For an \'etale 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 \ref{definition-tor-independent}. 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 \ref{spaces-properties-lemma-geometric-lift-to-usual} 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. \medskip\noindent Having made these initial points, it is clear that (4) comes down to the statement that Definition \ref{definition-tor-independent} agrees with Derived Categories of Schemes, Definition \ref{perfect-definition-tor-independent} when $X$, $Y$, and $B$ are schemes. \medskip\noindent Let $\overline{x}, \overline{b}, \overline{y}$ be as in Definition \ref{definition-tor-independent} lying over the points $x, y, b$. Recall that $\mathcal{O}_{X, \overline{x}} = \mathcal{O}_{X, x}^{sh}$ (Properties of Spaces, Lemma \ref{spaces-properties-lemma-describe-etale-local-ring}) and similarly for the other two. By Algebra, Lemma \ref{algebra-lemma-strictly-henselian-functorial-improve} 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 \ref{more-algebra-lemma-dumb-properties-henselization}). By More on Algebra, Lemma \ref{more-algebra-lemma-tor-independent-flat} 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 $B$. \medskip\noindent 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 \ref{more-algebra-lemma-tor-independent} 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 $B$.