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In Centazzo and Vitale's A Duality Relative to a Limit Doctrine (TAC, 2002, abstract), early on, they make the above claim and cite Kelly's Basic Concepts in Enriched Category Theory (TAC reprints). I am having difficulty finding exactly where these general reasons are.

Two questions:

(1) What exactly in Basic Concepts... are they referring to?

(2) What is an explicit formula for this specific left adjoint?

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This answer does not meet any of the questions, yet it provides the reason for which $i_A$ has a left adjoint. The whole story is contained in Categories of continuous functors I by Kelly and Freyd. Nowadays maybe we have more sophisticated ways of phrasing this result, but the core of the reason is still in the paper by Freyd and Kelly, which offers also a historical tour of all the partial results that led to the theorem.

Coming to your second question, it is very hard in general to provide an explicit formula. Since the left adjoint $L_A$ has to coincide by abstract nonsense with $\mathsf{ran}_{i_A}(1),$ there exists a somewhat obscure integral expression coming from (co)end calculus.

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    $\begingroup$ Freyd–Kelly: doi.org/10.1016/0022-4049(72)90001-1, with erratum doi.org/10.1016/0022-4049(74)90033-4 $\endgroup$
    – David Roberts ♦
    Commented Aug 16, 2019 at 11:12
  • $\begingroup$ Thank you @DavidRoberts, you aways provide links for my answers, one day I will learn to embed them every time! $\endgroup$
    – Ivan Di Liberti
    Commented Aug 16, 2019 at 11:14
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    $\begingroup$ It is worth to mention that the two perspectives (using FAFT and computing the Kan extension) are equivalent in that FAFT gives a way (the solution set condition) to compute $\text{lim } \big((1\downarrow i_A) \to \mathbb{D}\text{-cont}[A, {\bf Set}]\big)$. Also, Freyd and Kelly's paper gives sufficiently sharp conditions in order to apply AFT, but in concrete examples (eg for very concrete $\mathbb D$), you might find a easier way to find the left adjoint. $\endgroup$
    – fosco
    Commented Aug 16, 2019 at 17:21
  • $\begingroup$ The problem with the usual end formula is that $\mathbb D$-cont(A,Set)$ isn't small, right? Which is why one needs to find solution sets? It's pretty straightforward to show that it is complete. $\endgroup$
    – Daniel Satanove
    Commented Aug 16, 2019 at 18:55
  • $\begingroup$ I guess it really depends on what you consider an explicit formula, the end formalism is not very explicit to me. $\endgroup$
    – Ivan Di Liberti
    Commented Aug 17, 2019 at 8:41

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