Show that \mathcal {F}(X) \triangleq \mathbf {1} + X and \mathcal {F}(f) \triangleq \mathsf {id}_\mathbf {1} + f (using our + operator for functions) define an endofunctor \mathbf {Cpo} \rightarrow \mathbf {Cpo}, i.e. that they satisfy the functor laws.

Solution.

• Liam O'Connor

1. We must show \mathcal {F}(\mathsf {id}_A : A \twoheadrightarrow A) = \mathsf {id}_{\mathcal {F}(A)} : \mathcal {F}(A) \twoheadrightarrow \mathcal {F}(A). \begin {array}{lcl} \mathcal {F}(\mathsf {id}_A)& =& \mathsf {id}_\mathbf {1} + \mathsf {id}_A\\ &=& \mathsf {id}_{\mathbf {1} + A}\\ &=& \mathsf {id}_{\mathcal {F}(A)} \end {array}
2. We must show \mathcal {F}(f \circ g) = \mathcal {F}(f) \circ \mathcal {F}(g). \begin {array}{lcl} \mathcal {F}(f \circ g) &=& \mathsf {id}_\mathbf {1} + (f \circ g) \\ &=& (\mathsf {id}_\mathbf {1} + f) \circ (\mathsf {id}_\mathbf {1} + g) \\ & = & \mathcal {F}(f) \circ \mathcal {F}(g) \end {array}