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}) \\)... ???

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}) \\)... ???