Let's consider in an $\infty$-topos, we have an object $X$ of homotopy dimension $\leq n$ (in the sense of Lurie HTT), let $f: A\to B$ be an $n$-equivalence morphism. Can we conclude that $f$ induces bijection on homotopy sets $$ f_*: [X, A]\xrightarrow{\cong}[X, B]? $$ This is a finite analog of Whitehead Theorem, see e.g. May's A concise course in algebraic topology Section 10.3 in the concrete setting.

Perhaps somewhere already has an answer which I don't know.

  • $\begingroup$ Does it follow from HTT. maybe? $\endgroup$ – Dylan Wilson Nov 28 '18 at 22:13
  • $\begingroup$ You mean cohomological dimension ≤ n is enough to conclude? $\endgroup$ – Lao-tzu Nov 28 '18 at 22:17
  • $\begingroup$ Yeah like, wouldn't you try to factor A-->B with a Postnikov tower and lift a map from X to B up the tower? The obstructions at each stage are maps to EM-objects which start above the dimension of X... some ad hoc stuff is probably required when n<2, as usual $\endgroup$ – Dylan Wilson Nov 28 '18 at 22:21
  • $\begingroup$ Aha, Thanks! I also thought about Moore-Postnikov, but I always try to avoid it whenever possible and I believe there should be some argument can avoid it. $\endgroup$ – Lao-tzu Nov 28 '18 at 22:24

Let $\mathcal{X}$ be the $\infty$-topos in question containing an object $X \in \mathcal{X}$. I assume that by $X$ having homotopy dimension $\leq n$ you mean that the $\infty$-topos $\mathcal{X}_{/X}$ has homotopy dimension $\leq n$ in the sense of Def. of HTT, and that your notion of an $n$-equivalence is the same as the notion of being an $(n+1)$-connective map in the sense of HTT Def. If this is indeed ths case then the answer is yes.

To see this, observe that $X$ having homotopy dimension $\leq n$ means exactly that every $n$-connective map $Y \to X$ admits a section. Now suppose that $A \to B$ is an $(n+1)$-connective map and let $g: X \to B$ be any map. Then the pullback $X \times_B A \to X$ is $(n+1)$-connective, in particular $n$-connective, and hence admits a section $X \to X \times_B A$. Equivalently, the map $g: X \to B$ lifts to $\tilde{g}:X \to A$. It then follows that the map $[X,A] \to [X,B]$ is surjective. To show that it is also injective suppose that $\tilde{g},\tilde{g}': X \to A$ are such that $f\tilde{g}:X \to B$ is homotopic to $f\tilde{g}':X \to B$. Then any choice of homotopy $f\tilde{g} \sim f\tilde{g}'$ determines a map $h:X \to A \times_{B } A$. Using the fact that $A \to B$ is $(n+1)$-connective one can prove that the diagonal map $A \to A \times_{B} A$ is $n$-connective. Arguing as before we see that $h$ lifts to $\tilde{h}:X \to A$. Unwinding the definitions, this exactly means that the homotopy $f\tilde{g} \sim f\tilde{g}'$ lifts to a homotopy $\tilde{g} \sim \tilde{g}'$ and hence $[\tilde{g}] = [\tilde{g}']$ in $[X,A]$.

  • $\begingroup$ You are right, that's exactly what I mean. Thanks for your answer! I was considering some situation in algebraic geometry, where $X$ is a noetherian finite dimensional scheme, which satisfies that the category over it has homotopy dimension ≤dim X by a result of Lurie in DAG XI. $\endgroup$ – Lao-tzu Nov 29 '18 at 21:29

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