5
$\begingroup$

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).

$\endgroup$
4
$\begingroup$

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

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.