$F^{op}$ functor I'm trying to make sens of category theory. In my course it is stated that a functor is a map $$F:{\cal{C}}\rightarrow{\cal{D}}:X\mapsto F(X)$$ and which sends $f\in \text{Hom}_{\cal{C}}(X,Y)$ to $Ff\in \text{Hom}_{\cal{D}}(FX,FY)$ with two compatibility conditions. A little later it defines the functor $F^{op}:{\cal{C}}^{op}\rightarrow {\cal{D}}$ from the functor $F$. How is this functor defined? 1. Since the objects of $\cal{C}$ are equal to the objects of $\cal{C}^{op}$ I suppose that $F^{op}$ maps $X$ to $F^{op}(X) := F(X)$. 2. However for the morphisms it is less clear to me. It must send $f\in \text{Hom}_{{\cal{C}}^{op}}(X,Y)$ to $F^{op}f\in \text{Hom}_{\cal{D}}(F^{op}X,F^{op}Y) = \text{Hom}_{\cal{D}}(FX,FY)$. Simply applying $F$ will not work because we will not end in the right set. Is there such a thing as $f^{op}$ ($f\in \text{Hom}_{{\cal{C}}^{op}}(X,Y)$ would give $f^{op}\in \text{Hom}_{\cal{C}}(X,Y))$? Then we could define $F^{op}f:=F(f^{op})$

You can define a functor $\mathcal C^\text{op}\to \mathcal D^\text{op}$ by just reversing all morphisms in sight. But a functor $\mathcal C^\text{op}\to \mathcal D$ is in general not possible.

For example, consider the category with two objects and one non-trivial map $X\to Y$. Let $F:\mathcal C\to \mathcal C$ be the identity functor on this category.

Now your concern is valid -- there's no functor $\mathcal C^\text{op}\to \mathcal C$ that remains the identity on objects but reverses the arrow, simply since there are no maps $Y\to X$.