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. $\square$

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).