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text: '[summary: The [-5z9] may be described by a [-600] which is very similar to that of the [5zv universal property of the product]. Specifically, the disjoint union of sets $A$ and $B$ is a set denoted $A \\sqcup B$ together with maps $i_A : A \\to A \\sqcup B$ and $i_B: B \\to A \\sqcup B$, such 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 $\\gamma: A \\sqcup B \\to X$ such that $\\gamma \\circ i_A = f_A$ and $\\gamma \\circ i_B = f_B$.]\n\n[summary(Technical): The [-5z9] may be described as the [-coproduct] in the [4c7 category] of [3jz sets]. Specifically, the disjoint union of sets $A$ and $B$ is a set denoted $A \\sqcup B$ together with maps $i_A : A \\to A \\sqcup B$ and $i_B: B \\to A \\sqcup B$, such 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 $\\gamma: A \\sqcup B \\to X$ such that $\\gamma \\circ i_A = f_A$ and $\\gamma \\circ i_B = f_B$.]\n\nWhere could we start if we were looking for a nice "easy" [-600] describing the [5s8 union of sets]?\n\nThe first thing to notice is that universal properties only identify objects up to [-4f4], but there's a sense in which the union is *not* [-5ss] [-65y]: 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 $A' \\cup B'$. %%note: In this case, when we're talking about [3jz sets], an isomorphism is just a [499 bijection].%%\n\n%%hidden(The union is not well-defined up to isomorphism in this sense):\nConsider $A = \\{ 1 \\}$ and $B = \\{ 1 \\}$.\nThen the union $A \\cup B$ is equal to $\\{1\\}$.\n\nOn the other hand, consider $X = \\{1\\}$ and $Y = \\{2\\}$.\nThen the union $X \\cup Y = \\{1,2\\}$.\n\nSo by replacing $A$ with the isomorphic $X$, and $B$ with the isomorphic $Y$, we have obtained $\\{1,2\\}$ which is *not* isomorphic to $\\{1\\} = A \\cup B$.\n%%\n\nSo it's not obvious whether the union could even in principle be defined by a universal property.%%note: Actually [union_universal_property it *is* possible].%%\n\nBut we can make our job easier by taking the next best thing: the *[5z9 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.\n\nBefore reading this, you should make sure you grasp the [-5z9] well enough to know the difference between the two sets $\\{ 2, 3, 5 \\} \\cup \\{ 2, 6, 7 \\}$ and $\\{ 2, 3, 5 \\} \\sqcup \\{ 2, 6, 7 \\}$.\n\n%%hidden(Brief recap of the definition):\nGiven 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 $B' = \\{ (b, 2) : b \\in B \\}$.\nThat 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".\n%%\n\nThe important fact about the disjoint union here is that if $A \\cong A'$ and $B \\cong B'$, then $A \\sqcup B \\cong A' \\sqcup B'$.\nThis 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.\n(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.)\n\n[toc:]\n\n# Universal property of disjoint union\n\nHow can we look at the disjoint union solely in terms of the maps to and from it?\nFirst of all, how can we look at the disjoint union *at all* in terms of those maps?\n\nThere are certainly two maps $A \\to A \\sqcup B$ and $B \\to A \\sqcup B$: namely, the [inclusion_map inclusions] $i_A : a \\mapsto a$ and $i_B : b \\mapsto b$.\nBut 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 $A \\sqcup B$. %%note: If 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 $A \\sqcup B$.%%\nSo we need to find some way to limit ourselves to $A \\sqcup B$.\n\nThe 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.\nThe following discussion shows us how.\n\n- If we know $f: A \\sqcup B \\to X$, then we know the [restriction_map restriction maps] $f \\big|_A : A \\to X$ %%note: This is defined by $f \\big|_A (a) = f(a)$; that is, it is the function we obtain by "restricting" $f$ so that its [3js domain] is $A$ rather than the larger $A \\sqcup B$.%% and $f \\big|_B : B \\to X$.\n- 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 $x \\in B$.\n\nThe 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 $\\beta$:\n\n- 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.\n- 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 $B$ %%note: If 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 [3js 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 $\\beta(x)$.\n\n## The property\n\nOne 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**:\n\n> 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 $f \\circ i_B = f_B$.\n\n[comment: Imgur was down when I made this picture, so I'm hosting it instead.]\n\n\n## Aside: relation to the product\n\nRecall the [-5zv]:\n\n> 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 $\\pi_B \\circ f = f_B$.\n\nNotice that this is just the same as the universal property of the disjoint union, but we have reversed the [3js domain] and [3lg codomain] of every function, and we have correspondingly reversed any function compositions.