## 4.32 Categories over categories

In this section we have a functor $p : \mathcal{S} \to \mathcal{C}$. We think of $\mathcal{S}$ as being on top and of $\mathcal{C}$ as being at the bottom. To make sure that everybody knows what we are talking about we define the $2$-category of categories over $\mathcal{C}$.

Definition 4.32.1. Let $\mathcal{C}$ be a category. The $2$-category of categories over $\mathcal{C}$ is the $2$-category defined as follows:

1. Its objects will be functors $p : \mathcal{S} \to \mathcal{C}$.

2. Its $1$-morphisms $(\mathcal{S}, p) \to (\mathcal{S}', p')$ will be functors $G : \mathcal{S} \to \mathcal{S}'$ such that $p' \circ G = p$.

3. Its $2$-morphisms $t : G \to H$ for $G, H : (\mathcal{S}, p) \to (\mathcal{S}', p')$ will be morphisms of functors such that $p'(t_ x) = \text{id}_{p(x)}$ for all $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$.

In this situation we will denote

$\mathop{\mathrm{Mor}}\nolimits _{\textit{Cat}/\mathcal{C}}(\mathcal{S}, \mathcal{S}')$

the category of $1$-morphisms between $(\mathcal{S}, p)$ and $(\mathcal{S}', p')$

In this $2$-category we define horizontal and vertical composition exactly as is done for $\textit{Cat}$ in Section 4.28. The axioms of a $2$-category are satisfied for the same reason that the hold in $\textit{Cat}$. To see this one can also use that the axioms hold in $\textit{Cat}$ and verify things such as “vertical composition of $2$-morphisms over $\mathcal{C}$ gives another $2$-morphism over $\mathcal{C}$”. This is clear.

Analogously to the fibre of a map of spaces, we have the notion of a fibre category, and some notions of lifting associated to this situation.

Definition 4.32.2. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S} \to \mathcal{C}$ be a category over $\mathcal{C}$.

1. The fibre category over an object $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ is the category $\mathcal{S}_ U$ with objects

$\mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U) = \{ x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}) : p(x) = U\}$

and morphisms

$\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}_ U}(x, y) = \{ \phi \in \mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(x, y) : p(\phi ) = \text{id}_ U\} .$
2. A lift of an object $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ is an object $x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ such that $p(x) = U$, i.e., $x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. We will also sometime say that $x$ lies over $U$.

3. Similarly, a lift of a morphism $f : V \to U$ in $\mathcal{C}$ is a morphism $\phi : y \to x$ in $\mathcal{S}$ such that $p(\phi ) = f$. We sometimes say that $\phi$ lies over $f$.

There are some observations we could make here. For example if $F : (\mathcal{S}, p) \to (\mathcal{S}', p')$ is a $1$-morphism of categories over $\mathcal{C}$, then $F$ induces functors of fibre categories $F : \mathcal{S}_ U \to \mathcal{S}'_ U$. Similarly for $2$-morphisms.

Here is the obligatory lemma describing the $2$-fibre product in the $(2, 1)$-category of categories over $\mathcal{C}$.

Lemma 4.32.3. Let $\mathcal{C}$ be a category. The $(2, 1)$-category of categories over $\mathcal{C}$ has 2-fibre products. Suppose that $F : \mathcal{X} \to \mathcal{S}$ and $G : \mathcal{Y} \to \mathcal{S}$ are morphisms of categories over $\mathcal{C}$. An explicit 2-fibre product $\mathcal{X} \times _\mathcal {S}\mathcal{Y}$ is given by the following description

1. an object of $\mathcal{X} \times _\mathcal {S} \mathcal{Y}$ is a quadruple $(U, x, y, f)$, where $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, $x\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_ U)$, $y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{Y}_ U)$, and $f : F(x) \to G(y)$ is an isomorphism in $\mathcal{S}_ U$,

2. a morphism $(U, x, y, f) \to (U', x', y', f')$ is given by a pair $(a, b)$, where $a : x \to x'$ is a morphism in $\mathcal{X}$, and $b : y \to y'$ is a morphism in $\mathcal{Y}$ such that

1. $a$ and $b$ induce the same morphism $U \to U'$, and

2. the diagram

$\xymatrix{ F(x) \ar[r]^ f \ar[d]^{F(a)} & G(y) \ar[d]^{G(b)} \\ F(x') \ar[r]^{f'} & G(y') }$

is commutative.

The functors $p : \mathcal{X} \times _\mathcal {S}\mathcal{Y} \to \mathcal{X}$ and $q : \mathcal{X} \times _\mathcal {S}\mathcal{Y} \to \mathcal{Y}$ are the forgetful functors in this case. The transformation $\psi : F \circ p \to G \circ q$ is given on the object $\xi = (U, x, y, f)$ by $\psi _\xi = f : F(p(\xi )) = F(x) \to G(y) = G(q(\xi ))$.

Proof. Let us check the universal property: let $p_\mathcal {W} : \mathcal{W}\to \mathcal{C}$ be a category over $\mathcal{C}$, let $X : \mathcal{W} \to \mathcal{X}$ and $Y : \mathcal{W} \to \mathcal{Y}$ be functors over $\mathcal{C}$, and let $t : F \circ X \to G \circ Y$ be an isomorphism of functors over $\mathcal{C}$. The desired functor $\gamma : \mathcal{W} \to \mathcal{X} \times _\mathcal {S} \mathcal{Y}$ is given by $W \mapsto (p_\mathcal {W}(W), X(W), Y(W), t_ W)$. Details omitted; compare with Lemma 4.31.4. $\square$

Example 4.32.4. The constructions of $2$-fibre products of categories over categories given in Lemma 4.32.3 and of categories in Lemma 4.31.4 (as in Example 4.31.3) produce non-equivalent outputs in general. Namely, let $\mathcal{S}$ be the groupoid category with one object and two arrows, and let $\mathcal{X}$ be the discrete category with one object. Taking the $2$-fibre product $\mathcal{X} \times _\mathcal {S} \mathcal{X}$ as categories yields the discrete category with two objects. However, if we view all of these as categories over $\mathcal{S}$, the $2$-fiber product $\mathcal{X} \times _\mathcal {S} \mathcal{X}$ as categories over $\mathcal{S}$ is the discrete category with one object. The difference is that (in the notation of Lemma 4.32.3), we were allowed to choose any comparison isomorphism $f$ in the first situation, but could only choose the identity arrow in the second situation.

Lemma 4.32.5. Let $\mathcal{C}$ be a category. Let $f : \mathcal{X} \to \mathcal{S}$ and $g : \mathcal{Y} \to \mathcal{S}$ be morphisms of categories over $\mathcal{C}$. For any object $U$ of $\mathcal{C}$ we have the following identity of fibre categories

$\left(\mathcal{X} \times _\mathcal {S}\mathcal{Y}\right)_ U = \mathcal{X}_ U \times _{\mathcal{S}_ U} \mathcal{Y}_ U$

Proof. Omitted. $\square$

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