# Universal property of the disjoint union

Where could we start if we were look­ing for a nice “easy” uni­ver­sal prop­erty de­scribing the union of sets?

The first thing to no­tice is that uni­ver­sal prop­er­ties only iden­tify ob­jects up to iso­mor­phism, but there’s a sense in which the union is not well-defined up to iso­mor­phism: it is pos­si­ble to find sets $$A$$ and $$B$$, and sets $$A'$$ and $$B'$$, with $$A$$ iso­mor­phic to $$A'$$ and $$B$$ iso­mor­phic to $$B'$$, but where $$A \cup B$$ is not iso­mor­phic to $$A' \cup B'$$. noteIn this case, when we’re talk­ing about sets, an iso­mor­phism is just a bi­jec­tion.

Con­sider $$A = \{ 1 \}$$ and $$B = \{ 1 \}$$. Then the union $$A \cup B$$ is equal to $$\{1\}$$.

On the other hand, con­sider $$X = \{1\}$$ and $$Y = \{2\}$$. Then the union $$X \cup Y = \{1,2\}$$.

So by re­plac­ing $$A$$ with the iso­mor­phic $$X$$, and $$B$$ with the iso­mor­phic $$Y$$, we have ob­tained $$\{1,2\}$$ which is not iso­mor­phic to $$\{1\} = A \cup B$$. <div><div>

So it’s not ob­vi­ous whether the union could even in prin­ci­ple be defined by a uni­ver­sal prop­erty.noteAc­tu­ally it is pos­si­ble.

But we can make our job eas­ier by tak­ing the next best thing: the dis­joint union $$A \sqcup B$$, which is well-defined up to iso­mor­phism in the above sense: the defi­ni­tion of $$A \sqcup B$$ is con­structed so that even if $$A$$ and $$B$$ over­lap, the in­ter­sec­tion still doesn’t get treated any differ­ently.

Be­fore read­ing this, you should make sure you grasp the dis­joint union of sets well enough to know the differ­ence be­tween the two sets $$\{ 2, 3, 5 \} \cup \{ 2, 6, 7 \}$$ and $$\{ 2, 3, 5 \} \sqcup \{ 2, 6, 7 \}$$.

Given sets $$A$$ and $$B$$, we define their dis­joint union to be $$A \sqcup B = A' \cup B'$$, where $$A' = \{ (a, 1) : a \in A \}$$ and $$B' = \{ (b, 2) : b \in B \}$$. That is, “tag each el­e­ment of $$A$$ with the la­bel $$1$$, and tag each el­e­ment of $$B$$ with the la­bel $$2$$; then take the union of the tagged sets”.

The im­por­tant fact about the dis­joint union here is that if $$A \cong A'$$ and $$B \cong B'$$, then $$A \sqcup B \cong A' \sqcup B'$$. This is the fact that makes the dis­joint union a fairly ac­cessible idea to bring un­der the um­brella of cat­e­gory the­ory, and it also means we are jus­tified in us­ing a con­ven­tion that sim­plifies a lot of the no­ta­tion: we will as­sume from now on that $$A$$ and $$B$$ are dis­joint. (Since we only care about them up to iso­mor­phism, this is fine to do: we can re­place $$A$$ and $$B$$ with some dis­joint pair of sets of the same size if nec­es­sary, to en­sure that they are dis­joint.)

# Univer­sal prop­erty of dis­joint union

How can we look at the dis­joint union solely in terms of the maps to and from it? First of all, how can we look at the dis­joint union at all in terms of those maps?

There are cer­tainly two maps $$A \to A \sqcup B$$ and $$B \to A \sqcup B$$: namely, the in­clu­sions $$i_A : a \mapsto a$$ and $$i_B : b \mapsto b$$. But that’s not enough to pin down the dis­joint union pre­cisely, be­cause those maps ex­ist not just to $$A \sqcup B$$ but also to any su­per­set of $$A \sqcup B$$. noteIf this is not ob­vi­ous, stop and think for a cou­ple of min­utes about the case $$A = \{ 1 \}$$, $$B = \{ 2 \}$$, $$A \sqcup B = \{1,2\}$$, and the su­per­set $$\{1,2,3\}$$ of $$A \sqcup B$$. So we need to find some way to limit our­selves to $$A \sqcup B$$.

The key limit­ing fact about the dis­joint union is that any map from the dis­joint union can be ex­pressed in terms of two other maps (one from $$A$$ and one from $$B$$), and vice versa. The fol­low­ing dis­cus­sion shows us how.

