Kerodon

I think there's a subtelty not addressed in Remark 2.1.2.22. In light of Corollary 2.1.2.21 (particularly, the fact that $\lambda_1 = v$), the proof of Proposition 2.1.2.9 reduces to showing that there's a unique morphism $u: 1\rightarrow 1'$ such that the depicted outer rectangle commutes. And for this last part of the proof, it is indeed true that only the functors $X \mapsto 1 \otimes X$ and $X \mapsto X \otimes 1'$ are required to be fully faithful. However, the proof of Corollary 2.1.2.21 (more precisely, of its second half, i.e. the identity $\lambda_1 = v$) uses the triangle identity (Proposition 2.1.2.19) and in particular the existence of the right unit constraint $\rho_X$, while the weaker hypotheses of Remark 2.1.2.22 aren't enough to construct $\rho$ (the construction of $\rho$ requires that the functor $X \mapsto X \otimes 1$ should be fully faithful as well). Am I missing something?

The assumption that tensoring with $1$ on the left is fully faithful is enough to construct the left unit constraint $\lambda_{1'}$, and the assumption that tensoring on the right with $1'$ is fully faithful guarantees that there is a unique morphism $u: 1 \rightarrow 1'$ which makes the lower half of the diagram commutes, and that $u$ is an isomorphism. Once we know that $1$ and $1'$ are isomorphic, we also learn that tensoring with $1$ on the right is fully faithful and that tensoring with $1'$ on the left is fully faithful.