Proof.
For an étale morphism $\varphi : U \to X$ of algebraic spaces and geometric point $\overline{u}$ the map of local rings $\mathcal{O}_{X, \varphi (\overline{u})} \to \mathcal{O}_{U, \overline{u}}$ is an isomorphism. Hence the equivalence of (1) and (2) follows. So does the implication (1) $\Rightarrow $ (3). Assume (3) and pick a diagram of geometric points as in Definition 75.20.2. The assumptions imply that we can first lift $\overline{b}$ to a geometric point $\overline{w}$ of $W$, then lift the geometric point $(\overline{x}, \overline{b})$ to a geometric point $\overline{u}$ of $U$, and finally lift the geometric point $(\overline{y}, \overline{b})$ to a geometric point $\overline{v}$ of $V$. Use Properties of Spaces, Lemma 66.19.4 to find the lifts. Using the remark on local rings above we conclude that the condition of the definition is satisfied for the given diagram.
Having made these initial points, it is clear that (4) comes down to the statement that Definition 75.20.2 agrees with Derived Categories of Schemes, Definition 36.22.2 when $X$, $Y$, and $B$ are schemes.
Let $\overline{x}, \overline{b}, \overline{y}$ be as in Definition 75.20.2 lying over the points $x, y, b$. Recall that $\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{X, x}^{sh}$ (Properties of Spaces, Lemma 66.22.1) and similarly for the other two. By Algebra, Lemma 10.155.12 we see that $\mathcal{O}_{X, \overline{x}}$ is a strict henselization of $\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{B, b}} \mathcal{O}_{B, \overline{b}}$. In particular, the ring map
\[ \mathcal{O}_{X, x} \otimes _{\mathcal{O}_{B, b}} \mathcal{O}_{B, \overline{b}} \longrightarrow \mathcal{O}_{X, \overline{x}} \]
is flat (More on Algebra, Lemma 15.45.1). By More on Algebra, Lemma 15.61.3 we see that
\[ \text{Tor}_ i^{\mathcal{O}_{B, b}}(\mathcal{O}_{X, x}, \mathcal{O}_{Y, y}) \otimes _{\mathcal{O}_{X, x} \otimes _{\mathcal{O}_{B, b}} \mathcal{O}_{Y, y}} (\mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{B, \overline{b}}} \mathcal{O}_{Y, \overline y}) = \text{Tor}_ i^{\mathcal{O}_{B, \overline{b}}}( \mathcal{O}_{X, \overline{x}}, \mathcal{O}_{Y, \overline{y}}) \]
Hence it follows that if $X$ and $Y$ are Tor independent over $B$ as schemes, then $X$ and $Y$ are Tor independent as algebraic spaces over $B$.
For the converse, we may assume $X$, $Y$, and $B$ are affine. Observe that the ring map
\[ \mathcal{O}_{X, x} \otimes _{\mathcal{O}_{B, b}} \mathcal{O}_{Y, y} \longrightarrow \mathcal{O}_{X, \overline{x}} \otimes _{\mathcal{O}_{B, \overline{b}}} \mathcal{O}_{Y, \overline y} \]
is flat by the observations given above. Moreover, the image of the map on spectra includes all primes $\mathfrak s \subset \mathcal{O}_{X, x} \otimes _{\mathcal{O}_{B, b}} \mathcal{O}_{Y, y}$ lying over $\mathfrak m_ x$ and $\mathfrak m_ y$. Hence from this and the displayed formula of Tor's above we see that if $X$ and $Y$ are Tor independent over $B$ as algebraic spaces, then
\[ \text{Tor}_ i^{\mathcal{O}_{B, b}} (\mathcal{O}_{X, x}, \mathcal{O}_{Y, y})_\mathfrak s = 0 \]
for all $i > 0$ and all $\mathfrak s$ as above. By More on Algebra, Lemma 15.61.6 applied to the ring maps $\Gamma (B, \mathcal{O}_ B) \to \Gamma (X, \mathcal{O}_ X)$ and $\Gamma (B, \mathcal{O}_ B) \to \Gamma (X, \mathcal{O}_ X)$ this implies that $X$ and $Y$ are Tor independent over $B$.
$\square$
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