Proposition 5.3.23. The two functors \[ \CMon (-)\colon \Cat ^{\mathrm {prod}}_{\infty } \to \Cat ^{\sadd }_{\infty } \qquadtext { and } \CGrp (-)\colon \Cat ^{\mathrm {prod}}_{\infty } \to \Cat ^{\add }_{\infty } \] are right adjoint to the respective inclusion functors.
Proof. We will check the (dual version of the) criterion from Proposition 21.8.9. We will treat the case for \(\CMon (-)\); the case for \(\CGrp (-)\) is identical. The forgetful functor defines a natural transformation \(\epsilon \colon \CMon (-) \to \id _{\Cat ^{\mathrm {prod}}}\), and we must check that for every \(\infty \)-category \(C\) with finite products the two forgetful functors \[ \hspace {-10pt} \epsilon _{\CMon (C)} \colon \CMon (\CMon (C)) \to \CMon (C) \qquadtext { and } \CMon (\epsilon _C)\colon \CMon (\CMon (C)) \to \CMon (C) \] are equivalences. Since \(\CMon (C)\) is semiadditive by Proposition 5.3.20, the claim for \(\epsilon _{\CMon (C)}\) is an instance of Proposition 5.3.17. But now the case for \(\CMon (\epsilon _C)\) follows immediately: the swap map \(\Span (\Fin ) \times \Span (\Fin ) \iso \Span (\Fin ) \times \Span (\Fin )\) determines an equivalence \[ \CMon (\CMon (C)) \iso \CMon (\CMon (C)), \qquad (M_n)_m \mapsto (M_m)_n, \] and under this equivalence the forgetful functor \(\epsilon _{\CMon (C)}((M_n)_m) = (M_n)_1\) corresponds to the forgetful functor \(\CMon (\epsilon _C)((M_m)_n) = (M_1)_n\). □
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