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9.10.5 Classical Grothendieck Construction for Lax Functors

Definition 9.10.5.1. Let $\operatorname{\mathcal{C}}$ be a category and let $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ be a lax functor of $2$-categories (Definition 2.2.4.5). We define a category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ as follows:

  • The objects of $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ are pairs $(C, X)$, where $C$ is an object of $\operatorname{\mathcal{C}}$ and $X$ is an object of the category $\mathscr {F}(C)$.

  • Let $(C,X)$ and $(D,Y)$ be objects of $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. Then a morphism from $(C,X)$ to $(D,Y)$ in the category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ is a pair $(f,u)$, where $f: C \rightarrow D$ is a morphism in the category $\operatorname{\mathcal{C}}$ and $u: X \rightarrow \mathscr {F}(f)(Y)$ is a morphism in the category $\mathscr {F}(C)$.

  • Let $(f,u): (C,X) \rightarrow (D,Y)$ and $(g,v): (D,Y) \rightarrow (E,Z)$ be morphisms in the category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. Then the composition $(g,v) \circ (f,u)$ is the pair $(g \circ f, w)$, where $w: X \rightarrow \mathscr {F}( g \circ f)(Z)$ is the morphism of $\mathscr {F}(C)$ given by the composition

    \[ X \xrightarrow { u } \mathscr {F}(f)(Y) \xrightarrow { \mathscr {F}(f)(v) } (\mathscr {F}(f) \circ \mathscr {F}(g))(Z) \xrightarrow { \mu _{f,g}(Z) } \mathscr {F}(g \circ f)(Z), \]

    where $\mu _{f,g}: \mathscr {F}(f) \circ \mathscr {F}(g) \rightarrow \mathscr {F}( g \circ f)$ denotes the composition constraint for the lax functor $\mathscr {F}$.

We will refer to $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ as the Grothendieck construction on the lax functor $\mathscr {F}$.

Proposition 9.10.5.2. Let $\operatorname{\mathcal{C}}$ be a category and let $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ be a lax functor of $2$-categories. Then the Grothendieck construction $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ is well-defined: that is, the composition law described in Construction 9.5.0.52 is unital and associative.

Proof. We proceed as in the proof of Proposition 5.6.1.3, with some minor variations in notation. Let $(D,Y)$ be an object of $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. We let $\operatorname{id}_{ (D,Y)}$ denote the morphism from $(D,Y)$ to itself given by the pair $(\operatorname{id}_{D}, \epsilon _{D}(Y) )$, where $\epsilon _{D}: \operatorname{id}_{ \mathscr {F}(D) } \rightarrow \mathscr {F}( \operatorname{id}_{D} )$ is the identity constraint for the lax functor $\mathscr {F}$. We first show that $\operatorname{id}_{(D,Y)}$ is a (two-sided) unit for the composition law on $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. We consider two cases:

  • Let $(C,X)$ be another object of $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ and let $(f,u): (C,X) \rightarrow (D,Y)$ be a morphism in $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. We wish to show that the composition $\operatorname{id}_{(D,Y)} \circ (f,u)$ is equal to $(f,u)$ (as a morphism from $(C,X)$ to $(D,Y)$). Unwinding the definitions, this is equivalent to the assertion that the morphism $u: X \rightarrow \mathscr {F}(f)(Y)$ is equal to the composition

    \[ X \xrightarrow {u} \mathscr {F}(f)(Y) \xrightarrow { \mathscr {F}(f)( \epsilon _ D(Y) ) } (\mathscr {F}(f) \circ \mathscr {F}(\operatorname{id}_ D))(Y) \xrightarrow { \mu _{f, \operatorname{id}_{D} }(Y) } \mathscr {F}(f)(Y). \]

    This follows from the observation that the composite transformation

    \[ \mathscr {F}(f) \xrightarrow { \epsilon _ D } \mathscr {F}(f) \circ \operatorname{id}_ D \xrightarrow { \mu _{f, \operatorname{id}_{D} } } \mathscr {F}(f) \]

    is the identity, which follows from axiom $(b)$ of Definition 2.2.4.5.

  • Let $(E,Z)$ be another object of $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$, and let $(g,v): (D,Y) \rightarrow (E,Z)$ be a morphism in $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. We wish to show that the composition $(g,v) \circ \operatorname{id}_{(D,Y)}$ is equal to $(g,v)$ (as a morphism from $(D,Y)$ to $(E,Z)$). Unwinding the definitions, this is equivalent to the assertion that the morphism $v: Y \rightarrow \mathscr {F}(g)(Z)$ is given by the upper cycle in the diagram

    \[ \xymatrix@R =50pt@C=50pt{ Y \ar [r]^-{ \epsilon _{D}(Y) } \ar [d]^-{v} & \mathscr {F}(\operatorname{id}_ D)(Y) \ar [d]^-{ \mathscr {F}(\operatorname{id}_ D)(v) } & \\ \mathscr {F}(v)(Z) \ar [r]^-{\epsilon _ D( \mathscr {F}(v)(Z) )} & (\mathscr {F}(\operatorname{id}_ D) \circ \mathscr {F}(v))(Z) \ar [r]^-{ \mu _{\operatorname{id}_{D}, v)}(Z)} & \mathscr {F}(v)(Z). } \]

    Since $\epsilon _{D}$ is a natural transformation, this diagram commutes. It will therefore suffice to show that the lower composition is equal to $v$, which follows from axiom $(a)$ of Definition 2.2.4.5.

