Assume (2). Let $k$ be an algebraically closed field over $B$. We are going to show that $\sim_R$ is an equivalence relation. Suppose that $\overline{u}_i : \Spec(k) \to U$, $i = 1, 2$ are $k$-valued points of $U$. Suppose that $(\overline{u}_1, \overline{u}_2)$ is the image of a $K$-valued point $r \in R(K)$. Consider the solid commutative diagram $$ \xymatrix{ \Spec(K') \ar@{..>}[r] \ar@{..>}[d] & \Spec(k) \ar[d]_{(\overline{u}_2, \overline{u}_1)} & \Spec(K) \ar[d] \ar[l] \\ R \ar[r]^-j & U \times_B U & R \ar[l]_-{j_{flip}} } $$ We also denote $r \in |R|$ the image of $r$. By assumption the image of $|j_{flip}|$ is contained in the image of $|j|$, in other words there exists a $r' \in |R|$ such that $|j|(r') = |j_{flip}|(r)$. But note that $(\overline{u}_2, \overline{u}_1)$ is in the equivalence class that defines $|j|(r')$ (by the commutativity of the solid part of the diagram). This means there exists a field extension $k \subset K'$ and a morphism $r' : \Spec(K) \to R$ (abusively denoted $r'$ as well) with $j \circ r' = (\overline{u}_2, \overline{u}_1) \circ i$ where $i : \Spec(K') \to \Spec(K)$ is the obvious map. In other words the dotted part of the diagram commutes. This proves that $\sim_R$ is a symmetric relation on $U(k)$. In the similar way, using that the image of $|j_{diag}|$ is contained in the image of $|j|$ we see that $\sim_R$ is reflexive (details omitted). \medskip\noindent To show that $\sim_R$ is transitive assume given $\overline{u}_i : \Spec(k) \to U$, $i = 1, 2, 3$ and field extensions $k \subset K_i$ and points $r_i : \Spec(K_i) \to R$, $i = 1, 2$ such that $j(r_1) = (\overline{u}_1, \overline{u}_2)$ and $j(r_1) = (\overline{u}_2, \overline{u}_3)$. Then we may choose a commutative diagram of fields $$ \xymatrix{ K & K_2 \ar[l] \\ K_1 \ar[u] & k \ar[l] \ar[u] } $$ and we may think of $r_1, r_2 \in R(K)$. We consider the commutative solid diagram $$ \xymatrix{ \Spec(K') \ar@{..>}[r] \ar@{..>}[d] & \Spec(k) \ar[d]_{(\overline{u}_1, \overline{u}_3)} & \Spec(K) \ar[d]^{(r_1, r_2)} \ar[l] \\ R \ar[r]^-j & U \times_B U & R \times_{s, U, t} R \ar[l]_-{j_{comp}} } $$ By exactly the same reasoning as in the first part of the proof, but this time using that $|j_{comp}|((r_1, r_2))$ is in the image of $|j|$, we conclude that a field $K'$ and dotted arrows exist making the diagram commute. This proves that $\sim_R$ is transitive and concludes the proof that (2) implies (1). \medskip\noindent Assume (1) and let $k$ be an algebraically closed field over $B$ whose cardinality is larger than $\lambda(R)$, see Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-large-enough}. Suppose that $\overline{u} \sim_R \overline{u}'$ with $\overline{u}, \overline{u}' \in U(k)$. By assumption there exists a point in $|R|$ mapping to $(\overline{u}, \overline{u}') \in |U \times_B U|$. Hence by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-large-enough} we conclude there exists an $\overline{r} \in R(k)$ with $j(\overline{r}) = (\overline{u}, \overline{u}')$. In this way we see that (1) implies (3). \medskip\noindent Assume (3). Let us show that $\Im(|j_{comp}|) \subset \Im(|j|)$. Pick any point $c \in |R \times_{s, U, t} R|$. We may represent this by a morphism $\overline{c} : \Spec(k) \to R \times_{s, U, t} R$, with $k$ over $B$ having sufficiently large cardinality. By assumption we see that $j_{comp}(\overline{c}) \in U(k) \times U(k) = (U \times_B U)(k)$ is also the image $j(\overline{r})$ for some $\overline{r} \in R(k)$. Hence $j_{comp}(c) = j(r)$ in $|U \times_B U|$ as desired (with $r \in |R|$ the equivalence class of $\overline{r}$). The same argument shows also that $\Im(|j_{diag}|) \subset \Im(|j|)$ and $\Im(|j_{flip}|) \subset \Im(|j|)$ (details omitted). In this way we see that (3) implies (2). At this point we have shown that (1), (2) and (3) are all equivalent. \medskip\noindent It is clear that (4) implies (3) (without any assumptions on $s$, $t$). To finish the proof of the lemma we show that (1) implies (4) if $s, t$ are locally of finite type. Namely, let $k$ be an algebraically closed field over $B$. Suppose that $\overline{u} \sim_R \overline{u}'$ with $\overline{u}, \overline{u}' \in U(k)$. By assumption the algebraic space $Z = R \times_{j, U \times_B U, (\overline{u}, \overline{u}')} \Spec(k)$ is nonempty. On the other hand, since $j = (t, s)$ is locally of finite type the morphism $Z \to \Spec(k)$ is locally of finite type as well (use Morphisms of Spaces, Lemmas \ref{spaces-morphisms-lemma-permanence-finite-type} and \ref{spaces-morphisms-lemma-base-change-finite-type}). Hence $Z$ has a $k$ point by Morphisms of Spaces, Lemma \ref{spaces-morphisms-lemma-locally-finite-type-surjective-geometric-points} and we conclude that $(\overline{u}, \overline{u}') \in j(R(k))$ as desired. This finishes the proof of the lemma.