\$$F^\mathrm{op}\$$ should do the same thing as \$$F\$$ on objects (i.e., \$$\forall c\in\mathcal{C}\,\,F(c)=F^\mathrm{op}(c)\$$). On morphisms, I imagine it should take \$$f^\mathrm{op}\$$ to \$$(F(f))^\mathrm{op}\$$. This works trivially on identities. \$$F^\mathrm{op}(f^\mathrm{op}\circ g^\mathrm{op})=F^\mathrm{op}((g\circ f)^\mathrm{op})\$$ \$$=(F(g\circ f))^\mathrm{op}=(F(g)\circ F(f))^\mathrm{op}\$$ \$$=(F(f))^\mathrm{op}\circ(F(g))^\mathrm{op}\$$ \$$=F^\mathrm{op}(f)\circ F^\mathrm{op}(g)\$$, so \$$F^\mathrm{op}\$$ defined in this way is a functo from \$$\mathcal{C}^\mathrm{op}\$$ to \$$\mathcal{D}^\mathrm{op}\$$.