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\$$.