# A finite Whitehead Theorem for $\infty$-topos

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.

• Does it follow from HTT.7.2.2.30 maybe? – Dylan Wilson Nov 28 '18 at 22:13
• You mean cohomological dimension ≤ n is enough to conclude? – Lao-tzu Nov 28 '18 at 22:17
• 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 – Dylan Wilson Nov 28 '18 at 22:21
• 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. – 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. 7.2.1.1 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. 6.5.1.10. 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 $$[X,A]$$.
• 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. – Lao-tzu Nov 29 '18 at 21:29