We now introduce a special class of $2$-categories.
Definition 2.2.7.1. Let $\operatorname{\mathcal{C}}$ be a $2$-category. We will say that $\operatorname{\mathcal{C}}$ is strictly unitary if, for each $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, the left and right unit constraints are identity $2$-morphisms of $\operatorname{\mathcal{C}}$.
\[ \lambda _{f}: \operatorname{id}_{Y} \circ f \xRightarrow {\sim } f \quad \quad \rho _{f}: f \circ \operatorname{id}_{X} \xRightarrow {\sim } f \]
Proposition 2.2.7.2. Let $\operatorname{\mathcal{C}}$ be a $2$-category. Then $\operatorname{\mathcal{C}}$ is strictly unitary if and only if the following conditions are satisfied:
For each $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, we have $\operatorname{id}_{Y} \circ f = f = f \circ \operatorname{id}_{X}$.
For each object $X$ of $\operatorname{\mathcal{C}}$, the unit constraint $\upsilon _{X}: \operatorname{id}_{X} \circ \operatorname{id}_{X} \xRightarrow {\sim } \operatorname{id}_ X$ is the identity morphism from $\operatorname{id}_{X} \circ \operatorname{id}_{X} = \operatorname{id}_{X}$ to itself.
For every $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, the associativity constraints $\alpha _{ \operatorname{id}_{Y}, \operatorname{id}_{Y}, f}$ and $\alpha _{ f, \operatorname{id}_ X, \operatorname{id}_ X}$ are equal to the identity (as $2$-morphisms from $f$ to itself).
Proof. If $\operatorname{\mathcal{C}}$ is strictly unitary, then $(a)$ is clear and $(b)$ follows from Corollary 2.2.1.15. Assume that $(a)$ and $(b)$ are satisfied. For any $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, the left unit constraint $\lambda _{f}$ is characterized by the commutativity of the diagram
and is therefore the identity $2$-morphism if and only if $\alpha _{\operatorname{id}_ Y, \operatorname{id}_ Y, f}$ is an identity $2$-morphism (from $f$ to itself). Similarly, the right unit constraint $\rho _{f}$ is an identity $2$-morphism if and only if $\alpha _{ f, \operatorname{id}_ X, \operatorname{id}_ X }$ is an identity $2$-morphism in $\operatorname{\mathcal{C}}$. $\square$
Remark 2.2.7.3. Let $\operatorname{\mathcal{C}}$ be a strictly unitary $2$-category. Then $\operatorname{\mathcal{C}}$ satisfies the following stronger versions of conditions $(a)$ and $(c)$ of Proposition 2.2.7.2:
For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, the functors
are equal to the identity.
For every pair of $1$-morphisms $X \xrightarrow {f} Y \xrightarrow {g} Z$ in $\operatorname{\mathcal{C}}$, the associativity constraints $\alpha _{g, f, \operatorname{id}_ X}$, $\alpha _{ g, \operatorname{id}_ Y, f}$, and $\alpha _{\operatorname{id}_ Z, g, f}$ are equal to the identity (as $2$-morphisms from $g \circ f$ to itself).
\[ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) \quad \quad f \mapsto \operatorname{id}_{Y} \circ f \]
\[ \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) \quad \quad f \mapsto f \circ \operatorname{id}_ X \]
Here $(a')$ follows from the naturality of the left and right unit constraints (Remark 2.2.1.13), and $(c')$ follows from Propositions 2.2.1.14 and 2.2.1.16.
Example 2.2.7.4. 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). The following conditions are equivalent:
The $3$-cocycle $\alpha $ is normalized: that is, it satisfies the equations
for every pair of elements $x,y \in G$.
The $2$-category $B\operatorname{\mathcal{C}}$ is strictly unitary, in the sense of Definition 2.2.7.1.
\[ \alpha _{x,y,1} = \alpha _{x,1,y} = \alpha _{1,x,y} = 0 \]
Remark 2.2.7.5. Let $\operatorname{\mathcal{C}}$ and $\operatorname{\mathcal{D}}$ be strictly unitary $2$-categories (Definition 2.2.7.1). Then a strictly unitary lax functor $F: \operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{D}}$ is given by the following data:
For each object $X \in \operatorname{\mathcal{C}}$, an object $F(X) \in \operatorname{\mathcal{D}}$.
