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\\).