this definition would be helpful if we knew what an object was. But we don't, and the definition stands anyways.
Objects are axiomatic, like points in a topological space. Sometimes we can assign coordinates to the points in a topological space, but sometimes we can't. They are just points, and we leave it at that. In the same way, objects are just objects.
Even this analogy doesn't go far enough. The points of a topological space are at least sets, and topology is born of zf set theory. The objects in a category don't even have to be sets. They don't have to be anything at all. This is truly abstract.
A morphism is a labeled, directed connection from one object to another, like a directed arc in a digraph. But the digraph is not a perfect analogy. Morphisms can connect an object to itself, and many different morphisms can connect one object to another. Most digraphs don't allow loops and multiple edges. Also, morphisms are transitive. If we have A → B → C, then there is a specific morphism, determined by the two morphisms in question, that connects A to C. Digraphs don't usually require transitivity.
The word morphism is derived from the concept of a homomorphism, a group homomorphism or a ring homomorphism or a module homomorphism etc. Yet a morphism is much more general. It could be a continuous map from one topological space into another, or an arbitrary function between sets. In all these cases the composition of two morphisms gives another morphism. This is not a given; it's something we need to prove, as we did with continuous functions in Rn, in metric spaces, and in topological spaces.
If there is but one object, the category is called a monoid. There could be many morphisms from that object into itself, like the endomorphisms of a given ring.
Why is it called a monoid? Because the morphisms form a monoid under composition. Realize, however, that there is also a category of monoids, with many objects, one object for each monoid, and monoid homomorphisms become the connecting morphisms. I'll try to be clear - whether I mean a one object monoid or the category of monoids.
If there is at most one morphism between any two objects, the category is a pre ordering of objects.
A → B → C → D
The implied morphism A → C, combined with C → D, gives the same morphism as A → B composed with B → D.
When the objects are based on sets, such as groups rings etc, the composition of morphisms is uniquely determined by the underlying functions. If the resulting function satisfies the criteria of the category, we have our morphism. And morphism composition is always associative. If the functions of A → B → C → D are f g and h respectively, then for any x in A, the resulting function from A to D, no matter how you compose them, is, and has to be, h(g(f(x))).
If there are two identity morphisms f and g, compose them, and the result must equal f and g simultaneously. Hence the identity morphism on a given object is unique.
Every identity morphism is an equivalence. It acts as its own inverse.
Two objects that participate in an equivalence are equivalent.
In groups rings and fields, an equivalence is an isomorphism, and the objects on either side of the equivalence are called isomorphic. An equivalence betweeen sets is a bijection, and an equivalence between topological spaces is a homeomorphism.
Verify that equivalent objects are reflexive, symmetric, and transitive. Only the latter requires a little thought. If f takes A into B and g takes B into C, then consider the composition of f, g, g inverse, and f inverse. If h is f compose g, then our chain of morphisms implements h followed by h inverse, and that's supppose to be the identity morphism. Regroup, and g compose g inverse is the identity morphism. It can be combined with f, and the result is still f. That leaves f compose f inverse, which is, happily, the identity morphism. apply the same reasoning for h inverse followed by h, and A and C are equivalent.
As the name suggests, equivalent elements form equivalence classes. Sometimes the entire equivalence class collapses into one object. this is reflected in statements like, "There is one group of order p." Of course there are lots of groups of order p, but they are all isomorphic, and at the structural level, there is but one group of order p. All these groups are faithfully represented by Zp, the integers 0 through p-1 under addition mod p. This is also the canonical representative for the unique ring of order p, and the unique field of order p, in their respective categories.
If you accept the axiom of choice, or something like it, you can select a representative for each equivalence classe, and build a category on these representatives. If there is a morphism between two equivalence classes, prepend a morphism in one class, and apend a morphism in the other, to get a morphism between the two representatives. The morphism exists, and this is the one we will keep when we restrict our category to class representatives. These morphisms are part of the original category, so they follow all the rules, and the category of class representatives is valid.
We already gave an example of a subcategory, albeit abstract. Select a representative from each equivalence class and build a subcategory on these representatives alone.
We have gone beyond sets; even the objects themselves might not be sets, although they usually are.
The morphism is completely described by the function on the underlying set, hence the identity map on a set has to be the identity morphism on the corresponding object, and a bijection between sets is usually an equivalence, provided the function and its inverse follow the rules for a morphism in the category. Often the forward direction implies the reverse, but this is not a given; it has to be proved. If a group homomorphism is a bijection on sets it is an isomorphism, and the inverse map is another group isomorphism, giving an equivalence.
As mentioned earlier, the morphisms of a concrete category are automatically associative, because function composition is associative.
Describing a category from morphisms on up may be superior in some situations, but I find it rather confusing, so I'm not going to pursue it here. I just wanted you to know about it.