# Functors and “null” objects and arrows

I have a two-part question concerning functors and the existence of "null" objects and arrows in categories in general. I understand that a functor can be regarded as a pair of functions: one mapping objects to objects, and the other mapping arrows to arrows. Unclear (to me) is whether a functor can map an object to a “null object” and whether a functor can map an arrow to a "null arrow.” In Sets this would mean mapping sets to the null set, and functions to the null function, which would not seem to pose any problem. However in my reading so far, I have not seen any reference to "null objects" or "null arrows" in categories in general, much less in the context of functors.

Just for example, consider two categories $\textbf{C}\$ and $\textbf{D}\$. Let both categories contain two objects (among others) called A and B. Now the two categories are identical except that $\textbf{D}\$ contains the product AxB and the requisite arrows, while $\textbf{C}\$ does not.

Can we define a functor from $\textbf{D}\$ to $\textbf{C}\$ by mapping each object to itself and each arrow to itself, except that we map AxB to a null object, and map each of the arrows into or out of AxB to a null arrow? Does this work?

This part doesn't really make sense:

Now the two categories are identical except that D contains the product AxB and the requisite arrows, while C does not.

You can't just remove some objects and arrows from a category, leaving everything else the same, and still assume you have a category. That is like removing some elements from a group; what's left probably isn't a group any more.

To address your question more directly, if a category has a "null object" and "null arrows", then you can use them. But they need to be actual objects and arrows, like in Set. There is no such thing as a default "null object" or "null arrows" for all categories. For more information in this direction, you might look up initial objects, terminal objects, and zero objects.

• Thanks for your comment concerning null objects and arrows. That helps. Let me clarify my question. I am assuming C is a category. Then I create the new object AxB via the definition of a product, and call the resulting category D. Then, both C and D are categories. (Is there then a functor from D to C?) – MPitts Oct 21 '17 at 23:48
• Maybe there's a construction I'm not seeing, but to me it's not as easy as saying "create the new object AxB via the definition of product." For example, in the category of fields, the product of two fields doesn't exist. How do you add it to get a bigger category? And even if somehow you do, I don't see the functor; which field would be the image of $\mathbb{F}_p \times \mathbb{F}_q$ for distinct primes $p$, $q$? – Ted Oct 22 '17 at 0:33
• I am not claiming that my example makes sense in every case. My question concerns those cases where is does make sense. – MPitts Oct 23 '17 at 21:34

I think there is a confusion in what is meant by a null object:

In Sets this would mean mapping sets to the null set, and functions to the null function, which would not seem to pose any problem.

The category of Sets does not have a null object. It has the empty set $\varnothing$ as initial element and the point $\{\varnothing\}$ as final element, isomorphic to $\{x\}$ for any set $x$. This means that for any set $A$, you can always define an empty application $f:\varnothing \rightarrow A$ and a constant application $g: A \rightarrow \{ \varnothing \}$.

A null object $0$ is an object that is both an initial and a final object. For any objects $A$ and $B$, a null arrow $0_{AB}$ may thus be defined by factorisation through the null object $0$: $$0_{BA}: A \rightarrow 0 \rightarrow B$$

When a category $\mathcal{C}$ has a null object, there is a null functor mapping every elements to the null object. As an example, you may also decompose $\mathcal{C} \setminus \{ 0 \}$ in connected components, and define a functor as being null outside of a given connected component.

Can we define a functor from D to C by mapping each object to itself and each arrow to itself, except that we map AxB to a null object, and map each of the arrows into or out of AxB to a null arrow? Does this work?

Not really, as $A$ and $B$ will have to be mapped to the null object as well.

Suppose you have a functor $F$ mapping the product $A \times B$ to the null object $F(0)$. Then for every object $X$ with an arrow $f_{AX}: X \rightarrow A$, you have the other arrow $0_{BX}: X \rightarrow B$. The universal property of the product implies that both arrows factorise through $A \times B$, and functoriality of $F$ thus implies that $F(f_{AX})$ is null. If you now take $X = A$ and $f_{AX} = 1_A$, your functor $F$ maps the identity of $A$ to the null arrow so that $F(A)$ is isomorphic to the null object $F(0)$.