Lemma 8.3.6. In the situation of Definition 8.3.5 part (2) the maps $can_{ij} : \text{pr}_0^*f_ i^*X \to \text{pr}_1^*f_ j^*X$ are equal to $(\alpha _{\text{pr}_1, f_ j})_ X \circ (\alpha _{\text{pr}_0, f_ i})_ X^{-1}$ where $\alpha _{\cdot , \cdot }$ is as in Categories, Lemma 4.33.7 and where we use the equality $f_ i \circ \text{pr}_0 = f_ j \circ \text{pr}_1$ as maps $U_ i \times _ U U_ j \to U$.

Proof. Omitted. $\square$

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