= Solution
Identities are self-adjoint. If $f\dashv g$ and $f'\dashv g'$, then
$$
f'f\,gg'=f'(fg)g'\leq f'g'\leq1,
\qquad
1\leq gf\leq gg'f'f,
$$
with the second inequality written after inserting the two units in the appropriate order. Hence $f'f\dashv gg'$, so left adjoints form a subcategory.
In $\mathbf{Poset}$, these are exactly monotone maps possessing right adjoints, equivalently <Galois connection>[lower adjoints]; when all joins exist, they are precisely the arbitrary-join-preserving maps.
In the inclusion-ordered <category of relations>, a relation $R:A\rightsquigarrow B$ is left adjoint exactly when it is total and single-valued. It is therefore the graph of a function, and its right adjoint is $R^{\mathrm{op}}$. This is the <left adjoint relation is a function> criterion.
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