Lemma 4.32.5. Let $\mathcal{C}$ be a category. Let $f : \mathcal{X} \to \mathcal{S}$ and $g : \mathcal{Y} \to \mathcal{S}$ be morphisms of categories over $\mathcal{C}$. For any object $U$ of $\mathcal{C}$ we have the following identity of fibre categories

$\left(\mathcal{X} \times _\mathcal {S}\mathcal{Y}\right)_ U = \mathcal{X}_ U \times _{\mathcal{S}_ U} \mathcal{Y}_ U$

Proof. Omitted. $\square$

## Comments (2)

Comment #7143 by Hao Peng on

I think this is an equivalence of categories rather than an identity, because the final objects in the $(2, 1)$-categories of commtative diagrams is only unique up to "homotopy", which means their fibre categories is unique up to equivalence.

Comment #7295 by on

Yes we shouldn't use an equal sign and we shouldn't say "identity" in the statement of the lemma. This needs to be fixed eventually as well as the two similar problems pointed out by Manuel Hoff in comment #5394 on Section 4.31.

There are also:

• 2 comment(s) on Section 4.32: Categories over categories

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