1
$\begingroup$

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?

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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 $C$.

$\endgroup$
  • $\begingroup$ 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) $\endgroup$ – Mano Plizzi Jul 14 '16 at 22:04
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Mano Plizzi Jul 14 '16 at 22:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.