# Kerodon

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Corollary 2.1.2.21. Let $\operatorname{\mathcal{C}}$ be a monoidal category with unit object $\mathbf{1}$. Then the left and right unit constraints $\lambda _{ \mathbf{1} }, \rho _{ \mathbf{1} }: \mathbf{1} \otimes \mathbf{1} \xrightarrow {\sim } \mathbf{1}$ are equal to the unit constraint $\upsilon : \mathbf{1} \otimes \mathbf{1} \xrightarrow {\sim } \mathbf{1}$.

Proof. Let $X$ be any object of $\operatorname{\mathcal{C}}$. Then the left unit contraint $\lambda _{X}$ is characterized by the commutativity of the diagram

$\xymatrix@R =50pt@C=50pt{ \mathbf{1} \otimes (\mathbf{1} \otimes X) \ar [rr]^-{\alpha _{\mathbf{1}, \mathbf{1}, X} }_{\sim } \ar [dr]_{ \operatorname{id}_\mathbf {1} \otimes \lambda _ X}^{\sim } & & (\mathbf{1} \otimes \mathbf{1}) \otimes X \ar [dl]^{ \upsilon \otimes \operatorname{id}_ X}_{\sim } \\ & \mathbf{1} \otimes X. & }$

Using Proposition 2.1.2.19, we deduce that $\upsilon \otimes \operatorname{id}_{X} = \rho _{ \mathbf{1} } \otimes \operatorname{id}_ X$ as morphisms from $( \mathbf{1} \otimes \mathbf{1}) \otimes X$ to $\mathbf{1} \otimes X$. In other words, the morphisms $\upsilon , \rho _\mathbf {1}: \mathbf{1} \otimes \mathbf{1} \rightarrow \mathbf{1}$ have the same image under the functor

$\operatorname{\mathcal{C}}\rightarrow \operatorname{\mathcal{C}}\quad \quad Y \mapsto Y \otimes X.$

In the case $X = \mathbf{1}$, this functor is fully faithful; it follows that $\upsilon = \rho _\mathbf {1}$. The equality $\upsilon = \lambda _{ \mathbf{1} }$ follows by a similar argument. $\square$