Lemma 4.33.10. Let $\mathcal{C}$ be a category. The $(2, 1)$-category of fibred categories over $\mathcal{C}$ has 2-fibre products, and they are described as in Lemma 4.32.3.

**Proof.**
Basically what one has to show here is that given $F : \mathcal{X} \to \mathcal{S}$ and $G : \mathcal{Y} \to \mathcal{S}$ morphisms of fibred categories over $\mathcal{C}$, then the category $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$ described in Lemma 4.32.3 is fibred. Let us show that $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$ has plenty of strongly cartesian morphisms. Namely, suppose we have $(U, x, y, \phi )$ an object of $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$. And suppose $f : V \to U$ is a morphism in $\mathcal{C}$. Choose strongly cartesian morphisms $a : f^*x \to x$ in $\mathcal{X}$ lying over $f$ and $b : f^*y \to y$ in $\mathcal{Y}$ lying over $f$. By assumption $F(a)$ and $G(b)$ are strongly cartesian. Since $\phi : F(x) \to G(y)$ is an isomorphism, by the uniqueness of strongly cartesian morphisms we find a unique isomorphism $f^*\phi : F(f^*x) \to G(f^*y)$ such that $G(b) \circ f^*\phi = \phi \circ F(a)$. In other words $(a, b) : (V, f^*x, f^*y, f^*\phi ) \to (U, x, y, \phi )$ is a morphism in $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$. We omit the verification that this is a strongly cartesian morphism (and that these are in fact the only strongly cartesian morphisms).
$\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like `$\pi$`

). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

## Comments (5)

Comment #59 by Quoc Ho on

Comment #60 by Quoc Ho on

Comment #64 by Johan on

Comment #4997 by SDIGR on

Comment #5237 by Johan on