I wrote:

> 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?

Peter wrote:

> I think in my diagram I have defined it at least for the case where \$$g\$$ is the identity. Due to the fact that f is in the opposite category, we can define the map from \$$\mathcal{A}(F(a),b)\$$ to \$$\mathcal{A}(F(a'),b)\$$ by the composite \$$h \mapsto F(f) \circ h\$$.

Yes, right!

Thanks to Dan's remark about Steve Awodey's book, expanded on [here](https://forum.azimuthproject.org/discussion/comment/19552/#Comment_19552), we can get the desired map

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

as follows. First use \$$f: a' \to a\$$ to cook up a map

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

Then use \$$g: b \to b'\$$ to cook up a map

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

Finally, compose these two maps to get the job done:

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

You explained the first step, where we use \$$f : a' \to a\$$. This gives a map

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

that maps any morphism \$$h\$$ to \$$h \circ F(f)\$$.