Subsection 2.2.6 (0097): Isomorphisms of Bicategories

2.2.6 Isomorphisms of Bicategories

We now study isomorphisms between bicategories.

Definition 2.2.6.1. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be bicategories. We will say that a functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is an isomorphism if it is an isomorphism in the category $\operatorname{2Cat}$ of Definition 2.2.5.5. That is, $F$ is an isomorphism if there exists a functor $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ such that $GF = \operatorname{id}_{\operatorname{\mathcal{C}}}$ and $FG = \operatorname{id}_{\operatorname{\mathcal{C}}}$. We say that bicategories $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are isomorphic if there exists an isomorphism from $\operatorname{\mathcal{C}}$ to $\operatorname{\mathcal{D}}$.

Remark 2.2.6.2. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be an isomorphism of bicategories, and let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be the inverse isomorphism. Then:

The functor $F$ is strictly unital if and only if $G$ is strictly unital. In this case, we say that $F$ is a strictly unital isomorphism.
The functor $F$ is strict if and only if $G$ is strict. In this case, we say that $F$ is a strict isomorphism.

We say that bicategories $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are strictly isomorphic if there is a strict isomorphism from $\operatorname{\mathcal{C}}$ to $\operatorname{\mathcal{D}}$.

Warning 2.2.6.3. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be bicategories which are strictly isomorphic. Then $\operatorname{\mathcal{C}}$ is strict if and only if $\operatorname{\mathcal{D}}$ is strict. If we assume only that $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ are isomorphic (rather than strictly isomorphic), then we cannot draw the same conclusion. In other words, the condition that a bicategory $\operatorname{\mathcal{C}}$ is strict is invariant under strict isomorphism, but not under isomorphism.

Warning 2.2.6.4. The notions of isomorphism and strict isomorphism of bicategory are somewhat artificial. As in classical category theory, there is notion of equivalence of bicategory (Definition ) which is more general than isomorphism and more appropriate for describing articulating the idea that bicategories are “the same.”

Remark 2.2.6.5. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of bicategories. Then $F$ is an isomorphism (in the sense of Definition 2.2.6.1) if and only if it satisfies the following conditions:

The functor $F$ induces a bijection from the set of objects of $\operatorname{\mathcal{C}}$ to the set of objects of $\operatorname{\mathcal{D}}$.
For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the functor $F$ induces an isomorphism of categories $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( F(X), F(Y) )$.

One might be tempted to consider a more liberal version of Definition 2.2.6.1 working with lax functors rather than functors. However, the resulting notion of isomorphism turns out to be the same.

Proposition 2.2.6.6. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be bicategories, and let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a lax functor which is an isomorphism in the category $\operatorname{2Cat}_{\operatorname{Lax}}$. Then $F$ is a functor.

Proof. We will show that, for every pair of composable $1$-morphisms $X \xrightarrow {f} Y \xrightarrow {g} Z$ in the bicategory $\operatorname{\mathcal{C}}$, the composition constraint $\mu ^{F}_{g,f}: F(g) \circ F(f) \Rightarrow F(g \circ f)$ is an isomorphism (in the ordinary category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( F(X), F(Z) )$); the analogous statement for the identity constraints $\epsilon _{X}^{F}: \operatorname{id}_{F(X)} \Rightarrow F(\operatorname{id}_ X)$ follows by a similar (but easier) argument.

Let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be a lax functor which is an inverse of $F$ in the category $\operatorname{2Cat}_{\operatorname{Lax}}$. For any pair of composable $1$-morphisms $X' \xrightarrow {f'} Y' \xrightarrow {g'} Z'$ in the bicategory $\operatorname{\mathcal{D}}$, the composition constraint $\mu ^{F \circ G}_{g',f'}$ for the lax functor $F \circ G$ is given by the vertical composition

\[ (F \circ G)(g') \circ (F \circ G)(f') \xRightarrow { \mu ^{F}_{G(g'),G(f')} } F( G(g') \circ G(f') ) \xRightarrow {F(\mu ^{G}_{g',f'} )} (F \circ G)(g' \circ f'). \]

Since $F \circ G$ coincides with $\operatorname{id}_{\operatorname{\mathcal{D}}}$ as a lax functor, this composition is the identity $2$-morphism from $g' \circ f'$ to itself. In particular, we see that $F( \mu ^{G}_{g',f'} )$ has a right inverse in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( X', Z' )$. It follows that $\mu ^{G}_{g',f'} = G( F(\mu ^{G}_{g',f'} ) )$ has a right inverse in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}( G(X'), G(Z') )$.

