How does a category have to be connected

Question

A multi question here: Could I have an isolated object in a category that is neither a source or target (identities excluded)? I have a suspicion that everything in a category has to be connected, and therefore that any two things can be connected in theory. Could I still make it isolated if I were to know I could make a morphism that goes to/from it (ie if there does exist a morphism do I need to put it in the category)?

This leads to a question that was answered by another user in comments, and they didn’t have time to check their claim so I figured I’d ask it here just incase: can any cancellable monoid be extended to a group, or more generally does cancellability in an algebra (here, just a collection with an operation) imply the existence of an inverse, even if we don’t include it in our set? (Eg used was naturals under addition.)

Answer 1

1. Yes, there exist categories with isolated objects, e.g. in the discrete category on any set of objects, which only has the identity morphisms, every object is isolated.
   For a 'real' example, take the full subcategory of prime fields ($\Bbb Z/p\Bbb Z$ and $\Bbb Q$) in the category of rings. There are no ring homomorphisms between distinct prime fields.
2. We can also consider not full subcategories, that is, we are allowed to collect only some of the morphisms (as long as all their compositions are collected too). In particular, nothing prevents us to be able to take the discrete subcategory of any given category, by deleting all nonidentity arrows..
3. The possibility to embed a cancellative monoid into a group is an interesting question on its own, and is not trivial. The answer is positive for commutative or finite monoids, but negative in general. Look up cancellative semigroups in Wikipedia.
4. However, it is always possible to adjoin a (formal) inverse to a single cancellable element in a ring (or associative vector algebra), this is called localization.