Example 11.6.0.134 (009M)

Example 11.6.0.134. The contents of this tag are now contained in Remark 2.2.7.6 and Proposition 2.2.7.7.

Let $\operatorname{\mathcal{C}}$ be any $2$-category. Then the left and right unit constraints on $\operatorname{\mathcal{C}}$ determine a twisting cochain $\{ \mu _{g,f} \} $, given concretely by the formula

\[ \mu _{g,f} = \begin{cases} \lambda _{f}: g \circ f \Rightarrow f & \text{ if } g = \operatorname{id}_{Y} \\ \rho _{g}: g \circ f \Rightarrow g & \text{ if } f = \operatorname{id}_ Y \\ \operatorname{id}_{g \circ f}: g \circ f \Rightarrow g \circ f & \text{ otherwise. } \end{cases} \]

Note that this prescription is consistent, since $\lambda _{f} = \rho _{g}$ in the special case where $f = g = \operatorname{id}_{Y}$ (Corollary 2.2.1.15).

Let $\operatorname{\mathcal{C}}'$ be the twist of $\operatorname{\mathcal{C}}$ with respect to the cocycle $\{ \mu _{g,f} \} $. Then $\operatorname{\mathcal{C}}'$ is a strictly unitary $2$-category (in the sense of Variant 11.6.0.130), and Exercise 2.2.6.9 supplies a strictly unitary isomorphism of $2$-categories $\operatorname{\mathcal{C}}\simeq \operatorname{\mathcal{C}}'$ In particular, for every $2$-category $\operatorname{\mathcal{C}}$, there exists a strictly unitary isomorphism of $2$-categories $\operatorname{\mathcal{C}}\simeq \operatorname{\mathcal{C}}'$, where $\operatorname{\mathcal{C}}'$ is strictly unitary.