Applying the same argument with the roles of $F$ and $G$ reversed, we see that the composition constraint $\mu ^{G \circ F}_{g,f} = \operatorname{id}_{g \circ f}$ factors as a vertical composition

\[ (G \circ F)(g) \circ (G \circ F)(f) \xRightarrow { \mu ^{G}_{F(g),F(f)} } G( F(g) \circ F(f) ) \xRightarrow {G(\mu ^{F}_{g,f} )} (G \circ F)(g \circ f). \]

In particular, this shows that $\mu ^{G}_{F(g), F(f)}$ has a left inverse (in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Z)$). Applying the preceding argument in the case $g' = F(g)$ and $f' = F(f)$, we see that $\mu ^{G}_{F(g), F(f)}$ also has a right inverse. It follows that $\mu ^{G}_{F(g), F(f)}$ is an isomorphism in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Z)$. Since $G( \mu ^{F}_{g,f} )$ is a left inverse of $\mu ^{G}_{F(g), F(f)}$, it must also be an isomorphism. It follows that $F( G( \mu ^{F}_{g,f} ) ) = \mu ^{F}_{g,f}$ is an isomorphism in the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( F(X), F(Z) )$, as desired. $\square$

We now construct some examples of non-strict isomorphisms of bicategories.

Notation 2.2.6.7. Let $\operatorname{\mathcal{C}}$ be a bicategory. A twisting cochain for $\operatorname{\mathcal{C}}$ is a datum which assigns, to every pair of composable $1$-morphisms $X \xrightarrow {f} Y \xrightarrow {g} Z$, a $1$-morphism $(g \circ ' f): X \rightarrow Z$ and an invertible $2$-morphism $\mu _{g,f}: g \circ ' f \xRightarrow {\sim } g \circ f$. In this case, we will (slightly) abuse notation by identifying the twisting cochain with the collection of $2$-morphisms $\{ \mu _{g,f} \} $.

Construction 2.2.6.8. Let $\operatorname{\mathcal{C}}$ be a bicategory equipped with a twisting cochain

\[ \{ \mu _{g,f} \} = \{ \mu _{g,f}: (g \circ ' f) \Rightarrow (g \circ f) \} . \]

We define a new bicategory $\operatorname{\mathcal{C}}'$ as follows:

The objects of $\operatorname{\mathcal{C}}'$ are the objects of $\operatorname{\mathcal{C}}$.
For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, we define $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}'}(X,Y)$ to be the category $\underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y)$. In particular, we can identify $1$-morphisms of $\operatorname{\mathcal{C}}'$ with $1$-morphisms of $\operatorname{\mathcal{C}}$, $2$-morphisms of $\operatorname{\mathcal{C}}'$ with $2$-morphisms of $\operatorname{\mathcal{C}}$, and the vertical composition of $2$-morphisms in $\operatorname{\mathcal{C}}'$ with the vertical composition of $2$-morphisms in $\operatorname{\mathcal{C}}$.
For every object $X \in \operatorname{\mathcal{C}}$, the identity $1$-morphism from $X$ to itself in the bicategory $\operatorname{\mathcal{C}}'$ is the same as the identity morphism from $X$ to itself in the bicategory $\operatorname{\mathcal{C}}$.
For every triple of objects $X,Y,Z \in \operatorname{\mathcal{C}}$, the composition functor

\[ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}'}( Y,Z) \times \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}'}(X,Y) \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}'}( X, Z) \]

is given on objects by $(g,f) \mapsto g \circ ' f$ and on morphisms by the construction

\[ ( \delta : g \Rightarrow g', \gamma : f \Rightarrow f') \mapsto \mu _{g',f'}^{-1} (\delta \circ \gamma ) \mu _{g,f}. \]
For every object $X \in \operatorname{\mathcal{C}}$, the unit constraint $\upsilon '_{X}: \operatorname{id}_{X} \circ ' \operatorname{id}_{X} \xRightarrow {\sim } \operatorname{id}_{X}$ for the bicategory $\operatorname{\mathcal{C}}'$ is given by the composition

