# Is every algebraic space a 1-geometric stack?

In many references (Toen, Higher and derived stacks: a global overview, Toen, Vezzosi, Homotopical algebraic geometry II, and so on), the definition of $n$-geometric stack appears.

In the non-derived case, the definition starts by declaring affine schemes as $(-1)$-geometric stacks and inductively defines $n$-geometric stacks by some procedure.

Toen-Vezzosi, HAG II, Remark 2.1.1.5 says that algebraic spaces and schemes are automatically 1-geometric stacks. I can check that schemes are 1-geometric. (It does not depend on whether a scheme is separated or not.) But I can't check easily that algebraic spaces are 1-geometric stacks.

Always check the definitions being used in the reference - there are even significant differences between different arXiv versions of HAG2. Once you know you have an epimorphism from a union of affines, saying that $X$ is $n$-geometric in the HAG2 v7 sense basically amounts to saying that the higher diagonal $$X \to \mathrm{map}(S^{n},X)$$ is affine. Thus $0$-geometric is equivalent to semi-separated, and any algebraic space $X$ is $1$-geometric because $\mathrm{map}(S^1,X)\cong X$.
EDIT: in response to keaton's comment, the condition isn't quite equivalent to $n$-geometricity, as there are some epimorphism conditions to check, but arises inductively because $$X\times^h_{\mathrm{map}(S^{n},X)}X \cong \mathrm{map}(S^{n+1},X).$$
If you take affines $U,V$ etale over $X$, you want to show that $U\times_XV$ is $0$-geometric, and you already know that it is a scheme. Since the map $U\times_XV \to U \times V$ is a pullback of the diagonal $X \to X \times X$ of $X$, the relative diagonal $U\times_XV \to (U\times_XV)\times^h_{U \times V}(U\times_XV)$ is a pullback of $X \to \mathrm{map}(S^1,X)$, so is an isomorphism. The (absolute) diagonal of $U\times_XV$ is then a pullback of the diagonal of $U\times V$, hence affine.
• Thank you. I think you know something and have intuition. Since I'm a just beginner of derived algebraic geometry, I cannot understand that n-geometric is equivalent to that higher diagonal $X \to map(S^n, X)$ is affine. May I ask you the reason why? Is there a theorem saying this in HAG2? – keaton Feb 20 '18 at 10:19
• Sorry I'm little confusing. What is a definition of $\mathrm{map}(S^n,X)$? I think $X$ lives in the category of functors from $\mathrm{Ho}(sComm)^{op}$ to $\mathrm{SSet}$. Is it make sence to consider a map space from $S^n$ to $X$? – keaton Feb 21 '18 at 4:29
• On the other hand, why $\mathrm{map}(S^1,X)$ is isomorphic to $X$? – keaton Feb 21 '18 at 4:40
• By $\mathrm{map}(S^n,X)$, I just mean the functor $A \mapsto \mathrm{map}(S^n,X(A))$. For set-valued functors $X$, you then have $\mathrm{map}(S^n,X) \simeq X$. – Jon Pridham Feb 21 '18 at 7:56