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Let $B$ be a type (or space). A type (or space) is $B$-null if the canonical map $X \to X^B$ is an equivalence. The $B$-nullification of an arbitrary type $X$ is a $B$-null type $\bigcirc_B X$ together with a map $\eta_X: X \to \bigcirc_B X$ such that for any $B$-null type $Y$ the map $Y^{\bigcirc_B X} \to Y^X$ given by composition with $\eta_X$ is an equivalence.

In particular being $\mathbb{S}^2$-null is the same as being 1-truncated, and $\mathbb{S}^2$-nullification is exactly 1-truncation.

$\mathbb{S}^2$ has non trivial homotopy groups above dimension 2. We can eliminate these higher homotopy groups by taking the 2-truncation of $\mathbb{S}^2$. This gives us the Eilenberg-MacLane space $K(\mathbb{Z}, 2)$, which has $\pi_2$ equal to $\mathbb{Z}$ and all other homotopy groups trivial.

I would like to know how $K(\mathbb{Z}, 2)$-nullification compares to 2-truncation. Certainly, $K(\mathbb{Z}, 2)$ is 1-connected, which tells us some other things like $K(\mathbb{Z}, 2)$-connected implies 1-connected, and that 1-truncated implies $K(\mathbb{Z}, 2)$-null, but there's not much else I can see at the moment, including some basic things:

  1. Is $K(\mathbb{Z}, 2)$-nullification the same as 1-truncation?
  2. Is $\mathbb{S}^2$ $K(\mathbb{Z}, 2)$-null?
  3. Is $\mathbb{S}^2$ $K(\mathbb{Z}, 2)$-connected (i.e. has trivial nullification)?

NB: I had homotopy type theory in mind for this question, because that's what I'm more familiar with. See Rijke, Shulman, Spitters, Modalities in homotopy type theory for details on nullification in this setting and Licata, Finster, Eilenberg-MacLane Spaces in Homotopy Type Theory for Eilenberg-MacLane spaces.

However, I think that the question also makes sense for $\infty$-toposes and other approaches to homotopy theory, where the answer might be more well known and I would also be interested to hear such answers in those settings.

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    $\begingroup$ This modality is briefly discussed in Example 6.1 in Cellular properties of nilpotent spaces where they show that $S^3$ is not $K(\mathbb{Z},2)$-connected. $\endgroup$
    – Ulrik Buchholtz
    Commented Jun 8, 2023 at 8:41
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    $\begingroup$ I wonder how this interacts with the Sullivan Conjecture (which was proved by Miller)? That implies that if $X$ is a space with $p$-complete homotopy groups and $G$ is a finite $p$-group then $X$ is $BG$-null. Up to $p$-completion, the space $K(\mathbb{Z},2)=\mathbb{C}P^\infty$ is the colimit of the spaces $B(\mathbb{Z}/p^k)$. Also, there is a cofibration $B(\mathbb{Z}/p^k)\to \mathbb{C}P^\infty\to T$, where $T$ is the Thom space of the $p^k$th tensor power of the tautological bundle over $\mathbb{C}P^\infty$. I don't know how Thomification interacts with nullity. $\endgroup$
    – Neil Strickland
    Commented Jun 8, 2023 at 11:10
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    $\begingroup$ @UlrikBuchholtz I think that answers the question, right? Because if $\mathbb{S}^2$ was $K(\mathbb{Z},2)$-connected, then every 1-connected space would be $K(\mathbb{Z},2)$-connected, including $\mathbb{S}^3$. For the same reason, it seems that showing $\mathbb{S}^2$ is not $K(\mathbb{Z},2)$-connected should be easier than $\mathbb{S}^3$, so I'm wondering if there's a more direct proof? $\endgroup$
    – aws
    Commented Jun 8, 2023 at 14:40

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