\[ \operatorname{id}_{X} \circ ' \operatorname{id}_{X} \xRightarrow { \mu _{ \operatorname{id}_ X, \operatorname{id}_ X} } \operatorname{id}_ X \circ \operatorname{id}_ X \xRightarrow { \upsilon _ X } \operatorname{id}_ X. \]
For every triple of composable $1$-morphisms $W \xrightarrow {f} X \xrightarrow {g} Y \xrightarrow {h} Z$ of $\operatorname{\mathcal{C}}$, the associativity constraint of $\operatorname{\mathcal{C}}'$ is given by the composition

\begin{eqnarray*} h \circ ' (g \circ ' f) & \xRightarrow { \mu _{h, g \circ ' f} } & h \circ (g \circ ' f) \\ & \xRightarrow { \operatorname{id}_ h \circ \mu _{g,f}} & h \circ (g \circ f) \\ & \xRightarrow { \alpha _{h,g,f} } & (h \circ g) \circ f \\ & \xRightarrow { \mu _{h,g}^{-1} \circ \operatorname{id}_ f} & (h \circ ' g) \circ f \\ & \xRightarrow { \mu _{h \circ ' g, f}^{-1} } & (h \circ ' g) \circ ' f. \end{eqnarray*}

We will refer to $\operatorname{\mathcal{C}}'$ as the twist of $\operatorname{\mathcal{C}}$ with respect to $\{ \mu _{g,f} \} $.

Exercise 2.2.6.9. Let $\operatorname{\mathcal{C}}$ be a bicategory equipped with a twisting cochain $\{ \mu _{g,f} \} $. Show that the bicategory $\operatorname{\mathcal{C}}'$ of Construction 2.2.6.8 is well-defined. Moreover, there is a strictly unital isomorphism of bicategories $\operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}'$ which carries each object, $1$-morphism, and $2$-morphism of $\operatorname{\mathcal{C}}$ to itself, where the composition constraints are given by $\{ \mu _{g,f} \} $.

Exercise 2.2.6.10. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a strictly unital isomorphism of bicategories. Show that there is a unique twisting cochain $\{ \mu _{g,f} \} $ on the bicategory $\operatorname{\mathcal{C}}$ such that $F$ factors as a composition $\operatorname{\mathcal{C}}\xrightarrow {G} \operatorname{\mathcal{C}}' \xrightarrow {H} \operatorname{\mathcal{D}}$, where $G$ is the strictly unital isomorphism of Exercise 2.2.6.9 and $H$ is a strict isomorphism of bicategories. In other words, the notion of twisting cochain (in the sense of Notation 2.2.6.7) measures the difference between strictly unital isomorphisms and strict isomorphisms in the setting of bicategories.

Remark 2.2.6.11. It is possible to consider a generalization of the twisting procedure of Construction 2.2.6.8 in which one modifies not only the composition law for $1$-morphisms of $\operatorname{\mathcal{C}}$, but also the choice of identity $1$-morphisms of $\operatorname{\mathcal{C}}$. Since we will not need this generalization, we leave the details to the reader.

Example 2.2.6.12. Let $G$ be a group with identity element $1 \in G$, let $\Gamma $ be an abelian group on which $G$ acts by automorphisms, let $\alpha : G \times G \times G \rightarrow \Gamma $ be a $3$-cocycle, let $\operatorname{\mathcal{C}}$ be the monoidal category of Example 2.1.3.3, and let $B\operatorname{\mathcal{C}}$ be the bicategory obtained by delooping $\operatorname{\mathcal{C}}$ (Example 2.2.2.5). A twisting cochain for the bicategory $B \operatorname{\mathcal{C}}$ (in the sense of Notation 2.2.6.7) can be identified with a map of sets

\[ \mu : G \times G \rightarrow \Gamma \quad \quad (g,f) \mapsto \mu _{g,f}. \]

Let $(B\operatorname{\mathcal{C}})'$ denote the twist of $B \operatorname{\mathcal{C}}$ with respect to $\mu $. Unwinding the definitions, we see that $(B\operatorname{\mathcal{C}})'$ is obtained by delooping the same category $\operatorname{\mathcal{C}}$ with respect to a different monoidal structure: namely, the monoidal structure supplied by the $3$-cocycle $\alpha ': G \times G \times G \rightarrow \Gamma $ given by the formula

\[ \alpha '_{h,g,f} = \alpha _{h,g,f} + h( \mu _{g,f} ) - \mu _{hg,f} + \mu _{h,gf} - \mu _{h,g}. \]

