The generalized pullback lemma [frct-0014]
The generalized pullback lemma [frct-0014]
In light of our discussion of the fundamental self-indexing, the following result for displayed categories generalizes the ordinary “pullback lemma.”
Lemma. Generalized pullback lemma [frct-001H]
Lemma. Generalized pullback lemma [frct-001H]
Let \({\bar {x}}\xrightarrow [f]{\bar {f}}{\bar {y}}\), and suppose that \({\bar {y}}\xrightarrow [g]{\bar {g}}{\bar {z}}\) is cartesian over \(g\). Then \(\bar {f};\bar {g}\) is cartesian over \(f;g\) if and only if \(\bar {f}\) is cartesian over \(f\).
Proof.
Proof.
Suppose first that \(\bar {f}\) is cartesian. To see that \(\bar {f};\bar {g}\) is cartesian, we must construct a unique factorization as follows:
Because \(\bar {g}\) is cartesian, we can factor \(\bar {h} = i;\bar {g}\) for a unique \({\bar {u}}\xrightarrow [m;f]{i}{\bar {y}}\). Then, because \(\bar {f}\) is cartesian, we can further factor \(i = j;\bar {f}\) for a unique \({\bar {u}}\xrightarrow [m]{j}{\bar {x}}\). We conclude that there is a unique \({\bar {u}}\xrightarrow [m]{j}{\bar {x}}\) for which \(\bar {h} = j;\bar {f};\bar {g}\), as required.
Conversely, suppose that \(\bar {f};\bar {g}\) is cartesian. To see that \(\bar {f}\) is cartesian, we must construct a unique factorization as follows:
Because \(\bar {f};\bar {g}\) is cartesian, we can factor \(\bar {h};\bar {g} = i;\bar {f};\bar {g}\) for a unique \({\bar {u}}\xrightarrow [m]{i}{\bar {x}}\). On the other hand, because \(\bar {g}\) is cartesian, there is a unique \({\bar {u}}\xrightarrow [m;f]{j}{\bar {y}}\) for which \(\bar {h};\bar {g} = j;\bar {g}\); as both \(\bar {h}\) and \(i;\bar {f}\) satisfy this condition, we conclude \(\bar {h}=i;\bar {f}\). Therefore, there is a unique \({\bar {u}}\xrightarrow [m]{i}{\bar {x}}\) for which \(\bar {h} = i;\bar {f}\), as required.