The Stacks project

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
\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. $\square$

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 04Z2. Beware of the difference between the letter 'O' and the digit '0'.



View Section 4.34 as pdf