• If we know $$f: A \sqcup B \to X$$, then we know the re­stric­tion maps $$f \big|_A : A \to X$$ noteThis is defined by $$f \big|_A (a) = f(a)$$; that is, it is the func­tion we ob­tain by “re­strict­ing” $$f$$ so that its do­main is $$A$$ rather than the larger $$A \sqcup B$$. and $$f \big|_B : B \to X$$.

• If we know two maps $$\alpha: A \to X$$ and $$\beta: B \to X$$, then we can uniquely con­struct a map $$f: A \sqcup B \to X$$ which is ”$$\alpha$$ and $$\beta$$ glued to­gether”. For­mally, the map is defined as $$f(x) = \alpha(x)$$ if $$x \in A$$, and $$f(x) = \beta(x)$$ if $$x \in B$$.

The rea­son this pins down the dis­joint union ex­actly (up to iso­mor­phism) is be­cause the dis­joint union is the only set which leaves us with no choice about how we con­struct $$f$$ from $$\alpha$$ and $$\beta$$:

• We rule out strict sub­sets of $$A \sqcup B$$ be­cause we use the fact that ev­ery el­e­ment of $$A$$ is in $$A \sqcup B$$ and ev­ery el­e­ment of $$B$$ is in $$A \sqcup B$$, to en­sure that the re­stric­tion maps make sense.

• We rule out strict su­per­sets of $$A \sqcup B$$ be­cause we use the fact that any el­e­ment in $$A \sqcup B$$ is ei­ther in $$A$$ or it is in $$B$$ noteIf we re­placed $$A \sqcup B$$ by any strict su­per­set, this stops be­ing true: by pass­ing to a su­per­set, we in­tro­duce an el­e­ment which is nei­ther in $$A$$ nor in $$B$$., to en­sure that $$f$$ is defined ev­ery­where on its do­main. More­over, we use the fact that ev­ery el­e­ment of $$A \sqcup B$$ is in ex­actly one of $$A$$ or $$B$$, be­cause if $$x$$ is in both $$A$$ and $$B$$, then we can’t gen­er­ally de­cide whether we should define $$f(x)$$ by $$\alpha(x)$$ or $$\beta(x)$$.

## The property

One who is well-versed in cat­e­gory the­ory and uni­ver­sal prop­er­ties would be able to take the above dis­cus­sion and con­dense it into the fol­low­ing state­ment, which is the uni­ver­sal prop­erty of the dis­joint union:

Given sets $$A$$ and $$B$$, we define the dis­joint union to be the fol­low­ing col­lec­tion of three ob­jects:

$$\text{A set labelled }\ A \sqcup B \\ i_A : A \to A \sqcup B \\ i_B : B \to A \sqcup B$$
with the re­quire­ment that for ev­ery set $$X$$ and ev­ery pair of maps $$f_A: A \to X$$ and $$f_B: B \to X$$, there is a unique map $$f: A \sqcup B \to X$$ such that $$f \circ i_A = f_A$$ and $$f \circ i_B = f_B$$.

Imgur was down when I made this pic­ture, so I’m host­ing it in­stead.

## Aside: re­la­tion to the product

Re­call the uni­ver­sal prop­erty of the product:

Given ob­jects $$A$$ and $$B$$, we define the product to be the fol­low­ing col­lec­tion of three ob­jects, if it ex­ists:

$$A \times B \\ \pi_A: A \times B \to A \\ \pi_B : A \times B \to B$$
with the re­quire­ment that for ev­ery ob­ject $$X$$ and ev­ery pair of maps $$f_A: X \to A, f_B: X \to B$$, there is a unique map $$f: X \to A \times B$$ such that $$\pi_A \circ f = f_A$$ and $$\pi_B \circ f = f_B$$.

No­tice that this is just the same as the uni­ver­sal prop­erty of the dis­joint union, but we have re­versed the do­main and codomain of ev­ery func­tion, and we have cor­re­spond­ingly re­versed any func­tion com­po­si­tions. This shows us that the product and the dis­joint union are, in some sense, “two sides of the same coin”. If you re­mem­ber the uni­ver­sal prop­erty of the empty set, we saw that the empty set’s uni­ver­sal prop­erty also had a “flip side”, and this flipped prop­erty char­ac­ter­ises the one-el­e­ment sets. So in the same sense, the empty set and the one-point sets are “two sides of the same coin”. This is an in­stance of the con­cept of du­al­ity, and it turns up all over the place.

# Examples

## {1} ⊔ {2}

Let’s think about the very first ex­am­ple that came from the top of the page: the dis­joint union of $$A=\{1\}$$ and $$B=\{2\}$$. (This is, of course, $$\{1,2\}$$, but we’ll try and tie the uni­ver­sal prop­erty point of view in with the el­e­ments point of view.)

