The Stacks project

Lemma 4.31.11. Let

\[ \xymatrix{ \mathcal{C}_3 \ar[r] \ar[d] & \mathcal{S} \ar[d]^\Delta \\ \mathcal{C}_1 \times \mathcal{C}_2 \ar[r]^{G_1 \times G_2} & \mathcal{S} \times \mathcal{S} } \]

be a $2$-fibre product of categories. Then there is a canonical isomorphism $\mathcal{C}_3 \cong \mathcal{C}_1 \times _{G_1, \mathcal{S}, G_2} \mathcal{C}_2$.

Proof. We may assume that $\mathcal{C}_3$ is the category $(\mathcal{C}_1 \times \mathcal{C}_2)\times _{\mathcal{S} \times \mathcal{S}} \mathcal{S}$ constructed in Example 4.31.3. Hence an object is a triple $((X_1, X_2), S, \phi )$ where $\phi = (\phi _1, \phi _2) : (G_1(X_1), G_2(X_2)) \to (S, S)$ is an isomorphism. Thus we can associate to this the triple $(X_1, X_2, \phi _2^{-1} \circ \phi _1)$. Conversely, if $(X_1, X_2, \psi )$ is an object of $\mathcal{C}_1 \times _{G_1, \mathcal{S}, G_2} \mathcal{C}_2$, then we can associate to this the triple $((X_1, X_2), G_2(X_2), (\psi , \text{id}_{G_2(X_2)}))$. We claim these constructions given mutually inverse functors. We omit describing how to deal with morphisms and showing they are mutually inverse. $\square$

Comments (6)

Comment #7690 by Anonymous on

Is it clear that this lemma (and other such lemmas in this section) carry over to the -category of categories over categories, as in the next section (Section 4.32)? Or is there an implicit "by the same proof" assertion when the result (for categories over categories) is invoked in Lemma 4.42.6?

Comment #7691 by on

I think so. It seems to me the only thing to check is that the -fibre product constructed in Lemma 4.32.3 is canonically equivalent to the one constructed in Lemma 4.31.4. Or can you see other problems? Presumably there is a meta version of each of the lemmas here for any category with -fibre products (and certain products).

Comment #7692 by Anonymous on

Are these fiber products actually equivalent? Let be the groupoid category with one object and two arrows, and let be the discrete category with one object. Taking the -fibre product as categories yields the discrete category with two objects. However, if we view all of these as categories over , the -fiber product as categories over is the discrete category with one object.

The difference is that (in the notation of Lemma 4.32.3), we were allowed to choose any comparison isomorphism in the first situation, but could only choose the identity arrow in the second situation. I think this reflects the fact that -morphisms in the category of categories over are a little different.

Also agreed that this is probably some general fact about -fibre products...

Comment #7693 by on

Agreed: argh! Nice simple example! I intend to add this in the future. Thanks!

Interestingly, in your example the category is not fibred in groupoids over , right? So, maybe what is true: -fibre products in the -category of categories fibred in groupoids "agree" with the simlpe one in Example 4.31.3? Then we could add that as a lemma and use it whenever we use the lemmas in this section for categories fibred in groupoids. What do you think?

Obviously, I didn't put enough time into the previous comment and I am not putting a lot of time into this one either. So thank you for helping Anonymous!

Comment #7694 by Anonymous on

No problem! Yes I think something for categories fibred in groupoids would be good. Maybe what you're saying is true, although it still seems to me that the 2-morphisms are a bit different. I tried for a bit and didn't find a counterexample.

There are also:

  • 6 comment(s) on Section 4.31: 2-fibre products

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.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 02XE. Beware of the difference between the letter 'O' and the digit '0'.