Functors $1 \to C, 2 \to C, 3 \to C$ (McLane exercise 1.3.2)

Question

This is exercise 2 in Mclane's Categories For the Working Mathematician, chapter 1.3.

Show that functors $1 \to C, 2 \to C, 3 \to C$ correspond respectively to objects, arrows and composable pairs of arrows in C.

(1 is the category of one object with just the identity morphism, 2 the category of two objects a,b with a morphism $a \to b$ and 3 the category with three objects $a$, $b$, $c$ with morphisms $a \to b,b \to c, a \to c$)

My question is perhaps very simple: When defining a functor T between two categories $C$, $D$ you define an object function from objects to objects and a morphism function. The object funtion is said to assign to each object $c$ an object $Tc$. So if $C$ has less objects than $D$ (as, for example happens in the category $1$), how does the functor work? In other words, the object funtion in the exercise from $1$ to $C$ where $C$ has more than one object, is not surjective, so there are objects in $C$ that are not represented by the functor.
I know there must be a mistake in my syllogism but can you clarify it a little, maybe with an example?

Answer 1

The 'corresponds' just means that there is a correspondence between e.g. the class $\text{ Ob}(C)$ of objects of $C$ and the class $[1,C]$ of all functors from $1$ to $C$.

Likewise for the other cases.

In fact, you probably have already seen a similar statement in the special case where the category $C$ has only identity morphisms: elements of a class correspond to maps from a singleton to that class.

Answer 2

Well, yes, one functor $1\to C$ determines exactly one object in $C$, but this can be any object.

So, the functors $1\to C$ correspond to objects of $C$.

Answer 3

Hint: functions don't need to be surjective. A functor $C \to D$ does not have to "represent" every object in $D$: there is a functor that maps a field $K$ to the abelian group $K^*$ of invertible elements of $K$, but not every abelian group is the group of invertible elements of a field.