Processing math: 100%

The Stacks project

Lemma 7.5.1. Let u : \mathcal{C} \to \mathcal{D} be a functor between categories. Suppose that \mathcal{C} has fibre products and equalizers, and that u commutes with them. Then the categories (\mathcal{I}_ V)^{opp} satisfy the hypotheses of Categories, Lemma 4.19.8.

Proof. There are two conditions to check.

First, suppose we are given three objects \phi : V \to u(U), \phi ' : V \to u(U'), and \phi '' : V \to u(U'') and morphisms a : U' \to U, b : U'' \to U such that u(a) \circ \phi ' = \phi and u(b) \circ \phi '' = \phi . We have to show there exists another object \phi ''' : V \to u(U''') and morphisms c : U''' \to U' and d : U''' \to U'' such that u(c) \circ \phi ''' = \phi ', u(d) \circ \phi ''' = \phi '' and a \circ c = b \circ d. We take U''' = U' \times _ U U'' with c and d the projection morphisms. This works as u commutes with fibre products; we omit the verification.

Second, suppose we are given two objects \phi : V \to u(U) and \phi ' : V \to u(U') and morphisms a, b : (U, \phi ) \to (U', \phi '). We have to find a morphism c : (U'', \phi '') \to (U, \phi ) which equalizes a and b. Let c : U'' \to U be the equalizer of a and b in the category \mathcal{C}. As u commutes with equalizers and since u(a) \circ \phi = u(b) \circ \phi = \phi ' we obtain a morphism \phi '' : V \to u(U''). \square


Comments (2)

Comment #1119 by Olaf Schnürer on

Typos in first part of proof: , , , .

There are also:

  • 7 comment(s) on Section 7.5: Functoriality of categories of presheaves

Your email address will not be published.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.