Definition 8.3.1. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a fibred category. Make a choice of pullbacks as in Categories, Definition 4.33.6. Let $\mathcal{U} = \{ f_ i : U_ i \to U\} _{i \in I}$ be a family of morphisms of $\mathcal{C}$. Assume all the fibre products $U_ i \times _ U U_ j$, and $U_ i \times _ U U_ j \times _ U U_ k$ exist.

1. A descent datum $(X_ i, \varphi _{ij})$ in $\mathcal{S}$ relative to the family $\{ f_ i : U_ i \to U\}$ is given by an object $X_ i$ of $\mathcal{S}_{U_ i}$ for each $i \in I$, an isomorphism $\varphi _{ij} : \text{pr}_0^*X_ i \to \text{pr}_1^*X_ j$ in $\mathcal{S}_{U_ i \times _ U U_ j}$ for each pair $(i, j) \in I^2$ such that for every triple of indices $(i, j, k) \in I^3$ the diagram

$\xymatrix{ \text{pr}_0^*X_ i \ar[rd]_{\text{pr}_{01}^*\varphi _{ij}} \ar[rr]_{\text{pr}_{02}^*\varphi _{ik}} & & \text{pr}_2^*X_ k \\ & \text{pr}_1^*X_ j \ar[ru]_{\text{pr}_{12}^*\varphi _{jk}} & }$

in the category $\mathcal{S}_{U_ i \times _ U U_ j \times _ U U_ k}$ commutes. This is called the cocycle condition.

2. A morphism $\psi : (X_ i, \varphi _{ij}) \to (X'_ i, \varphi '_{ij})$ of descent data is given by a family $\psi = (\psi _ i)_{i\in I}$ of morphisms $\psi _ i : X_ i \to X'_ i$ in $\mathcal{S}_{U_ i}$ such that all the diagrams

$\xymatrix{ \text{pr}_0^*X_ i \ar[r]_{\varphi _{ij}} \ar[d]_{\text{pr}_0^*\psi _ i} & \text{pr}_1^*X_ j \ar[d]^{\text{pr}_1^*\psi _ j} \\ \text{pr}_0^*X'_ i \ar[r]^{\varphi '_{ij}} & \text{pr}_1^*X'_ j \\ }$

in the categories $\mathcal{S}_{U_ i \times _ U U_ j}$ commute.

3. The category of descent data relative to $\mathcal{U}$ is denoted $DD(\mathcal{U})$.

Comment #7454 by Hao Peng on

I am confused by the cocycle condition. My worry is that the composition doesn't make sence: While $pr^\star_{01}\phi_{ij}$ is a morphism from $(U_{ijk}\to U_{ij})^\star(U_{ij}\to U_i)^\star X_i$ to $(U_{ijk}\to U_{ij})^\star(U_{ij}\to U_j)^\star X_j$, $pr^\star_{12}\phi_{jk}$ is a morphism from $(U_{ijk}\to U_{jk})^\star(U_{jk}\to U_j)^\star X_j$ to $(U_{ijk}\to U_{jk})^\star(U_{jk}\to U_k)^\star X_k$. Thus it is preassumed that $(U_{ijk}\to U_{ij})^\star (U_{ij}\to U_j)^\star X_j=(U_{ijk}\to U_{jk})^\star (U_{jk}\to U_j)^\star X_j$, which is not in general true. Is this a mistake or we insect a natural isomorphism between them, or we can choose the cleavage such that this is true for any fiber products?

Comment #7606 by on

Dear Hao Peng, a computer couldn't parse this definition. However, we assume the human reader has read Section 8.2 where it is explained that we use the morphisms $\alpha_{g, f}$ from Lemma 4.33.7 to identify these things. I do agree the text could be improved somewhat and I am happy to get explicit suggestions.

Comment #8881 by Siyuan Zheng on

$\varphi_{ij}$ is defined to be a map from $\mathrm{pr}_0^\star X_i$ to $\mathrm{pr}_1^\star X_j$. However, in the diagram, the map from $\mathrm{pr}_0^\star X_i$ to $\mathrm{pr}_{1}^\star X_j$ is labeled as $\mathrm{pr}_{ij}^\star\varphi_{ij}$.

Comment #8882 by Siyuan Zheng on

Sorry, I realized the map in the diagram comes from the triple fiber product which is different from the map coming from the double fiber product.

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• 4 comment(s) on Section 8.3: Descent data in fibred categories

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