# Does profinite completion commute with mapping spaces?

Does there exist a prime number $$p$$ and a smooth complex projective variety $$X$$ such that $$F_{\infty p}\mathrm{Map}(B\mathbb{Z}/p\mathbb{Z}, X)$$ is not weakly homotopy equivalent to $$\mathrm{Map}(B\mathbb{Z}/p\mathbb{Z}, F_{\infty p}X)$$?

Here $$F_{\infty p}$$ stands either for Bousfield--Kan $$p$$-completion or Sullivan $$p$$-profinite completion (so this question consists of two sub-questions).

I am not a real expert on the extreme aspects of completions, but I think that, in the Sullivan completion case at least, these two things are always equivalent. The Sullivan $$p$$--profinite completion was explored quite carefully (with model category language, e.g.) in an early paper of Fabien Morel: "ensembles profinis simpliciaux et interpretation geometrique de foncteur $$T$$", Bull. Soc. Math. France, vol 124 (1994), 347-373. ($$T$$ here is Jean Lannes' fabulous $$T$$--functor. And yes, this paper is in French.)

In particular, in the middle of page 371, in parentheses, he says (I am roughly translating here): one should remark that if $$X$$ is a pro-$$p$$-space, so is $$\mathrm{Map}(BZ/p,X)$$, because if $$Y$$ is a finite-$$p$$--space [one with only a finite number of nonzero homotopy groups all of which are finite $$p$$-groups], the same is true for $$\mathrm{Map}(BZ/p,Y)$$.