Is a full functor not necessarily surjective in terms of objects?

Question

Categories for the Working Mathematicians says

A functor $T: C\to B$ is full when to every pair $c, c'$ of objects of $C$ and to every arrow $g: Tc\to Tc' $ of $B$, there is an arrow $f: c\to c'$ of $C$ with $g = Tf$.

It seems to be defined only in terms of morphisms. Why not also in terms of objects?

Does it imply or not further require that for each object $b$ in $B$, there must be an object $c$ in $C$, so that $b=Tc$?

Thanks.

Answer 1

A full functor is only defined in terms of the local behaviour on hom sets, so no, a full functor need not be surjective on object. In fact, the inclusion functor of the empty category into any category is automatically (vacuously) full, but certainly almost never surjective on objects.

As for the reason why one should not demand surjectivity on objects, well, that would be a very strong property resulting in an altogether different notions. Functors which are surjective on objects, or more commonly essentially surjective on objects (meaning every object in the codomain is isomorphic to the image of an object in the domain) are certainly important. Such functors may or may not be full. So, the terminology allows us the flexibility we need to speak of, e.g., essentially surjective full functors. Incidentally, fully faithful essentially surjective functors are precisely the notion of equivalence in category theory: Two categories are equivalent if there exists an essentially surjective fully faithful functor between them.

Answer 2

Full functors do not need to be surjective on object. To understand why this makes sense, you can think about posets (aka categories with at most one arrow between any two objects).

In this context, what is a full functor between two posets? It turns out that it is the same as a monotonous function (I let you work this out, it is essentially unwrapping the definitions to see that they are the same). If you had required the functor to be surjective on object, you would have gotten the surjective continuous maps (which are interesting as well). This example already shows that it makes sense to consider full functors without requiring surjectivity on objects.

To go even further, you can notice that anyway surjectivity of objects is somewhat an awkward thing to ask. I don't know if you are familiar with the notion of equivalence of categories yet, but if you are, you can consider the following "classical" toy example :

• Take the category $\mathcal{C}$ to have just one object $c$, and the only morphism from $c$ to $c$ is the identity
• Take the category $\mathcal{D}$ to have two objects $d_1$ and $d_2$, with two morphisms $f : d_1 \to d_2$ and $g : d_2 \to d_1$. These compose to the identity in both ways Then the categories $\mathcal{C}$ and $\mathcal{D}$ are equivalent, even though you will never find a surjective on objects functor from $\mathcal{C}$ to $\mathcal{D}$. This is because essentially the objects $d_1$ and $d_2$ are isomorphic in the category $\mathcal{D}$, even though they are not equal.

What this (toy) example shows is that categories are not really interested in their objects, but more is isomorhism class of objects. This makes sense since we usually reason about things "up to isomorphisms". Thus being surjective on object is not really a relevant property usually (don't get me wrong, it is used some times, but is always clunky, and is often used because we lack the correct setting to express the real hypothesis we want).

If you are interested, you can find less toy-ish examples of categories that are equivalent but do not have a functor surjective on objects between them, to convince you that this is a real problem and not just an edge case. For instance the category of finite sets, a category whose objects are natural numbers (and you can work out the correct arrows)