The double-braiding center is a symmetric monoidal subcategory

The tensor unit belongs to ${\mathcal{Z}}_2(\mathcal{C})$: for every $Y$, [L3] gives $c_{Y,\mathbf{1}}^{-1}=c_{\mathbf{1},Y}$, hence $c_{Y,\mathbf{1}}\circ c_{\mathbf{1},Y}=1_{\mathbf{1}\otimes Y}$.

Suppose $X$ and $Y$ lie in ${\mathcal{Z}}_2(\mathcal{C})$, and let $Z$ be arbitrary. By [L4], the canonical associators in the two hexagons may be suppressed after transporting both sides between the same parenthesised source and target. In that coherent notation, the hexagons give $c_{Z,X\otimes Y}\circ c_{X\otimes Y,Z}=(1_X\otimes c_{Z,Y})(c_{Z,X}\otimes 1_Y)(c_{X,Z}\otimes 1_Y)(1_X\otimes c_{Y,Z})\mathrm{.}$

Transparency of $X$ gives $(c_{Z,X}\otimes 1_Y)(c_{X,Z}\otimes 1_Y)=1_{X\otimes Z}\otimes 1_Y$. Substituting this into step 1.2 leaves $(1_X\otimes c_{Z,Y})(1_X\otimes c_{Y,Z})=1_X\otimes (c_{Z,Y}\circ c_{Y,Z}),$ which is the identity by transparency of $Y$. Hence $X\otimes Y$ lies in ${\mathcal{Z}}_2(\mathcal{C})$.

Because the subcategory is full and is closed under tensor product by step 2.1 and under the unit by step 1.1, tensor products of its morphisms and the ambient associator and unitors all remain in it. The ambient braiding also restricts to it. For central objects $X,Y$, their defining condition says $c_{Y,X}\circ c_{X,Y}=1_{X\otimes Y}$. By the definition in [L2], the restricted braiding is therefore a symmetry. Thus ${\mathcal{Z}}_2(\mathcal{C})$ is symmetric monoidal.