For every pair of objects $X,Y \in \operatorname{\mathcal{C}}$, a functor of ordinary categories
For every pair of composable $1$-morphisms $X \xrightarrow {f} Y \xrightarrow {g} Z$ in $\operatorname{\mathcal{C}}$, a composition constraint $\mu _{g,f}: F(g) \circ F(f) \Rightarrow F(g \circ f)$, depending functorially on $f$ and $g$.
\[ F_{X,Y}: \underline{\operatorname{Hom}}_{\operatorname{\mathcal{C}}}(X,Y) \rightarrow \underline{\operatorname{Hom}}_{\operatorname{\mathcal{D}}}( F(X), F(Y) ). \]
This data must be required to satisfy axiom $(c)$ of Definition 2.2.4.5, together with the identities $F( \operatorname{id}_ X) = \operatorname{id}_{F(X)}$ for each object $X \in \operatorname{\mathcal{C}}$ and $\mu _{\operatorname{id}_ Y, f} = \operatorname{id}_{ F(f)} = \mu _{f, \operatorname{id}_ X}$ for each $1$-morphism $f: X \rightarrow Y$ of $\operatorname{\mathcal{C}}$.
Remark 2.2.7.6. Let $\operatorname{\mathcal{C}}$ be a strictly unitary $2$-category, let $\{ \mu _{g,f} \} $ be a twisting cochain for $\operatorname{\mathcal{C}}$ (see Notation 2.2.6.7), and let $\operatorname{\mathcal{C}}'$ denote the twist of $\operatorname{\mathcal{C}}'$ with respect to $\{ \mu _{g,f} \} $ (Construction 2.2.6.8). The following conditions are equivalent:
The $2$-category $\operatorname{\mathcal{C}}'$ is strictly unitary.
For every $1$-morphism $f: X \rightarrow Y$ in $\operatorname{\mathcal{C}}$, both $\mu _{f, \operatorname{id}_ X}$ and $\mu _{\operatorname{id}_{Y},f}$ are identity $2$-morphisms (from $f \circ \operatorname{id}_{X} = f = \operatorname{id}_{Y} \circ f$ to itself).
If these conditions are satisfied, we will say that the twisting cochain $\{ \mu _{g,f} \} $ is normalized.
It is generally harmless to assume that a $2$-category $\operatorname{\mathcal{C}}$ is strictly unitary, by virtue of the following:
Proposition 2.2.7.7. Let $\operatorname{\mathcal{C}}$ be a $2$-category. Then there exists a strictly unitary isomorphism $\operatorname{\mathcal{C}}\simeq \operatorname{\mathcal{C}}'$, where $\operatorname{\mathcal{C}}'$ is a strictly unitary $2$-category.
Proof. Let $\mu = \{ \mu _{g,f} \} $ be the twisting cochain on $\operatorname{\mathcal{C}}$ given on composable $1$-morphisms $X \xrightarrow {f} Y \xrightarrow {g} Z$ by the formula
Note that this prescription is consistent, since $\lambda _{f} = \upsilon _{Y} = \rho _{g}$ in the special case where $f = \operatorname{id}_ Y = g$ (Corollary 2.2.1.15). Let $\operatorname{\mathcal{C}}'$ be the twist of $\operatorname{\mathcal{C}}$ with respect to the cocycle $\{ \mu _{g,f} \} $ (Construction 2.2.6.8). Then $\operatorname{\mathcal{C}}'$ is a strictly unitary $2$-category (in the sense of Definition 2.2.7.1), and Exercise 2.2.6.9 supplies a strictly unitary isomorphism of $2$-categories $\operatorname{\mathcal{C}}\simeq \operatorname{\mathcal{C}}'$ $\square$
Remark 2.2.7.8. Let $\operatorname{2Cat}'_{\operatorname{ULax}}$ denote the subcategory of $\operatorname{2Cat}_{\operatorname{Lax}}$ (and full subcategory of $\operatorname{2Cat}_{\operatorname{ULax}}$) whose objects are strictly unitary $2$-categories and whose morphisms are strictly unitary lax functors. It follows from Proposition 2.2.7.7 that the inclusion $\operatorname{2Cat}'_{\operatorname{ULax}} \hookrightarrow \operatorname{2Cat}_{\operatorname{ULax}}$ is an equivalence of categories.
Remark 2.2.7.9. Let $G$ be a group and let $\Gamma $ be an abelian group with an action of $G$. When applied to the $2$-categories described in Example 2.2.7.4, Proposition 2.2.7.7 reduces to the assertion that every $3$-cocycle $\alpha : G \times G \times G \rightarrow \Gamma $ is cohomologous to a normalized $3$-cocycle $\alpha ': G \times G \times G \rightarrow \Gamma $.