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Apr 21, 2021 at 10:58 comment added varkor If you don't want uniqueness of the factorisation, you can ask for $L$ just to be full. If it arises via a universal property (without uniqueness), then this is suggestive of a weak (2-)limit as @fosco points out.
Apr 21, 2021 at 5:58 comment added fosco I think it is a particular case of an inserter, but without uniqueness of the 1-cell; the non-uniqueness instead suggests some sort of (weak) orthogonality condition.
Apr 21, 2021 at 3:39 comment added Ben MacAdam Yeah it’s a bit weaker than that - I have a 2-monad and three strict algebras of it. The morphism $B$ to $C$ acts like an inserter for every parallel pair of strict algebra homomorphisms. My example certainly isn’t a fully faithful functor - it’s actually a reflector.
Apr 21, 2021 at 2:38 comment added David Roberts @varkor full faithfulness would mean a unique factorisation
Apr 20, 2021 at 23:13 comment added varkor Could you clarify what your question is? It sounds like you're asking for $L$ to be fully faithful, so that postcomposition by $L$ induces an isomorphism between 2-cells $L \circ F \Rightarrow L \circ G$ and 2-cells $F \Rightarrow G$.
Apr 20, 2021 at 22:37 history asked Ben MacAdam CC BY-SA 4.0