We now show that composition of morphisms in $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ is associative. Suppose we are given a composable sequence

\[ (B,W) \xrightarrow { (e,t) } (C,X) \xrightarrow { (f,u) } (D,Y) \xrightarrow { (g,v) } (E,Z) \]

of morphisms of $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$. Unwinding the definitions, we obtain equalities

\[ (g,v) \circ ( (f,u) \circ (e,t) ) = (g \circ f \circ e, w \circ \mathscr {F}(e)(u) \circ t) \]

\[ ( (g,v) \circ (f,u) ) \circ (e,t) ) = (g \circ f \circ e, w' \circ \mathscr {F}(e)(u) \circ t), \]

where $w,w': (\mathscr {F}(e) \circ \mathscr {F}(f))(Y) \rightarrow \mathscr {F}(g \circ f \circ e)(Z)$ are the morphisms in the category $\mathscr {F}(B)$ given by clockwise and counterclockwise composition in the diagram

\[ \xymatrix@R =50pt@C=50pt{ (\mathscr {F}(e) \circ \mathscr {F}(f))(Y) \ar [r]^-{ \mu _{e,f}(Y) } \ar [d]^-{ (\mathscr {F}(e) \circ \mathscr {F}(f))(v) } & \mathscr {F}(f \circ e)(Y) \ar [d]^-{ \mathscr {F}(f \circ e)(v)} \\ (\mathscr {F}(e) \circ \mathscr {F}(f) \circ \mathscr {F}(g))(Z) \ar [d]^-{\mathscr {F}(e)(\mu _{g,f}(Z)) } \ar [r]^-{ \mu _{e,f}( \mathscr {F}(g)(Z) )} & (\mathscr {F}(f \circ e) \circ \mathscr {F}(g) )(Z) \ar [d]^-{\mu _{g,f \circ e}(Z) } \\ (\mathscr {F}(e) \circ \mathscr {F}(g \circ f)) \ar [r]^-{ \mu _{e,g\circ f}(Z) } & \mathscr {F}( g \circ f \circ e)(Z), } \]

respectively. It will therefore suffice to show that this diagram commutes. For the upper square, this follows from the naturality of the composition constraint $\mu _{e,f}$. For the lower square, it follows from axiom $(c)$ of Definition 2.2.4.5. $\square$

Example 9.10.5.3 (Kleisli Categories). Let $\operatorname{\mathcal{C}}= [0]$ denote the category having a single object and a single morphism. By virtue of Example 2.2.4.10, we can identify lax functors $\mathscr {F}: \operatorname{\mathcal{C}}^{\operatorname{op}} \rightarrow \mathbf{Cat}$ with pairs $(\operatorname{\mathcal{E}}, T)$, where $\operatorname{\mathcal{E}}$ is a category and $T$ is a monad on $\operatorname{\mathcal{E}}$: that is, a functor from $\operatorname{\mathcal{E}}$ to itself which is equipped with unit and multiplication maps

\[ \epsilon : \operatorname{id}_{\operatorname{\mathcal{E}}} \rightarrow T \quad \quad \mu : T \circ T \rightarrow T \]

which endow $T$ with the structure of an associative algebra object of the category of endofunctors $\operatorname{Fun}(\operatorname{\mathcal{E}}, \operatorname{\mathcal{E}})$. In this case, the Grothendieck construction $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ can be identified with the Kleisli category of the monad $T$. Concretely, it can be described as follows:

  • The objects of $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ are the objects of the category $\operatorname{\mathcal{E}}$.

  • For every pair objects $X,Y \in \operatorname{\mathcal{E}}$, we have $\operatorname{Hom}_{ \int ^{\operatorname{\mathcal{C}}} \mathscr {F} }( X, Y) = \operatorname{Hom}_{\operatorname{\mathcal{E}}}( X, T(Y) )$.

  • Let $u: X \rightarrow Y$ and $v: Y \rightarrow Z$ be morphisms in the category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$, corresponding to a pair of morphisms $u': X \rightarrow T(Y)$ and $v': Y \rightarrow T(Z)$ in the category $\operatorname{\mathcal{E}}$. Then the composition $v \circ u$ in $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ corresponds to the composite map

    \[ X \xrightarrow {u} T(Y) \xrightarrow { T(v) } (T \circ T)(Z) \xrightarrow { \mu (Z) } T(Z) \]

    in the category $\operatorname{\mathcal{E}}$.

Remark 9.10.5.4. In the situation of Example 9.10.5.3, the Kleisli category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$ is equipped with a comparison functor $H: \operatorname{\mathcal{E}}\rightarrow \int ^{\operatorname{\mathcal{C}}} \mathscr {F}$, which carries each object of $\operatorname{\mathcal{E}}$ to itself and each morphism $u: X \rightarrow Y$ in $\operatorname{\mathcal{E}}$ to the composite morphism $X \xrightarrow {u} Y \xrightarrow { \epsilon (Y) } T(Y)$ (regarded as a morphism from $X$ to $Y$ in the category $\int ^{\operatorname{\mathcal{C}}} \mathscr {F}$). The following conditions are equivalent:

  • The functor $H$ is an equivalence of categories.

  • The functor $H$ is an isomorphism of categories.

  • The unit map $\epsilon : \operatorname{id}_{\operatorname{\mathcal{E}}} \rightarrow T$ is an isomorphism of functors.