We cannot get morphism mappings in the category \textbf {Cpo} as easily as § [dt-0050] when our object mapping makes use of the function operators \mathbin {\circ \hspace{-0.4em}\twoheadrightarrow } and \twoheadrightarrow , as they are contravariant in their first argument. This means the functor that they extend to is not \textbf {Cpo} \rightarrow \textbf {Cpo} but \textbf {Cpo}^\textsf {op} \rightarrow \textbf {Cpo}, where \textbf {Cpo}^\textsf {op} is the dual category of \textbf {Cpo}. The morphisms end up the wrong way around.
As an example, consider \mathcal {F}(X) = X\twoheadrightarrow \mathbb {Z}_\bot . Then, recalling the morphism mapping of the \twoheadrightarrow functor:
\dfrac {f : A \twoheadrightarrow B \quad g : C\twoheadrightarrow D}{f\twoheadrightarrow g : (B\twoheadrightarrow C)\twoheadrightarrow (A\twoheadrightarrow D)}
We can generate a morphism mapping for \mathcal {F}: \mathcal {F}(f)\; =\; f\twoheadrightarrow \mathsf {id}_{\mathbb {Z}_\bot }\; =\; (\lambda h.\ \mathsf {id}_{\mathbb {Z}_\bot }\circ h \circ f)\; = \;(\lambda h.\ h \circ f)
However, this mapping has the wrong type. For f : A\twoheadrightarrow B, then \mathcal {F}(f) : (B\twoheadrightarrow \mathbb {Z}_\bot ) \rightarrow (A\twoheadrightarrow \mathbb {Z}_\bot ), which is \mathcal {F}(B)\twoheadrightarrow \mathcal {F}(A), not the required \mathcal {F}(A)\twoheadrightarrow \mathcal {F}(B). This is because the generated functor is contravariant, not covariant.