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Oct 2 at 22:55 comment added varkor I agree that's an improvement.
Oct 2 at 22:21 comment added Ivan Di Liberti Finally, I am perfectly aware that I added the hypothesis of completeness when answering your question. At the moment I can't find a sharper argument than this one.
Oct 2 at 22:20 comment added Ivan Di Liberti The fact that $y(c)^y(a)$ is indeed $y(c^a)$ follows by direct inspection. Indeed $P(A)(F \times y(a), y(c)) = P(A)(\text{colim} y(d_i) \times y(a), y(c)) = \text{lim} P(A)( y(d_i) \times y(a), y(c)) = \text{lim} A( d_i \times a, c)...$.
Oct 2 at 22:02 comment added Ivan Di Liberti Sure. Let $A$ be a bicomplete cartesian closed category and consider the Yoneda embedding into small presheaves $y: A \to P(A)$. Now, the embedding preserves finite products and it is easy to see that because $A$ is cartesian closed, $A$ must be an exponential ideal in $P(A)$, indeed $y(c)^d = y(c)^{\text{colim} yc_i} = \text{lim} y(c)^{y(c_i)} = y(\text{lim} c^{c_i})$. So now we can apply Sketches A4.3.1.
Oct 2 at 19:01 comment added varkor @IvanDiLiberti: I don't see you obtain finite product preservation by the reflector in this case. Perhaps you could elaborate in a second answer?
Oct 2 at 18:51 comment added Ivan Di Liberti Better is true, no? Every cocomplete cartesian closed category is reflective in its category of small presheaves and the reflector preserves finite products (the argument is the same of Day's). Notice that finite products indeed do exists in the category of small presheaves by Remark 3.9 in Rosicky-Adamek "How nice are are free completions?"
Oct 2 at 18:31 history answered varkor CC BY-SA 4.0