Word count: 1677

Here’s a mathematical situation that comes up a lot more often than is
reasonable. Suppose we have some mathematical object $G$, generally
defined as some structure on a set: a group, a ring, a topological
space. These *all* have natural categorical structures, so we can - in
a uniform way - talk about their *subobjects*:

- A
**subgroup**$X$ of $G$ is a subset of $G$^{1}*closed under the group operations*: If $x, y \in X$ then you must also have $xy \in X$, the unit $1$ is in $X$, and if $x \in X$ then so is $x^{-1}$. Alternatively — this is the categorical phrasing — a subgroup $X$ of $G$ is a group $X$ equipped with an*injective*^{2}group homomorphism $f : X \hookrightarrow G$. Injective means “reflecting equality”: If $f(x) = f(y)$ then we must have $x = y$.

- A
**subring**$X$ of $R$ is a subset of $U(R)$ closed under the ring operations; Alternatively, it’s a ring $X$ equipped with an injective ring homomorphism $X \hookrightarrow R$.

For now, let us focus on groups. Fix a group $G$: It has an underlying set of elements $U(G)$ (keep in mind that the “underlying set” operation is really a faithful functor). Of course, not every subset of $U(G)$ is going to be a subgroup of $G$ — consider the case where $G = (\mathbb{Z},+)$ and the subset we have at hand is e.g. $\{ 1, 2, 3 \}$. But — suppose that we do have a subset, and we’re really keen on seeing it grow up to be an honest-to-god subgroup. Are we out of luck? Is there any hope for us?

# Generating subgroups

As it turns out — yes! Suppose that we have two subgroups $X$, $Y$ of
$G$: one can (and you probably should) check that their intersection $X \cap Y$ is also a subgroup of $G$. I’m going to phrase this with a lot
more abstract nonsense though: There is a canonical way of regarding
“arbitrary objects with maps into a fixed object” as a category: A
slice category, in this case the slice $\mathrm{Grp}/G$ (it’s read
“groups over $G$”). The *specific* case we’re interested in is referred
to as $\mathrm{Sub}(G)$ (the “poset of subobjects”) — it’s the full
subcategory of $\mathrm{Grp}/G$ on the monic maps.

Suppose that we have subgroups $(X, \iota_1)$ and $(Y, \iota_2)$, which,
by definition, are groups equipped with maps into $G$. Their
intersection, *if it exists*, is an object $X \times Y$ together with
maps $(X \times Y) \to X$ and $(X \times Y) \to Y$, universal among such
“cones” (see the diagram below): It’s a *product* of $X$ and $Y$ in the
category $\mathrm{Grp}/G$. Some abstract nonsense (by which I mean
category theory) tells us that the product of $(X, \iota_1)$ and $(Y, \iota_2)$ in $\mathrm{Grp}/G$ is given by the *pullback* $X \times_G Y$ in $\mathrm{Grp}$, equipped with the canonical projections and choice
of map into $G$.

Of course, there’s also the normal, set-theoretic, *boring* intersection
of subgroups $X$ and $Y$. I’ll (cheekily) leave it as an exercise to the
reader to show that, given material subsets $X$ and $Y$ of $G$, the
underlying set of the pullback group $X \times_G Y$ (where the
injections are canonical) is isomorphic to the intersection $X \cap Y$.
The abstract nonsense perspective here is to emphasise thinking of *the
subgroups of a given group* as forming their own little category.

Now, suppose that we have just some random subset $X$ of the underlying
set of $G$, which — keeping with the abstract nonsense — we will
consider as a set $X$ equipped with an injection $i : X \hookrightarrow U(G)$. We can now ask our question with slightly more precision: Is
there any object $X'$ of $\mathrm{Sub}(G)$ which we can
map $X$ into? Note that this question is still slightly misphrased: $X$
and *whatever $X'$ might be* are in different categories!

Fortunately, our functor $U : \mathrm{Grp} \to \mathrm{Sets}$ extends smoothly to a functor $\mathrm{Sub}(G) \to \mathrm{Sub}(U(G))$, patching up the last paragraph and letting us ask the question “is there any hope for us?” in a precise way:

**Question**. Given an object $X$ of $\mathrm{Sub}(U(G))$, are there any
*interesting* objects $X' \in \mathrm{Sub}(G)$ with *cool* maps $X \to U(X')$?

Note the emphasis on *interesting* and *cool*: $\mathrm{Sub}(G)$ has a
terminal object (it’s $G$ itself, equipped with the identity map), so
there are plenty of *boring* answers to the question: any subset of $G$
embeds into the “greatest subgroup” — $G$ itself! Note that we still
haven’t answered the question either way: we’ve just rephrased it into
the language of category theory. This means that we can now attack it
with abstract nonsense!

# The nonsense

We’ll start by generalising our question away from the specific
situation of subgroups and subsets: If we have *any functor* $R : \mathcal{D} \to \mathcal{C}$, is there a way of assigning objects $X' \in \mathcal{D}$ to objects $X \in \mathcal{C}$, such that there are
*interesting* maps $X \to R(X')$? First, I’ll note that the *objects
$X'$ equipped with maps $X \to R(X')$* form their own category, which is
referred to as $X \swarrow R$ (the maps in *that* category are a
surprise tool that will help us later). But let’s ping back to the group
example — hell, let’s consider a *concrete* example, for a big change,
and yes, right after I said we’d attack it with abstract nonsense
— to figure out what *interesting* maps might mean.

