Suppose $C$ is a category, $X\in C$. I want to find minimal conditions on $C$ for which the forgetful functor $U:C/X\rightarrow C$ has a left adjoint.
edit: As pointed out in the comments, $U$ has no left adjoint unless $X$ is terminal which then says that $C$ and $C/X$ are isomorphic.
I did manage to produce a right adjoint by defining $G:C\rightarrow C/X$ on objects: $A\rightarrow (\pi _{2}:A\times X\rightarrow X)$ and on morphisms $\overline f:A\rightarrow S$ by $G(\overline f)=\overline f\times 1_{X}$, where $\overline f\times 1_{X}$ is the arrow in $C/X)$ from $\pi _{2}:A\times X\rightarrow X$ to $p _{2}:S\times X\rightarrow X$ So if \begin{array}{&&} A & \stackrel{f}{\to} & S\times X \\ & h \searrow & \downarrow p_{2} \\ & & X \end{array} is an arrow in $C/X$, then we obtain a unique $\overline f:U(A,h)=A\rightarrow S$ using the UMP of the product $A\times S$. More specifically, $\overline f$ is the unique morphism such that $p_{1}f=\overline f$ and $p_{2}f=h$. The unit $\eta :1_{C/X}\rightarrow GU$ is determined uniquely, again, by the UMP of the product $A\times X$: $\pi {_1}\eta _{(A,h)}=1_{A}$ and $\pi {_2}\eta _{(A,h)}=h$ . So $U$ has a right adjoint if and only if $C$ has products of type $A\times X$. My question is: is there an easier way to show this? I did it basically from scratch. I wonder if there is a slick way to do it.