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

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?

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.

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 $C$.

• So, "corresponds to objects" means that every object can be determined? This can also happen with other, more "complex" categories (like 2 and 3 for example). Does it mean that it is the "least complex" category that can determine each property (objects, arrows, composable arrows) – Mano Plizzi Jul 14 '16 at 22:04

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.

• So, what does it actually mean to "correspond to " as in the exercise? That there is the possibility to "reach" from 1 every object in C, from 2 every morphism etc? – Mano Plizzi Jul 14 '16 at 22:15