Let $\mathcal{C}$ be a category with finite coproducts. Fix some object $A$ in $\mathcal{C}$. We have a forgetful functor $U: A/\mathcal{C} \to \mathcal{C}$ which sends $f: A \to X$ to $X$ and morphisms to themselves. Does this functor have left and right adjoints?
It does not have a right adjoint, because if it had, then it must preserve colimits, in particular the empty coproduct (= initial object) which is just the identity on $A$. In general $A$ will not be the initial object in $\mathcal{C}$, so in general $U$ does not have a right adjoint.
The left adjoint is more difficult. Since it is forgetful, I am guessing it has some free functor as left adjoint, but cannot find a meaningful way to assign a morphism with domain $A$ to an object $C$ in $\mathcal{C}$.
Best Answer
Sometimes it helps to unpack the definition. In this case, for the forgetful functor $A/\mathcal{C} \to \mathcal{C}$ to have a left adjoint, you'd need to assign to each object $X$ an object $FX \in \mathcal{C}$ and a morphism $f_X : A \to FX$ such that $$\mathrm{Hom}_{A/\mathcal{C}}(A \xrightarrow{f_X} FX, A \xrightarrow{g} Y) \cong \mathrm{Hom}_{\mathcal{C}}(X, Y)$$ naturally in $X \in \mathcal{C}$ and all $A \xrightarrow{g} Y \in A/\mathcal{C}$.
Can you see how you might define such a natural isomorphism? If not, hover over the box below for an additional hint.