A formula for a right adjoint in terms of a left

Question

For those familiar with (covariant) Galois connections, you may have noticed that they can be viewed as categorical adjunctions. A Galois connection is a pair of maps between posets $X$ and $Y$ $$ f_{\bullet}: X \rightleftharpoons Y: f^{\bullet}$$ such that $f_{\bullet} x \preceq y$ if and only if $x \preceq f^{\bullet} y$. If we view $X$ and $Y$ as small categories, then order-preserving maps are simply functors between such categories as $$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$ is equivalent to the condition above (naturality is trivial).

In the case of posets, there is an explicit formula to calculate the right adjoint from the left adjoint if one exists (sim. left adjoint from right adjoint).

You can check that $$f^\bullet (y) = \bigvee f_{\bullet}^{-1}(\downarrow y)$$ is a right adjoint (if $f_\bullet$ preserves joins I believe??).

Now here is my question: Are there other areas of mathematics or other examples where an explicit formula for a right adjoint in terms of the left adjoint appears?

References most welcome. Thanks!

Answer 1

Just so that it is recorded here as an answer, here's the formula from the "naive" adjoint functor theorem that directly generalizes the one given in the post for posets:

$$f^\bullet(y) = \lim_{(x,\alpha)\in (f_\bullet \downarrow y)} x$$

where the limit is over the comma category $(f_\bullet \downarrow y)$, whose objects are pairs of $x\in X$ with $\alpha:f_\bullet(x)\to y$.

The subtlety in adjoint functor theorems comes from the fact that this comma category is of the same "size" as the categories $X$ and $Y$, while non-poset categories rarely have limits of the same size as themselves. Thus, the adjoint functor theorems used in practice impose various technical conditions enabling this limit to be replaced by a smaller one; see e.g. the nlab article for details.

Answer 2

It is common for the structure of objects in a category to be corepresentable, and so the formula $$Hom(f_\bullet x, y) \cong Hom(x, f^\bullet y)$$ is a formula for the right adjoint.

As an example of how this works, in the category of groups, you have the fact that $$ |G| \cong \hom(\mathbb{Z}, G) $$ where $|G|$ means the underlying set, and that the group operation is given by the natural transformation $$ \hom(\mathbb{Z}, G) \times \hom(\mathbb{Z}, G) \cong \hom(F_2, G) \to \hom(\mathbb{Z}, G) $$ induced by the map $\mathbb{Z} \to F_2$ sending $1 \mapsto xy$, where $F_2$ is the free group on two elements $x$ and $y$. (it is the coproduct of $\mathbb{Z}$ with itself, thus the first isomorphism)

So, if $f^\bullet$ is a group-valued functor, you can compute it as follows:

• The underlying set of $f^\bullet X$ is given by $\hom(f_\bullet \mathbb{Z}, X)$
• The group operation is $\hom(f_\bullet \mathbb{Z}, X) \times \hom(f_\bullet \mathbb{Z}, X) \cong \hom(f_\bullet F_2, X) \to \hom(f_\bullet \mathbb{Z}, X)$

and this information suffices to determine the group; the identity and inverse can be given directly by similar formulas if desired.