# The Universe of Discourse

Fri, 12 Nov 2021

The fact that singleton sets are terminal in the category of sets, and the empty set is initial, is completely elementary, so it's often passed over without discussion. But understanding it requires understanding the behavior of empty functions, and while there is nothing complex about that, novices often haven't thought it through, because empty functions are useless except for the important role they play in Set. So it's not unusual to see questions like this one:

I have trouble understanding the difference between initial and terminal objects in category theory. … Why there can be morphism from empty set to any other set? And why there is not morphism to empty set as well?

I'm happy with the following answer, which is of the “you already knew this, you only thought you didn't” type. It doesn't reveal any new information, it doesn't present any insights. All it does is connect together some things that the querent hasn't connected before.

This kind of connecting is an important part of pedagogy, one that Math Stack Exchange is uniquely well-suited to deal with. It is not well-handled by the textbook (which should not be spending time or space on such an elementary issue) or in lectures (likewise). In practice it's often handled by the TA (or the professor), during office hours, which isn't a good way to do it: the TA will get bored after the second time, and most students never show up to office hours anyway. It can be well-handled if the class has a recitation section where a subset of the students show up at a set time for a session with the TA, but upper-level classes like category theory don't usually have enough students to warrant this kind of organization. When I taught at math camp, we would recognize this kind of thing on the fly and convene a tiny recitation section just to deal with the one issue, but again, very few category theory classes take place at math camp.

Stack Exchange, on the other hand, is a great place to do this. There are no time or space limitations. One person can write up the answer, and then later querents can be redirected to the pre-written answer.

Your confusion seems to be not so much about initial and terminal objects, but about what those look like in the category of sets. Looking at the formal definition of “function” will help make clear some of the unusual cases such as functions with empty domains.

A function from !!A!! to !!B!! can be understood as a set of pairs $$\langle a,b\rangle$$ where !!a\in A!! and !!b\in B!!. And:

There must be exactly one pair !!\langle a,b\rangle!! for each element !!a!! of !!A!!.

Exactly one, no more and no less, or the set of pairs is not a function.

For example, the function that takes an integer !!n!! and yields its square !!n^2!! can be understood as the (infinite) set of ordered pairs:

$$\{ \ldots ,\langle -2, 4\rangle, \langle -1, 1\rangle, \langle 0, 0\rangle ,\langle 1, 1\rangle, \langle 2, 4\rangle\ldots\}$$

And for each integer !!n!! there is exactly one pair !!\langle n, n^2\rangle!!. Some numbers can be missing on the right side (for example, there is no pair !!\langle n, 3\rangle!!) and some numbers can be repeated on the right (for example the function contains both !!\langle -2, 4\rangle!! and !!\langle 2, 4\rangle!!) but on the left each number appears exactly once.

Now suppose !!A!! is some set !!\{a_1, a_2, \ldots\}!! and !!B!! is a set with only one element !!\{b\}!!. What does a function from !!A!! to !!B!! look like? There is only one possible function: it must be: $$\{ \langle a_1, b\rangle, \langle a_2, b\rangle, \ldots\}.$$ There is no choice about the left-side elements of the pairs, because there must be exactly one pair for each element of !!A!!. There is also no choice about the right-side element of each pair. !!B!! has only one element, !!b!!, so the right-side element of each pair must be !!b!!.

So, if !!B!! is a one-element set, there is exactly one function from !!A!! to !!B!!. This is the definition of “terminal”, and one-element sets are terminal.

Now what if it's !!A!! that has only one element? We have !!A=\{a\}!! and !!B=\{b_1, b_2, \ldots\}!!. How many functions are there now? Only one?

One function is $$\{\langle a, b_1\rangle\}$$ another is $$\{\langle a, b_2\rangle\}$$ and another is $$\{\langle a, b_3\rangle\}$$ and so on. Each function is a set of pairs where the left-side elements come from !!A!!, and each element of !!A!! is in exactly one pair. !!A!! has only one element, so there can only be one pair in each function. Still, the functions are all different.

You said:

I would find it more intuitive if one-element set would be initial object too.

But for a one-element set !!A!! to be initial, there must be exactly one function !!A\to B!! for each !!B!!. And we see above that usually there are many functions !!A\to B!!.

Now we do functions on the empty set. Suppose !!A!! is !!\{a_1, a_2, \ldots\}!! and !!B!! is empty. What does a function from !!A\to B!! look like? It must be a set of pairs, it must have exactly one pair for each element of !!a!!, and the right-side of each pair must be an element of !!B!!. But !!B!! has no elements, so this is impossible: $$\{\langle a_1, ?\rangle, \langle a_2, ?\rangle, \ldots\}.$$

There is nothing to put on the right side of the pairs. So there are no functions !!A\to\varnothing!!. (There is one exception to this claim, which we will see in a minute.)

What if !!A!! is empty and !!B!! is not, say !!\{b_1, b_2, \ldots\}!!? A function !!A\to B!! is a set of pairs that has exactly one pair for each element of !!A!!. But !!A!! has no elements. No problem, the function has no pairs! $$\{\}$$

A function is a set of pairs, and the set can be empty. This is called the “empty function”. When !!A!! is the empty set, there is exactly one function from !!A\to B!!, the empty function, no matter what !!B!! is. This is the definition of “initial”, so the empty set is initial.

Does the empty set have an identity morphism? It does; the empty function !!\{ \}!! is its identity morphism. This is the one exception to the claim that there are no functions from !!A\to\varnothing!!: if !!A!! is also empty, the empty function is such a function, the only one.

The issue for topological spaces is exactly the same:

• When !!B!! has only one element, there is exactly one continuous map !!A\to B!! for every !!A!!.
• When !!A!! is empty, there is exactly one continuous map !!A\to B!! for every !!B!!: the empty function is the homeomorphism.
• When !!A!! has only one element, there are usually many continuous maps !!A\to B!!, one different one for each element of !!B!!.

There are categories in which the initial and terminal objects are the same:

• In the category of groups, the trivial group (with one element) is both initial and terminal.

• A less important but simpler example is Set*, the category of pointed sets, whose objects are nonempty sets in which one element has been painted red. The morphisms of Set* are ordinary functions that map the red element in the domain to the red element of the codomain.

I hope this was some help.

[ Thanks to Rupert Swarbrick for pointing out that I wrote “homeomorphism” instead of “continuous map” ]