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2.2.6 Isomorphisms of $2$-Categories

We now study isomorphisms between $2$-categories.

Definition 2.2.6.1. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be $2$-categories. 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 $2$-categories $\operatorname{\mathcal{C}}$ 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 $2$-categories, and let $G: \operatorname{\mathcal{D}}\rightarrow \operatorname{\mathcal{C}}$ be the inverse isomorphism. Then:

  • The functor $F$ is strictly unitary if and only if $G$ is strictly unitary. In this case, we say that $F$ is a strictly unitary 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 $2$-categories $\operatorname{\mathcal{C}}$ 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 $2$-categories 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 $2$-category $\operatorname{\mathcal{C}}$ is strict is invariant under strict isomorphism, but not under isomorphism.

Warning 2.2.6.4. The notions of isomorphism and strict isomorphism of $2$-categories are somewhat artificial. As in classical category theory, there is notion of equivalence of $2$-categories (Definition ) which is more general than isomorphism and more appropriate for describing what it means for $2$-categories to be “the same.”

Remark 2.2.6.5. Let $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ be a functor of $2$-categories. 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 $2$-categories, 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 $2$-category $\operatorname{\mathcal{C}}$, 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 $2$-category $\operatorname{\mathcal{D}}$, 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 $2$-categories.

Notation 2.2.6.7. Let $\operatorname{\mathcal{C}}$ be a $2$-category. 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 $2$-category equipped with a twisting cochain

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

We define a new $2$-category $\operatorname{\mathcal{C}}'$ 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 $2$-category $\operatorname{\mathcal{C}}'$ is the same as the identity morphism from $X$ to itself in the $2$-category $\operatorname{\mathcal{C}}$.

  • 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 $2$-category $\operatorname{\mathcal{C}}'$ 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 $2$-category equipped with a twisting cochain $\{ \mu _{g,f} \} $. Show that the $2$-category $\operatorname{\mathcal{C}}'$ of Construction 2.2.6.8 is well-defined. Moreover, there is a strictly unitary isomorphism of $2$-categories $\operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}'$ 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 unitary isomorphism of $2$-categories. Show that there is a unique twisting cochain $\{ \mu _{g,f} \} $ on the $2$-category $\operatorname{\mathcal{C}}$ 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 unitary isomorphism of Exercise 2.2.6.9 and $H$ is a strict isomorphism of $2$-categories. In other words, the notion of twisting cochain (in the sense of Notation 2.2.6.7) measures the difference between strictly unitary isomorphisms and strict isomorphisms in the setting of $2$-categories.

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 $2$-category obtained by delooping $\operatorname{\mathcal{C}}$ (Example 2.2.2.5). A twisting cochain for the $2$-category $B \operatorname{\mathcal{C}}$ (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 $2$-category $B \operatorname{\mathcal{C}}$ 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 $2$-categories $\operatorname{\mathcal{C}}$ 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 unitary 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 $2$-categories $\mathrm{Bimod}$ 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 $2$-category $\mathrm{Bimod}$, 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 $2$-category $\mathrm{Bimod}$.

  • 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 $2$-category $\operatorname{Cospan}(\operatorname{\mathcal{C}})$, 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 $2$-category $\operatorname{Cospan}(\operatorname{\mathcal{C}})$.

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