3
$\begingroup$

I try to understand adjoint functors in category theory. I like the idea of thinking of a left adjoint as 'best approximation from above' and a right adjoint as 'best approximation from below.'

More formally, this corresponds to the universal arrow construction, if one want to find a left adjoint of $R:{\bf D}\rightarrow{\bf C}$, for any $c\in{\bf C}$ one needs to find an initial object $(Lc,i_c:c\rightarrow RLc)$ in the comma category $c\Downarrow R$.

My question is now, how should I think about the unique arrow demand for the initial object? Like, if I think of categories where all hom sets are trivial (empty or one element), the 'best approximation from above' metaphor works just fine because when I say things as 'from above' I usually don't think about multiple arrows between objects. Is there any way to enhance this metaphor in a nice way such that it also works for the general case?

EDIT: I shared some new thoughts in an answer below.

$\endgroup$
5
$\begingroup$

Take your favourite example of a universal arrow, say $\{a, b\} → U\{a, b\}^*$ (the free monoid generated by letters a, b). Now $\{a, b\}^* $ is certainly big enough to factor every arrow $\{a, b\} → UM$, and it will remain so even if you add some clutter, making it into eg. $\{a, b, c\}^*$. But now the factorization is not unique, because you can map $c$ to any element of $M$. So uniqueness seems to guarantee that the universal arrow is optimal in a sense.

This of course is very vague and imprecise: the biggest problem is that the size terminology is unpredictable, because an arrow $A → B$ can intuitively look as either proving that $A$ feels smaller (eg. $ℤ → ℝ$) or larger ($ℤ → ℤ/(2)$) than $B$, but it works fine for many basic examples, ie. free constructions, reflections, and (co)limits in familiar categories.

For initial/terminal objects, I would suggest remembering the classical cliché that it's the arrows that matter, not objects. If you are thinking about a terminal object as a sort of "best approximation from above", then every arrow $A → T$ to the terminal object needs to be "the best" one, which in the presence of parallel arrows to $T$ is impossible. In $\mathrm{Set}$, every non-empty set is "above", but only the singletons can be so in an optimal way.

Of course at the end of the day it's the formal properties of the uniqueness requirement that really matter. If $η : C → R(LC)$ is the universal arrow from $C$ to $R : \mathscr D → \mathscr C$, then the existence of factorization makes the function $\mathrm{Hom}(LC, D) → \mathrm{Hom}(C, RD)$ given by $g ↦ Rg ∘ η$ surjective, while the uniqueness makes it injective, and I would say that the resulting isomorphism of functors $\mathrm{Hom}(LC, -) ≅ \mathrm{Hom}(C, R-)$ is the most important formulation of universality.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

I thought of an example generalizing the partial order category which makes the requirement for unique arrows seem more reasonable.

Let $\cal L$ be a fixed set of labels and to a set $X$ with partial orders $\{\le_\ell\}_{\ell\in\cal L}$ associate the category $\cal C$ with object set $X$ and for all $A\subseteq \cal L$ an arrow ${\sf ar}(x,A,y):x\rightarrow y$ iff $x\le_\ell y$ for all $\ell\in A$ (we also write $x\rightarrow_A y$ in that case). Arrow composition is defined via ${\sf ar}(y,B,z)\circ{\sf ar}(x,A,y)={\sf ar}(x,A\cap B,z)$. So one has e.g. $x\rightarrow_\emptyset y$ and ${\sf id}_x={\sf ar}(x,\cal L,x)$ for all $x,y\in X$.

Now assume $X,Y$ are equipped with $\cal L$-labeled partial orders as above and let $\cal C,D$ denote the corresponding categories, further let a functor $F:\cal D\rightarrow C$ be given, this is just an order morphism in all given partial orders.

Assume $R$ has a left adjoint $R:\cal C\rightarrow D$, i.e. each $x\in X$ corresponds to an initial object $(Rx,i_x:x\rightarrow RLx)$ in the comma category. The unique arrow requirement guarantees that ${\sf id}_{Rx}$ is indeed the only automorphism on $(Rx,i_x)$, which means that for any $A\neq \cal L$, there is no composition $i_x={\sf ar}(RLx,A,RLx)\circ i_x$, which means that $i_x={\sf ar}(x,{\cal L},RLx)$, i.e. $RLx$ is indeed an approximation from above w.r.t. all labels $\ell\in\cal L$. Computation that $RLx$ is also the best approximation from above w.r.t. all $\ell\in\cal L$ is done analogously to the case of $|\cal L|=1$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.