The defi­ni­tion be­comes:

The dis­joint union of $$\{1\}$$ and $$\{2\}$$ is the fol­low­ing col­lec­tion of three ob­jects:

$$\text{A set labelled } \{1\} \sqcup \{2\} \\ i_A : \{1\} \to \{1\} \sqcup \{2\} \\ i_B : \{2\} \to \{1\} \sqcup \{2\}$$
with the re­quire­ment that for ev­ery set $$X$$ and ev­ery pair of maps $$f_A: \{1\} \to X$$ and $$f_B: \{2\} \to X$$, there is a unique map $$f: A \sqcup B \to X$$ such that $$f \circ i_A = f_A$$ and $$f \circ i_B = f_B$$.

This is still long and com­pli­cated, but re­mem­ber from when we dis­cussed the product that a map $$f_A: \{1\} \to X$$ is pre­cisely pick­ing out a sin­gle el­e­ment of $$X$$: namely $$f_A(1)$$. To ev­ery el­e­ment of $$X$$, there is ex­actly one such map; and for ev­ery map $$\{1\} \to X$$, there is ex­actly one el­e­ment of $$X$$ it picks out. (Of course, the same rea­son­ing goes through with $$B$$ as well: $$f_B$$ is just pick­ing out an el­e­ment of $$X$$, too.)

Also, $$i_A$$ and $$i_B$$ are just pick­ing out sin­gle el­e­ments of $$A \sqcup B$$, though we are cur­rently pre­tend­ing that we don’t know what $$A \sqcup B$$ is yet.

So we can rewrite this talk of maps in terms of el­e­ments, to make it eas­ier for us to un­der­stand this defi­ni­tion in the spe­cific case that $$A$$ and $$B$$ have just one el­e­ment each:

The dis­joint union of $$\{1\}$$ and $$\{2\}$$ is the fol­low­ing col­lec­tion of three ob­jects:

$$\text{A set labelled }\ \{1\} \sqcup \{2\} \\ \text{An element } i_A \text{ of } \{1\} \sqcup \{2\} \\ \text{An element }i_B \text{ of } \{1\} \sqcup \{2\}$$
with the re­quire­ment that for ev­ery set $$X$$ and ev­ery pair of el­e­ments $$x \in X$$ and $$y \in X$$, there is a unique map $$f: A \sqcup B \to X$$ such that $$f(i_A) = x$$ and $$f(i_B) = y$$.

From this, we can see that $$i_A$$ and $$i_B$$ can’t be the same, be­cause $$f$$ acts differ­ently on them (if we take $$X = \{x,y\}$$ where $$x \not = y$$, for in­stance). But also if there were any $$z$$ in $$\{1\} \sqcup \{2\}$$ other than $$i_A$$ and $$i_B$$, then for any at­tempt at our unique $$f: A \sqcup B \to X$$, we would be able to make a differ­ent $$f$$ by just chang­ing where $$z$$ is sent to.

So $$A \sqcup B$$ is pre­cisely the set $$\{i_A, i_B\}$$; and this has de­ter­mined it up to iso­mor­phism as “the set with two el­e­ments”.

### Aside: cardinality

No­tice that the dis­joint union of “the set with one el­e­ment” and “the set with one el­e­ment” yields “the set with two el­e­ments”. It’s ac­tu­ally true for gen­eral finite sets that the dis­joint union of “the set with $$m$$ el­e­ments” with “the set with $$n$$ el­e­ments” yields “the set with $$m+n$$ el­e­ments”; a few min­utes’ thought should be enough to con­vince you of this, but it is an ex­cel­lent ex­er­cise for you to do this from the cat­e­gory-the­o­retic uni­ver­sal prop­er­ties view­point as well as from the (much eas­ier) el­e­ments view­point. We will touch on this con­nec­tion be­tween “dis­joint union” and “ad­di­tion” a bit more later.

## The empty set

What is the dis­joint union of the empty set with it­self? (Of course, it’s just the empty set, but let’s pre­tend we don’t know that.) The ad­van­tage of this ex­am­ple is that we already know a lot about the uni­ver­sal prop­erty of the empty set, so it’s a good test­ing ground to see if we can do this with­out think­ing about el­e­ments at all.

Re­call that the empty set is defined as fol­lows:

The empty set $$\emptyset$$ is the unique set such that for ev­ery set $$X$$, there is pre­cisely one func­tion $$\emptyset \to X$$.