We can summarize the situation as follows:

To every $3$-cocycle $\alpha : G \times G \times G \rightarrow \Gamma $, we can associate a bicategory $B \operatorname{\mathcal{C}}$ in which the $1$-morphisms are the elements of $G$, the $2$-morphisms are the elements of $\Gamma $, and the associativity constraint is given by $\alpha $.
If $\alpha , \alpha ': G \times G \times G \rightarrow \Gamma $ are cohomologous $3$-cocycles on $G$ with values in $\Gamma $, then the associated bicategories $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{C}}'$ are isomorphic (though not necessarily strictly isomorphic). More precisely, every choice of $2$-cocycle $\mu : G \times G \rightarrow \Gamma $ satisfying $\alpha ' = \alpha + \partial (\mu )$ determines a strictly unital isomorphism from $\operatorname{\mathcal{C}}$ to $\operatorname{\mathcal{C}}'$. Here $\partial $ denotes the boundary operator from $2$-cochains to $3$-cocycles, given concretely by the formula

\[ (\partial \mu )_{h,g,f} =h( \mu _{g,f} ) - \mu _{hg,f} + \mu _{h,gf} - \mu _{h,g}. \]

Example 2.2.6.13. The bicategories $\mathrm{Bimod}$ and $\operatorname{Cospan}(\operatorname{\mathcal{C}})$ of Examples 2.2.2.4 and 2.2.2.1 both depend on certain auxiliary choices:

Let $A$, $B$, and $C$ be associative rings, and suppose we are given a pair of bimodules $M = {}_{A}^{}M_{B}$ and $N = {}_{B}^{}N_{C}$. Then we can regard $M$ and $N$ as $1$-morphisms in the bicategory $\mathrm{Bimod}$, whose composition is defined to be the relative tensor product $M \otimes _{B} N$. This tensor product is well-defined up to (unique) isomorphism: it is universal among abelian groups $P$ which are equipped with a $B$-bilinear map $M \times N \rightarrow P$. However, it is possible to give many different constructions of an abelian group with this universal property, each of which gives a (slightly) different composition law for the $1$-morphisms in the bicategory $\mathrm{Bimod}$.
Let $\operatorname{\mathcal{C}}$ be a category which admits pushouts, and suppose we are given a pair of cospans

\[ X \leftarrow B \rightarrow Y \quad \quad Y \leftarrow C \rightarrow Z \]

in $\operatorname{\mathcal{C}}$. Then $B$ and $C$ can be regarded as $1$-morphisms in the bicategory $\operatorname{Cospan}(\operatorname{\mathcal{C}})$, whose composition is given by the pushout $C \amalg _{Y} B$ (regarded as a cospan from $X$ to $Z$). This pushout is well-defined up to (unique) isomorphism as an object of $\operatorname{\mathcal{C}}$, but there is generally no preferred representative of its isomorphism class. Consequently, different choices of pushout lead to (slightly) different definitions for the composition of $1$-morphisms in the bicategory $\operatorname{Cospan}(\operatorname{\mathcal{C}})$.

By making a different choice of conventions in these examples, one can obtain bicategories $\mathrm{Bimod}'$ and $\operatorname{Cospan}'(\operatorname{\mathcal{C}})$ having the same objects, $1$-morphisms, and $2$-morphisms as the bicategories $\mathrm{Bimod}$ and $\operatorname{Cospan}(\operatorname{\mathcal{C}})$, but different composition laws for $1$-morphisms. In this case, the bicategories $\mathrm{Bimod}'$ and $\operatorname{Cospan}'(\operatorname{\mathcal{C}})$ can be obtained from $\mathrm{Bimod}$ and $\operatorname{Cospan}(\operatorname{\mathcal{C}})$ (respectively) by the twisting procedure of Construction 2.2.6.8. In particular, the resulting bicategories $\mathrm{Bimod}'$ and $\operatorname{Cospan}'(\operatorname{\mathcal{C}})$ are isomorphic (though not necessarily strictly isomorphic) to the bicategories $\mathrm{Bimod}$ and $\operatorname{Cospan}(\operatorname{\mathcal{C}})$, respectively.