Lemma 4.43.1. Let $(\mathcal{C}, \otimes , \phi )$ be as above. There is a 1-to-1 correspondence between units $(\mathbf{1}, l, r)$ in $\mathcal{C}$ and pairs $(\mathbf{1}, 1)$ where $\mathbf{1}$ is an object of $\mathcal{C}$ and $1 : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ is an isomorphism such that the functors $L : X \mapsto \mathbf{1} \otimes X$ and $R : X \mapsto X \otimes \mathbf{1}$ are equivalences.
Proof. Given a unit $(\mathbf{1}, l, r)$ we get an isomorpism $r : \mathbf{1} \otimes \mathbf{1} \to \mathbf{1}$ and $L$ and $R$ are equivalences as they are isomorphic to the identity functor. Conversely, suppose given $(\mathbf{1}, 1)$ such that $L$ and $R$ are equivalences. We obtain functorial isomorphisms $l_ X : \mathbf{1} \otimes X \to X$ and $r_ X : X \otimes \mathbf{1} \to X$ characterized by $L(l_ X) = 1 \otimes \text{id}_ X$ and $R(r_ X) = \text{id}_ X \otimes 1$. Then we have to show that the two arrows $\text{id}_ X \otimes l_ Y$ and $r_ X \otimes \text{id}_ Y$ from $X \otimes \mathbf{1} \otimes Y$ to $X \otimes Y$ are the same for all $X$ and $Y$. This property only depends on the isomorphism classes of $X$ and $Y$. Since $R$ and $L$ are equivalences, it suffices to do this for $X = Z \otimes \mathbf{1}$ and $Y = \mathbf{1} \otimes W$ for some objects $Z$ and $W$. In other words, we have to show that
\[ \text{id}_ Z \otimes \text{id}_\mathbf {1} \otimes l_{\mathbf{1} \otimes W} = r_{Z \otimes \mathbf{1}} \otimes \text{id}_\mathbf {1} \otimes \text{id}_ Y \]
By construction these maps are equal to $\text{id}_ Z \otimes 1 \otimes \text{id}_\mathbf {1} \otimes \text{id}_ W$ and $\text{id}_ Z \otimes \text{id}_\mathbf {1} \otimes 1 \otimes \text{id}_ W$. Thus it suffices to show that $1 \otimes \text{id}_\mathbf {1} = \text{id}_\mathbf {1} \otimes 1$.
We have $\text{id}_\mathbf {1} \otimes 1 = r_\mathbf {1} \otimes \text{id}_\mathbf {1}$ and $l_\mathbf {1} \otimes \text{id}_\mathbf {1} = \text{id}_\mathbf {1} \otimes 1$. We may write $r_\mathbf {1} = a \circ 1$ and $l_\mathbf {1} = b \circ 1$ for some $a, b$ automorphisms of $\mathbf{1}$. Thus we have $\text{id}_\mathbf {1} \otimes 1 = (a \otimes \text{id}_\mathbf {1}) \circ (1 \otimes \text{id}_\mathbf {1})$ and $1 \otimes \text{id}_\mathbf {1} = (\text{id}_\mathbf {1} \otimes b) \circ (\text{id}_\mathbf {1} \otimes 1)$. Then we can write
\begin{align*} (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes 1 \otimes \text{id}_\mathbf {1}) & = (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1} \otimes b) \circ (\text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1} \otimes 1) \\ & = (\text{id}_\mathbf {1} \otimes b) \circ (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1} \otimes 1) \\ & = (\text{id}_\mathbf {1} \otimes b) \circ (1 \otimes 1) \end{align*}
and we also have
\begin{align*} (1 \otimes \text{id}_\mathbf {1}) \circ (\text{id}_\mathbf {1} \otimes 1 \otimes \text{id}_\mathbf {1}) & = (\text{id}_\mathbf {1} \otimes b) \circ (\text{id}_\mathbf {1} \otimes 1) \circ (a \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \circ (1 \otimes \text{id}_\mathbf {1} \otimes \text{id}_\mathbf {1}) \\ & = (\text{id}_\mathbf {1} \otimes b) \circ (a \otimes \text{id}_\mathbf {1}) \circ (1 \otimes 1) \end{align*}
This proves that $a \otimes \text{id}_\mathbf {1}$ is the identity and hence $a$ is the identity as desired.
To finish the proof, we note that the rules above determine inverse equivalences of categories between the category of units (suitably defined) and the category of pairs $(\mathbf{1}, 1)$. $\square$
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