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What was the argument you were trying to give? Maybe it can be repaired, and anyhow it would be interesting to know how you were thinking about this!
One can probably for example say: a functor $f: A \rightarrow B$ is fully faithful if, for any arrow $g$ of $B$, viewed as a functor $I \rightarrow B$, where $I$ is the interval category, the pullback of $f$ and $g$ can be taken to be $I$ with the identity arrow as the projection to $I$ if the pullback of $f$ and the restriction of $g$ to the inclusion of $1 \sqcup 1$ into $I$ is non-empty (it is necessarily empty otherwise). Is that the kind of thing you had in mind?
One can probably also say: the pullback of the functor $B^I \rightarrow B^{1 \sqcup 1}$ and the functor $f^{1 \sqcup 1}: A^{1 \sqcup 1} \rightarrow B^{1 \sqcup 1}$ can be taken to be $A^I$ with the obvious projections.
As a random aside, the difference between #4 and #5 is a quite typical example of a translation of a weaker (foundationally, or alternatively with regard to the structure required of the ’doctrine’ one is working in ), in particular predicative, formulation into a concise but impredicative one!
Yes, #5 was the characterization I was thinking of; that $A^I$ is the pullback (in the $(\infty,1)$ category of $\infty$-categories) of $A \times A \to B \times B \leftarrow B^I$ in the manner you describe. Or maybe that you need to restrict to the core first. It’s clear that this implies full faithfulness, but I’ve confused myself over the details for the reverse implication and haven’t gotten around to looking at it again.
Ah, I hadn’t actually noticed that you were asking about $(\infty,1)$-functors! We should add something about this in the ordinary category theory setting to full and faithful functor too.
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