I have the following construction, which seems too easy. Could you review and comment? Thanks in advance.
Suppose $I$ is a set, and $P(I)$ is its power set, viewed as a category whose arrows are set inclusion. Then the functor $i:P(I)\rightarrow Set/I$ defined in the obvious way by $U(\subset I)\mapsto(U\hookrightarrow I)$ and $(U\subset V)\mapsto $ \begin{array}{&&} U & {\to} & V\\ & \searrow & \downarrow \\ & & I \end{array}
(where all the arrows are inclusions) has a left adjoint, $F$.
$F:Set/I\rightarrow P(I)$ just takes any $h:X\rightarrow I$ to $h(X)$. On arrows $\varphi :h\rightarrow h'$, we have \begin{array}{&&} X & \stackrel{\varphi}{\to} & X' \\ & h \searrow & \downarrow h' \\ & & I \end{array}
so that $h(X)\subset h'(X')$ as required. The unit, $\eta _{h}:h\rightarrow (i\circ F)h$ is then $h \rightarrow (h(X)\hookrightarrow I)$. If $f:h\rightarrow i(V)$ then $h(X)\subset V$ and so we get our (unique) $ \overline f:Fh\rightarrow V$. To show that \begin{array}{&&} h & \stackrel{\eta _{h}}{\to} & i(Fh) \\ & f \searrow & \downarrow i(\overline f) \\ & & i(V) \end{array}
commutes just amounts to observing that $f=h$.
The OP essentially answered their question in the course of posing it:
By definition $h = h'\circ \varphi,$ and so $h(X) = h'(\varphi(X)) \subset h'(X')$, as required.