Lemma 4.43.2. Let $(\mathcal{C}, \otimes , \phi )$ be as above. Let $(\mathbf{1}, 1)$ be a unit (see Lemma 4.43.1). Then
$1 \otimes \text{id}_\mathbf {1} = \text{id}_\mathbf {1} \otimes 1$
$\Gamma = \text{Mor}(\mathbf{1}, \mathbf{1})$ is a commutative monoid,
$a = 1 \circ (a \otimes \text{id}_\mathbf {1}) \circ 1^{-1} = 1 \circ (\text{id}_\mathbf {1} \otimes a) \circ 1^{-1}$ for all $a \in \Gamma $,
any other unit is isomorphic to $(\mathbf{1}, 1)$ by a unique isomorphism.
Proof.
Part (1) was shown in the proof of Lemma 4.43.1. For $a \in \Gamma $ we have
\begin{align*} (1 \circ (a \otimes \text{id}_\mathbf {1})) \otimes \text{id}_\mathbf {1} & = (1 \otimes \text{id}_\mathbf {1}) \circ (a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \\ & = (\text{id}_\mathbf {1} \otimes 1) \circ (a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \\ & = (a \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes 1) \\ & = (a \otimes \text{id}_\mathbf {1}) \circ (1 \otimes \text{id}_\mathbf {1}) \\ & = (a \circ 1) \otimes \text{id}_\mathbf {1} \end{align*}
Thus $1 \circ (a \otimes \text{id}_\mathbf {1}) = a \circ 1$ and this implies one half of (3). The other half follows in the same way. To see (2) we observe that for $a, b \in \Gamma $ we have
\[ a \circ b = 1 \circ (a \otimes \text{id}_\mathbf {1}) \circ 1^{-1} \circ 1 \circ (\text{id}_\mathbf {1} \otimes b) \circ 1^{-1} = 1 \circ (a \otimes b) \circ \circ 1^{-1} \]
and $b \circ a$ evaluates to the same expression. Thus (2) holds. To see (4) suppose that $(\mathbf{1}', 1')$ is a second unit. Using $r$ and $l'$ there are isomorphisms $\mathbf{1}' \otimes \mathbf{1} \to \mathbf{1}'$ and $\mathbf{1}' \otimes \mathbf{1} \to \mathbf{1}$. Thus there exists an isomorphism $t : \mathbf{1}' \to \mathbf{1}$. Then the diagram
\[ \xymatrix{ \mathbf{1}' \otimes \mathbf{1}' \ar[r]_{t \otimes t} \ar[d]_{1'} & \mathbf{1} \otimes \mathbf{1} \ar[d]^1 \\ \mathbf{1}' \ar[r]^ t & \mathbf{1} } \]
commutes up to an automorphism $a$ of $\mathbf{1}$. After replacing $t$ by $a \circ t$ the diagram will commute (hint: use (3) to see that $1 \circ (a \otimes a) = a^2 \circ 1$).
$\square$
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