## 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

1. objects are pairs $(x, \alpha )$ where $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ and $\alpha : x \to x$ is an automorphism with $F(\alpha ) = \text{id}$,

2. 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

3. the functor $\mathcal{I}_{\mathcal{S}/\mathcal{S}'} \to \mathcal{C}$ is given by $(x, \alpha ) \mapsto p(x)$.

Then

1. 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

2. $\mathcal{I}_{\mathcal{S}/\mathcal{S}'}$ 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.

1. 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.

2. 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
$$\label{categories-equation-inertia-structure-map} \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \longrightarrow \mathcal{S} \quad \text{and}\quad \mathcal{I}_\mathcal {S} \longrightarrow \mathcal{S}$$

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
$$\label{categories-equation-neutral-section} e : \mathcal{S} \to \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \quad \text{and}\quad e : \mathcal{S} \to \mathcal{I}_\mathcal {S}$$

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
$$\label{categories-equation-functorial} \mathcal{I}_{\mathcal{S}_1/\mathcal{S}'_1} \longrightarrow \mathcal{I}_{\mathcal{S}_2/\mathcal{S}'_2} \quad \text{and}\quad \mathcal{I}_{\mathcal{S}_1} \longrightarrow \mathcal{I}_{\mathcal{S}_2}$$

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
$$\label{categories-equation-comparison} \mathcal{I}_{\mathcal{S}/\mathcal{S}'} \longrightarrow \mathcal{I}_\mathcal {S}$$

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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