased as follows: $X \in \mathbf{C}$ is discrete iff it lives in the image of the unique functor $\mathbf{C} \leftarrow \mathbf{Set}$ that preserves $1$ and all small coproducts.

What I'd like to know is:

Question. What are some specific and interesting examples of concrete categories $\mathbf{C}$ with a terminal object $1$ such that the above two definitions don't actually agree?

I'm not sure why you say the definitions seem to agree "in most cases"; exceptions seem far more common than examples to me. For a simple counterexample, in the category of groups, no object is discrete in the first sense, and the trivial group is discrete in the second sense. More generally, any category of algebraic structures which include distinguished elements (such as the identity element of a group) will have no discrete objects by the first definition, since $FUX$ will always have a distinguished element which is different from the one $X$ already had. For an algebraic example where both kinds of discrete objects exist but are different, consider the category of sets with equipped with a unary operation. The only discrete object by the first definition is the empty set, while the discrete objects by the second definition are sets whose unary operation is the identity map.
• I don't think your statement about $\mathbb{N}$ is quite correct: under the second definition, the discrete objects in the category of monounary algebras ought to be precisely those pairs $(X,f)$ satisfying $f=\mathrm{id}_X$. (I haven't proved this, so I could be wrong.) – goblin Jan 26 '16 at 10:05