4.34 Inertia
Given fibred categories p : \mathcal{S} \to \mathcal{C} and p' : \mathcal{S}' \to \mathcal{C} over a category \mathcal{C} and a 1-morphism F : \mathcal{S} \to \mathcal{S}' we have the diagonal morphism
\Delta = \Delta _{\mathcal{S}/\mathcal{S}'} : \mathcal{S} \longrightarrow \mathcal{S} \times _{\mathcal{S}'} \mathcal{S}
in the (2, 1)-category of fibred categories over \mathcal{C}.
Lemma 4.34.1. Let \mathcal{C} be a category. Let p : \mathcal{S} \to \mathcal{C} and p' : \mathcal{S}' \to \mathcal{C} be fibred categories. Let F : \mathcal{S} \to \mathcal{S}' be a 1-morphism of fibred categories over \mathcal{C}. Consider the category \mathcal{I}_{\mathcal{S}/\mathcal{S}'} over \mathcal{C} whose
objects are pairs (x, \alpha ) where x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}) and \alpha : x \to x is an automorphism with F(\alpha ) = \text{id},
morphisms (x, \alpha ) \to (y, \beta ) are given by morphisms \phi : x \to y such that
\xymatrix{ x\ar[r]_\phi \ar[d]_\alpha & y\ar[d]^{\beta } \\ x\ar[r]^\phi & y \\ }
commutes, and
the functor \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \to \mathcal{C} is given by (x, \alpha ) \mapsto p(x).
Then
there is an equivalence
\mathcal{I}_{\mathcal{S}/\mathcal{S}'} \longrightarrow \mathcal{S} \times _{\Delta , (\mathcal{S} \times _{\mathcal{S}'} \mathcal{S}), \Delta } \mathcal{S}
in the (2, 1)-category of categories over \mathcal{C}, and
\mathcal{I}_{\mathcal{S}/\mathcal{S}'} is a fibred category over \mathcal{C}.
Proof.
Note that (2) follows from (1) by Lemmas 4.33.10 and 4.33.8. Thus it suffices to prove (1). We will use without further mention the construction of the 2-fibre product from Lemma 4.33.10. In particular an object of \mathcal{S} \times _{\Delta , (\mathcal{S} \times _{\mathcal{S}'} \mathcal{S}), \Delta } \mathcal{S} is a triple (x, y, (\iota , \kappa )) where x and y are objects of \mathcal{S}, and (\iota , \kappa ) : (x, x, \text{id}_{F(x)}) \to (y, y, \text{id}_{F(y)}) is an isomorphism in \mathcal{S} \times _{\mathcal{S}'} \mathcal{S}. This just means that \iota , \kappa : x \to y are isomorphisms and that F(\iota ) = F(\kappa ). Consider the functor
I_{\mathcal{S}/\mathcal{S}'} \longrightarrow \mathcal{S} \times _{\Delta , (\mathcal{S} \times _{\mathcal{S}'} \mathcal{S}), \Delta } \mathcal{S}
which to an object (x, \alpha ) of the left hand side assigns the object (x, x, (\alpha , \text{id}_ x)) of the right hand side and to a morphism \phi of the left hand side assigns the morphism (\phi , \phi ) of the right hand side. We claim that a quasi-inverse to that morphism is given by the functor
\mathcal{S} \times _{\Delta , (\mathcal{S} \times _{\mathcal{S}'} \mathcal{S}), \Delta } \mathcal{S} \longrightarrow I_{\mathcal{S}/\mathcal{S}'}
which to an object (x, y, (\iota , \kappa )) of the left hand side assigns the object (x, \kappa ^{-1} \circ \iota ) of the right hand side and to a morphism (\phi , \phi ') : (x, y, (\iota , \kappa )) \to (z, w, (\lambda , \mu )) of the left hand side assigns the morphism \phi . Indeed, the endo-functor of I_{\mathcal{S}/\mathcal{S}'} induced by composing the two functors above is the identity on the nose, and the endo-functor induced on \mathcal{S} \times _{\Delta , (\mathcal{S} \times _{\mathcal{S}'} \mathcal{S}), \Delta } \mathcal{S} is isomorphic to the identity via the natural isomorphism
(\text{id}_ x, \kappa ) : (x, x, (\kappa ^{-1} \circ \iota , \text{id}_ x)) \longrightarrow (x, y, (\iota , \kappa )).
Some details omitted.
\square
Definition 4.34.2. Let \mathcal{C} be a category.
Let F : \mathcal{S} \to \mathcal{S}' be a 1-morphism of fibred categories over \mathcal{C}. The relative inertia of \mathcal{S} over \mathcal{S}' is the fibred category \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \to \mathcal{C} of Lemma 4.34.1.
By the inertia fibred category \mathcal{I}_\mathcal {S} of \mathcal{S} we mean \mathcal{I}_\mathcal {S} = \mathcal{I}_{\mathcal{S}/\mathcal{C}}.
Note that there are canonical 1-morphisms
4.34.2.1
\begin{equation} \label{categories-equation-inertia-structure-map} \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \longrightarrow \mathcal{S} \quad \text{and}\quad \mathcal{I}_\mathcal {S} \longrightarrow \mathcal{S} \end{equation}
of fibred categories over \mathcal{C}. In terms of the description of Lemma 4.34.1 these simply map the object (x, \alpha ) to the object x and the morphism \phi : (x, \alpha ) \to (y, \beta ) to the morphism \phi : x \to y. There is also a neutral section
4.34.2.2
\begin{equation} \label{categories-equation-neutral-section} e : \mathcal{S} \to \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \quad \text{and}\quad e : \mathcal{S} \to \mathcal{I}_\mathcal {S} \end{equation}
defined by the rules x \mapsto (x, \text{id}_ x) and (\phi : x \to y) \mapsto \phi . This is a right inverse to (4.34.2.1). Given a 2-commutative square
\xymatrix{ \mathcal{S}_1 \ar[d]_{F_1} \ar[r]_ G & \mathcal{S}_2 \ar[d]^{F_2} \\ \mathcal{S}'_1 \ar[r]^{G'} & \mathcal{S}'_2 }
there are functoriality maps
4.34.2.3
\begin{equation} \label{categories-equation-functorial} \mathcal{I}_{\mathcal{S}_1/\mathcal{S}'_1} \longrightarrow \mathcal{I}_{\mathcal{S}_2/\mathcal{S}'_2} \quad \text{and}\quad \mathcal{I}_{\mathcal{S}_1} \longrightarrow \mathcal{I}_{\mathcal{S}_2} \end{equation}
defined by the rules (x, \alpha ) \mapsto (G(x), G(\alpha )) and \phi \mapsto G(\phi ). In particular there is always a comparison map
4.34.2.4
\begin{equation} \label{categories-equation-comparison} \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \longrightarrow \mathcal{I}_\mathcal {S} \end{equation}
and all the maps above are compatible with this.
Lemma 4.34.3. Let F : \mathcal{S} \to \mathcal{S}' be a 1-morphism of categories fibred over a category \mathcal{C}. Then the diagram
\xymatrix{ \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \ar[d]_{F \circ (042H)} \ar[rr]_{(04Z5)} & & \mathcal{I}_\mathcal {S} \ar[d]^{(04Z4)} \\ \mathcal{S}' \ar[rr]^ e & & \mathcal{I}_{\mathcal{S}'} }
is a 2-fibre product.
Proof.
Omitted.
\square
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