The dis­joint union of $$\emptyset$$ with it­self would be defined as fol­lows (where I’ve stuck to us­ing $$A$$ and $$B$$ in cer­tain places, be­cause oth­er­wise the whole thing just fills up with the $$\emptyset$$ sym­bol):

The fol­low­ing col­lec­tion of three ob­jects:

$$\text{A set labelled }\ A \sqcup B \\ i_A : \emptyset \to A \sqcup B \\ i_B : \emptyset \to A \sqcup B$$
with the re­quire­ment that for ev­ery set $$X$$ and ev­ery pair of maps $$f_A: \emptyset \to X$$ and $$f_B: \emptyset \to X$$, there is a unique map $$f: A \sqcup B \to X$$ such that $$f \circ i_A = f_A$$ and $$f \circ i_B = f_B$$.

Now, since we know that for ev­ery set $$X$$ there is a unique map $$!_X: \emptyset \to X$$ noteIn gen­eral, the ex­cla­ma­tion mark $$!$$ is used for a uniquely-defined map., we can re­place our talk of $$f_A$$ and $$f_B$$ by just the same $$!_X$$:

The fol­low­ing col­lec­tion of three ob­jects:

$$\text{A set labelled }\ A \sqcup B \\ i_A : \emptyset \to A \sqcup B \\ i_B : \emptyset \to A \sqcup B$$
with the re­quire­ment that for ev­ery set $$X$$, there is a unique map $$f: A \sqcup B \to X$$ such that $$f \circ i_A = (!_{X})$$ and $$f \circ i_B = (!_{X})$$.

More­over, we know that $$i_A$$ and $$i_B$$ are also uniquely defined, be­cause there is only one map $$\emptyset \to A \sqcup B$$ (no mat­ter what $$A \sqcup B$$ might end up be­ing); so we can re­move them from the defi­ni­tion be­cause they’re forced to ex­ist any­way. They’re both just $$!_{A \sqcup B}$$.

The set

$$A \sqcup B$$
with the re­quire­ment that for ev­ery set $$X$$, there is a unique map $$f: A \sqcup B \to X$$ such that $$f \circ (!_{A \sqcup B}) = (!_X)$$.

A di­a­gram is in or­der.

The uni­ver­sal prop­erty is pre­cisely say­ing that for ev­ery set $$X$$, there is a unique map $$f: A \sqcup B \to X$$ such that $$f \circ (!_{A \sqcup B})$$ is the map $$!_X : \emptyset \to X$$.

But since $$!_X$$ is the only map from $$\emptyset \to X$$, we have $$f \circ (!_{A \sqcup B}) = (!_X)$$ what­ever $$f$$ is. In­deed, $$f \circ (!_{A \sqcup B})$$ is a map from $$\emptyset \to X$$, and $$!_X$$ is the only such map. So we can drop that con­di­tion from the re­quire­ments, be­cause it au­to­mat­i­cally holds!

There­fore the dis­joint union is:

The set

$$A \sqcup B$$
with the re­quire­ment that for ev­ery set $$X$$, there is a unique map $$f: A \sqcup B \to X$$.

Do you recog­nise that defi­ni­tion? It’s just the uni­ver­sal prop­erty of the empty set!

So $$A \sqcup B$$ is the empty set in this ex­am­ple.

# More gen­eral concept

We saw ear­lier that the dis­joint union is char­ac­ter­ised by the “re­verse” (or “dual”) of the uni­ver­sal prop­erty of the product. In a more gen­eral cat­e­gory, we say that some­thing which satis­fies this prop­erty is a co­product. notePrepend­ing “co” to some­thing means “re­verse the maps”.

In gen­eral, the co­product of ob­jects $$A$$ and $$B$$ is writ­ten not as $$A \sqcup B$$ (which is spe­cific to the cat­e­gory of sets), but as $$A + B$$. Among other things, the co­product unifies the ideas of ad­di­tion of in­te­gers, least up­per bounds in posets, dis­joint union of sets, least com­mon mul­ti­ple of nat­u­rals, di­rect sum of vec­tor spaces, and free product of groups; all of these are co­prod­ucts in their re­spec­tive cat­e­gories.

Parents:

• Disjoint union of sets

One of the most ba­sic ways we have of join­ing two sets to­gether.

• Set

An un­ordered col­lec­tion of dis­tinct ob­jects.

• Universal property

A uni­ver­sal prop­erty is a way of defin­ing an ob­ject based purely on how it in­ter­acts with other ob­jects, rather than by any in­ter­nal prop­erty of the ob­ject it­self.