Peter wrote:

> 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} \; ?\$$

No, there's not a functor from \$$\mathcal{A}^{\text{ op}} \$$ to \$$\mathcal{B}\$$ in this game. We have a functor \$$F : \mathcal{A} \to \mathcal{B} \$$. There's no way to turn that into a functor \$$\mathcal{A}^{\text{op}} \$$ to \$$\mathcal{B}\$$.

The "op" comes in when we consider

$\mathcal{B}(F(-),-) : \mathcal{A}^{\mathrm{op}} \times \mathcal{B} \to \mathbf{Set}$

This functor sends any pair \$$a, b\$$ consisting of an object \$$a\$$ in \$$\mathcal{A}\$$ and an object \$$b\$$ in \$$\mathcal{B}\$$ to the set

$\mathcal{B}(F(a),b) .$

But what does this functor do to morphisms? This is where the "op" comes in! If we have a pair of morphisms \$$f: a' \to a\$$ and \$$g: b \to b'\$$ this functor gives us a map from

$\mathcal{B}(F(a),b)$

to

$\mathcal{B}(F(a'),b')$

**Puzzle.** Can somehow say how we define this map? It's my duty to explain this, but someone probably knows already.

The key point is that to get this map is built from a pair \$$(f,g) \$$ consisting of a morphism \$$g: b \to b'\$$ going _forwards_ from \$$b\$$ to \$$b'\$$, and a morphism \$$f: a' \to a\$$ going _backwards_ from \$$a'\$$ to \$$a\$$. That backwardsness is why we need \$$\mathcal{A}^\textrm{op}\$$.