\nThis shows us that the product and the disjoint union are, in some sense, "two sides of the same coin".\nIf you remember the [-5zr], we saw that the empty set's universal property also had a "flip side", and this flipped property characterises the one-element sets.\nSo in the same sense, the empty set and the one-point sets are "two sides of the same coin".\nThis is an instance of the concept of [duality_category_theory duality], and it turns up all over the place.\n\n# Examples\n\n## {1} ⊔ {2}\n\nLet's think about the very first example that came from the top of the page: the disjoint union of $A=\\{1\\}$ and $B=\\{2\\}$.\n(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.)\n\nThe definition becomes:\n\n> 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 $f \\circ i_B = f_B$.\n\nThis is still long and complicated, but remember from [5zv when we discussed the product] that a map $f_A: \\{1\\} \\to X$ is precisely picking out a single element of $X$: namely $f_A(1)$.\nTo 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.\n(Of course, the same reasoning goes through with $B$ as well: $f_B$ is just picking out an element of $X$, too.)\n\nAlso, $i_A$ 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.\n\nSo 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:\n\n> 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 $f(i_B) = y$.\n\nFrom 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).\nBut 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.\n\nSo $A \\sqcup B$ is precisely the set $\\{i_A, i_B\\}$; and this has determined it up to isomorphism as "the set with two elements".\n\n### Aside: [-4w5]\n\nNotice that the disjoint union of "the set with one element" and "the set with one element" yields "the set with two elements".\nIt'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.\nWe will touch on this connection between "disjoint union" and "addition" a bit more later.\n\n## The empty set\n\nWhat is the disjoint union of the [-5zc] with itself?\n(Of course, it's just the empty set, but let's pretend we don't know that.)\nThe advantage of this example is that we already know a lot about the [-5zr], so it's a good testing ground to see if we can do this without thinking about *elements* at all.\n\nRecall that the empty set is defined as follows:\n\n> The empty set $\\emptyset$ is the unique set such that for every set $X$, there is precisely one function $\\emptyset \\to X$.\n\nThe 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):\n\n> 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 $f \\circ i_B = f_B$.\n\nNow, since we know that for every set $X$ there is a *unique* map $!_X: \\emptyset \\to X$ %%note: In 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 $!_X$:\n\n> 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 $f \\circ i_B = (!_{X})$.\n\nMoreover, 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.\nThey're both just $!_{A \\sqcup B}$.\n\n> 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 $f \\circ (!_{A \\sqcup B}) = (!_X)$.\n\nA diagram is in order.\n\n\n\nThe 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 $!_X : \\emptyset \\to X$.\n\nBut since $!_X$ is the *only* map from $\\emptyset \\to X$, we have $f \\circ (!_{A \\sqcup B}) = (!_X)$ *whatever $f$ is*.\nIndeed, $f \\circ (!_{A \\sqcup B})$ is a map from $\\emptyset \\to X$, and $!_X$ is the only such map.\nSo we can drop that condition from the requirements, because it automatically holds!\n\nTherefore the disjoint union is: \n\n> The set $$A \\sqcup B$$ with the requirement that for every set $X$, there is a *unique* map $f: A \\sqcup B \\to X$.\n\nDo you recognise that definition?\nIt's just the universal property of the empty set!\n\nSo $A \\sqcup B$ is the empty set in this example.\n\n# More general concept\n\nWe saw earlier that the disjoint union is characterised by the "reverse" (or "[duality_category_theory dual]") of the universal property of the product.\nIn a more general [4c7 category], we say that something which satisfies this property is a *[-coproduct]*. %%note: Prepending "co" to something means "reverse the maps".%%\n\nIn general, the coproduct of objects $A$ and $B$ is written not as $A \\sqcup B$ (which is specific to the category of [3jz sets]), but as $A + B$.\nAmong other things, the coproduct unifies the ideas of addition of [48l integers], [least_upper_bound least upper bounds] in [3rb posets], disjoint union of sets, [-65x] of [45h naturals], [-3w3], and [-free_product] of [3gd groups]; all of these are coproducts in their respective categories.',
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