The Stacks project

Lemma 4.44.3. Let $\mathcal{C}$ be a $(2,1)$-category. Assume given a solid $2$-commutative diagram

\[ \xymatrix{ S \ar[r]_-x \ar[dd]_ j & X \ar[d]^ f \\ & Y \ar[d]^ g \\ T \ar[r]^-z \ar@{..>}[ruu] & Z } \]

in $\mathcal{C}$. Choose a $2$-isomorphism $\gamma \colon z \circ j \to g \circ f \circ x$. Let $\mathcal{D}$ be the category of dotted arrows for the outer rectangle and $\gamma $. Let $\mathcal{D}'$ be the category of dotted arrows for the solid square

\[ \xymatrix{ S \ar[r]_-{f \circ x} \ar[d]_ j & Y \ar[d]^ g \\ T \ar[r]^-z \ar@{..>}[ru] & Z } \]

and $\gamma $. Then $\mathcal{D}$ is equivalent to a category $\mathcal{D}''$ which has the following property: there is a functor $\mathcal{D}'' \to \mathcal{D}'$ which turns $\mathcal{D}''$ into a category fibred in groupoids over $\mathcal{D}'$ and whose fibre categories are isomorphic to categories of dotted arrows for certain solid squares of the form

\[ \xymatrix{ S \ar[r]_-x \ar[d]_ j & X \ar[d]^ f \\ T \ar[r]^-y \ar@{..>}[ru] & Y } \]

and some choices of $2$-isomorphism $y \circ j \to f \circ x$.

Proof. Construct the category $\mathcal{D}''$ whose objects are tuples $(a,\alpha ,\beta ,b,\eta )$ where $(a,\alpha ,\beta )$ is an object of $\mathcal{D}$ and $b \colon T \rightarrow Y$ is a $1$-morphism and $\eta \colon b \rightarrow f \circ a$ is a $2$-isomorphism. Morphisms $(a,\alpha ,\beta ,b,\eta ) \rightarrow (a',\alpha ',\beta ',b',\eta ')$ in $\mathcal{D}''$ are pairs $(\theta _1,\theta _2)$, where $\theta _1 \colon a \rightarrow a'$ defines an arrow $(a, \alpha , \beta ) \rightarrow (a', \alpha ', \beta ')$ in $\mathcal{D}$ and $\theta _2 \colon b \rightarrow b'$ is a $2$-isomorphism with the compatibility condition $\eta ' \circ \theta _2 = (\text{id}_ f \star \theta _1) \circ \eta $.

There is a functor $\mathcal{D}'' \rightarrow \mathcal{D}'$ which is $(a, \alpha , \beta , b, \eta ) \mapsto (b, (\text{id}_ f \star \alpha ) \circ (\eta \star \text{id}_ j), (\text{id}_ g \star \eta ^{-1}) \circ \beta )$ on objects and $(\theta _1,\theta _2) \mapsto \theta _2$ on arrows. Then $\mathcal{D}'' \rightarrow \mathcal{D}'$ is fibred in groupoids.

If $(y, \delta , \epsilon )$ is an object of $\mathcal{D}'$, write $\mathcal{D}_{y,\delta }$ for the category of dotted arrows for the last displayed diagram with $y \circ j \rightarrow f \circ x$ given by $\delta $. There is a functor $\mathcal{D}_{y,\delta } \rightarrow \mathcal{D}''$ given by $(a, \alpha , \eta ) \mapsto (a, \alpha , (\text{id}_ g \star \eta ) \circ \epsilon , y, \eta )$ on objects and $\theta \mapsto (\theta , \text{id}_ y)$ on arrows. This exhibits an isomorphism from $\mathcal{D}_{y,\delta }$ to the fibre category of $\mathcal{D}'' \rightarrow \mathcal{D}'$ over $(y,\delta ,\epsilon )$.

There is also a functor $\mathcal{D} \rightarrow \mathcal{D}''$ which is $(a,\alpha ,\beta ) \mapsto (a,\alpha ,\beta ,f \circ a, \text{id}_{f \circ a})$ on objects and $\theta \mapsto (\theta , \text{id}_ f \star \theta )$ on arrows. This functor is fully faithful and essentially surjective, hence an equivalence. Details omitted. $\square$

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