Suppose we set $G = (\mathbb{Z},+)$, and our subset is $X = \{2, 4\}$.
Consider the trivial solution, where we set $X' \subseteq \mathbb{Z}$
(all the inclusions in this paragraph are omitted — they’re
canonical). A *better* solution is to consider just the evens: $X' = \{ 2n : n \in \mathbb{Z} \}$. But how exactly is it better? Well - it’s
smaller! Literally! It is, in fact, the *smallest* such solution, in
that if you have a subgroup $S$ which includes $X$, then it includes
every even integer. Categorifying, we say that any other solution $X \to S$ *factors uniquely as* $X \to X' \to S$.

Now, conveniently, we can introduce the maps in $X \swarrow R$: A map $(A, f) \to (B, g)$ is a map $h : A \to B$ for which the diagram below commutes. A factoring of $g$ as $f$ followed by $R(h)$.

Finally - finally! - we can express what it means to “solve $R$” in an “interesting” way: we have an object $X'$, together with a map $\eta : X \to R(X')$, map such that $(X', \eta)$ is an initial object in the comma category $X \swarrow R$ — which is precisely the data of a left adjoint functor $L : \mathcal{C} \to \mathcal{D}$ to our functor $R$.

## Generating sub-thingies

Here’s a list of constructions, including the solution to the “subgroup problem” we’ve been pondering. Note that all the constructions below are done “relative to” a bigger structure, because (for instance) it makes no sense to consider subgroups without having a bigger group to embed them in!

The

*subgroup $F(S)$ generated by a subset $S$*is the**intersection of all**subgroups of $G$ containing $S$;The

*normal subgroup generated $F(S)$ generated by a subgroup $G$*is the**intersection of all**normal subgroups of $G$ containing $S$;The

*closure $\overline{S}$*of a subset $S$ (of a topological space $X$) is the**intersection of all**closed subsets of $X$ containing $S$;The

*interior $\mathrm{int}(S)$*of a subset $S$ (of a topological space $X$) is the**union of all**open subsets contained in $S$;

That last one is sneaky: it’s dual to all of the others (it’s happening
in an *opposite category*). Let’s focus on the first three; the interior
case is, well, dual. In all of them, we’re constructing an object —
the smallest object satisfying some property — by taking the
intersection of *all* objects satisfying that property. We already know
that intersections correspond to products (the “empty intersection”,
i.e. the maximal subset, is a terminal object): this generalises to
$n$-ary intersections corresponding to **limits**. This construction
is entirely formulaic! Does this generalise?

# The Adjoint Functor Theorem

Yes! All of the constructions above are instances of the **general
adjoint functor theorem**: For a functor $R : \mathcal{D} \to \mathcal{C}$, if:

- $\mathcal{D}$ is a small, complete preorder; and
- $R$ preserves all limits (meets) in $\mathcal{D}$, then

Then $R$ admits a left adjoint $L$, with $L$ mapping $x$ to the limit of
— pardon the strong wording — **the whole-ass category $x \swarrow R$**, which exists by our assumption that $D$ is complete. I
will warn you that, since $\mathcal{D}$ has limits the size of its class
of arrows, it must^{3} be a preorder. Please note that, since — and
I quote:

this generalises to $n$-ary intersections corresponding to

limits.

The construction of the left adjoint in the AFT *is given in the same
way* as our bullet list before, taking the intersection of all the
candidates! And the case with duals? That’s using the AFT to construct a
*right* adjoint: formally dual to the situation above. Let me spell out
an example in a tiny bit more detail:

Fix a group $G$; Consider the full subcategory of $\mathrm{Grp}/G$ on the normal monomorphisms, i.e. the poset of normal subgroups of G — let’s call it $\mathrm{Norm}(G)$. $\mathrm{Norm}(G)$ is small because, e.g., $\mathrm{Grp}$ is locally presentable: standard-issue abstract nonsense. This category admits a functor $U : \mathrm{Norm}(G) \to \mathrm{Sub}(U(G))$, to the poset of subsets of the underlying set of $G$. $\mathrm{Norm}(G)$ is also a preorder: it admits a fully faithful functor into the poset of subobjects of G.

Moreover, since $\mathrm{Grp}$ has limits, and normal monomorphisms are
closed under arbitrary intersections, we can conclude that
$\mathrm{Norm}(G)$ is complete; Since intersections of subgroups are
taken to intersections of *sets* by $U$, the AFT tells us it has a left
adjoint $F$: The “free normal subgroup” on a subset of $G$. Moreover,
$F(X)$ is computed by taking the limit of all normal subobjects which
include $X$, as promised.

## Conclusion

Okay, but then what? Well, then nothing. I just wanted to write about
something that clicked for me — I knew about the AFT and “big
intersection” constructions both, but not that they’re *the same thing*!
This blew my mind when I learned about it, and I hope that it’s helped
make at least one of these concepts seem less arbitrary.

Also: I hope you noticed some of those links! Yeah, I’ve been gone for a while. I’ve been working on the 1Lab: a formalised, cross-linked, explorable reference resource for univalent mathematics. I’m going back to playing with cubes now — take care!

Formally speaking, it’s a subset of the

*underlying set*of $G$, generally written $U(G)$.↩︎A stickler for details could point out that the right notion of “injectivity” in an arbitrary category is being a

*monomorphism*rather than reflecting equality. However, both $\mathrm{Grp}$ and $\mathrm{Ring}$ admit a faithful functor into $\mathrm{Sets}$, and faithful functors reflect monos; It suffices to check that the underlying function of the homomorphism is injective.↩︎Assuming excluded middle, or in any Grothendieck topos↩︎