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)\\).