So, how is that exactly? We are interested in the functor \$$F':\mathcal{A}^{op} \times \mathcal{B} \rightarrow \boldsymbol {\text{Set}}. \$$ For that we consider the functor \$$F: \mathcal{A} \rightarrow \mathcal{B}. \$$ Is this correct? There isn't a functor \$$F^{op}: \mathcal{A}^{op} \rightarrow \mathcal{B} \; ?\$$

A morphism \$$f: X \rightarrow Y \$$ in \$$\mathcal{A}^{op} \$$ then induces a map of \$$\mathcal{B} (F (\mathcal{A}^{op} ),\mathcal {B}) \$$ like following?
![How a morphism is maped on a morphism on the Hom-sets by a functor](https://svgshare.com/i/78r.svg)
Hmm, but actually we are looking for a map on \$$\mathcal{B} (F (\mathcal{A} ),\mathcal {B}) \$$ instead of \$$\mathcal{B} (F (\mathcal{A}^{op} ),\mathcal {B}) \$$... ???