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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 exist a somewhat obscure integral expression coming from (co)end calculus.

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.

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 exist a somewhat obscure integral expression coming from coend calculus.

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 exist a somewhat obscure integral expression coming from (co)end calculus.

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.

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 exist a somewhat obscure integral expression coming from coend calculus.