= Solution
A presheaf on $0\to1$ is an arrow $r:B=X(1)\to A=X(0)$ in sets. A morphism of presheaves is a commuting square. Define
$$
\Pi(r)=A,\qquad\Gamma(r)=B,\qquad\Delta(S)=(S\xrightarrow1S),\qquad\nabla(S)=(S\to1).
$$
A commuting square from $r$ to $\Delta S$ is determined by its map $A\to S$, while a square from $\Delta S$ to $r$ is determined by its map $S\to B$. A square from $r$ to $\nabla S$ is also determined by its map $B\to S$, because the other component is the unique map $A\to1$. Therefore these Hom-set bijections establish the <adjoint chain for evaluation in an arrow category>:
$$
\boxed{\Pi\dashv\Delta\dashv\Gamma\dashv\nabla.}
$$
There is a further left adjoint to $\Pi$: $\Lambda(S)=(\varnothing\to S)$. A square $\Lambda S\to r$ is exactly a map $S\to A$, proving $\Lambda\dashv\Pi$.
There is no right adjoint to $\nabla$. If there were, $\nabla$ would be a left adjoint and preserve the <initial object>. But it sends the empty set to $(\varnothing\to1)$, whereas the <initial object> of this arrow category is $(\varnothing\to\varnothing)$. They are not isomorphic. Hence \b[$\Pi$ does have a left adjoint, while $\nabla$ has no right adjoint].
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