The Stacks project

Lemma 4.44.2. Let $\mathcal{C}$ be a $(2,1)$-category. Assume given a $2$-commutative diagram

\[ \xymatrix{ S \ar[r]_-{x'} \ar[d]_ j & X' \ar[d]^ p \ar[r]_ q & X \ar[d]^ f \\ T \ar[r]^-{y'} & Y' \ar[r]^ g & Y } \]

in $\mathcal{C}$, where the right square is $2$-cartesian with respect to a $2$-isomorphism $\phi \colon g \circ p \to f \circ q$. Choose a $2$-arrow $\gamma ' : y' \circ j \to p \circ x'$. Set $x = q \circ x'$, $y = g \circ y'$ and let $\gamma : y \circ j \to f \circ x$ be the $2$-isomorphism $\gamma = (\phi \star \text{id}_{x'}) \circ (\text{id}_ g \star \gamma ')$. Then the category $\mathcal{D}'$ of dotted arrows for the left square and $\gamma '$ is equivalent to the category $\mathcal{D}$ of dotted arrows for the outer rectangle and $\gamma $.

Proof. There is a functor $\mathcal{D}' \to \mathcal{D}$ which is $(a, \alpha , \beta ) \mapsto (q \circ a, \text{id}_ q \star \alpha , (\phi \star \text{id}_ a) \circ (\text{id}_ g \star \beta ))$ on objects and $\theta \mapsto \text{id}_ q \star \theta $ on arrows. Checking that this functor $\mathcal{D}' \to \mathcal{D}$ is an equivalence follows formally from the universal property for $2$-fibre products as in Section 4.31. Details omitted. $\square$

Comments (0)

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Your email address will not be published.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 0H1A. Beware of the difference between the letter 'O' and the digit '0'.