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).


1 Answer 1


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 $\mathrm{Map}(BZ/p,Y)$.


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