Universal arrow example which doesn't come as an adjoint pair? Given $2$ locally small categories $\textsf{D}$ and $\textsf{C}$, consider a functor $K:\textsf{D}\to\textsf{C}$ and an object $C\in\textsf{C}$. A universal arrow from $K$ is a couple $D\in\textsf{D}$ together with an arrow $u:C\to KD$ such that for every $f:C\to KD'$ there exists a unique $\bar{f}:D\to D'$ which satisfies
$$K\bar{f}\circ u=f$$
If we consider a free algebraic structure such as a free group or a vector space, it is pretty clear this property holds true for the forgetful functor $U$, but that comes along since we have a left adjoint for $U$, namely the free functor $F$. Is there an example of functor with universal arrow which isn't explained by an adjoint pair?
 A: $K : \mathcal{D} \to \mathcal{C}$ is a right adjoint iff for every $C \in \mathcal{C}$ there is a universal arrow from $C$ to $K$. So it suffices to give an example of a functor which is "partially right adjoint" but not "totally right adjoint". Equivalently, when we see $K$ as a forgetful functor, it suffices to find a free $\mathcal{D}$-structure on some object of $\mathcal{C}$ such that however not every object of $\mathcal{C}$ has a free $\mathcal{D}$-structure.
Consider for example the forgetful functor from finite groups to sets. There is a free (finite group) on an empty set, but no free (finite group) on a non-empty set (see below).
Or consider the forgetful functor from complete boolean algebras to sets. For every finite set $X$ there is a free complete boolean algebra on that set (namely $P(P(X))$ with the usual operations), but this fails for every infinite set by a result of Hales.

Here is a proof that if $X$ is a non-empty set then there is no free (finite group) on $X$. (Notice that it is not enough to prove that the free group on $X$ is not finite.)  Assume that it exists, name it $F$. Choose some $x \in X$. Its image in $F$ is also denoted by $x \in F$. Since $F$ is finite, $x^n=1$ for some $n \in \mathbb{N}^+$. Now if $G$ is any finite group and $g \in G$, then there is a map $X \to G$, sending everything to $g$. It extends to a (finite group) homomorphism $F \to G$, sending in particular $x$ to $g$. It follows $g^n=1$. This is a contradiction, since there are finite groups with elements of order $n+1$.
