Proof.
Let K be a field, and let s : \mathop{\mathrm{Spec}}(K) \to S be a morphism. Giving a morphism x : \mathop{\mathrm{Spec}}(K) \to X such that f \circ x = s is the same as giving a section of the projection X_ K = \mathop{\mathrm{Spec}}(K) \times _ S X \to \mathop{\mathrm{Spec}}(K), which in turn is the same as giving a point x \in X_ K whose residue field is K. Hence we see that (2) implies (1).
Conversely, suppose that (1) holds. Assume that x, x' \in X_{S'} map to the same point s' \in S'. Choose a commutative diagram
\xymatrix{ K & \kappa (x) \ar[l] \\ \kappa (x') \ar[u] & \kappa (s') \ar[l] \ar[u] }
of fields. By Schemes, Lemma 26.13.3 we get two morphisms a, a' : \mathop{\mathrm{Spec}}(K) \to X_{S'}. One corresponding to the point x and the embedding \kappa (x) \subset K and the other corresponding to the point x' and the embedding \kappa (x') \subset K. Also we have f' \circ a = f' \circ a'. Condition (1) now implies that the compositions of a and a' with X_{S'} \to X are equal. Since X_{S'} is the fibre product of S' and X over S we see that a = a'. Hence x = x'. Thus (1) implies (2).
If there are two different points x, x' \in X mapping to the same point of s then (2) is violated. If for some s = f(x), x \in X the field extension \kappa (x)/\kappa (s) is not purely inseparable, then we may find a field extension K/\kappa (s) such that \kappa (x) has two \kappa (s)-homomorphisms into K. By Schemes, Lemma 26.13.3 this implies that the map \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(K), X) \to \mathop{\mathrm{Mor}}\nolimits (\mathop{\mathrm{Spec}}(K), S) is not injective, and hence (1) is violated. Thus we see that the equivalent conditions (1) and (2) imply f is radicial, i.e., they imply (3).
Assume (3). By Schemes, Lemma 26.13.3 a morphism \mathop{\mathrm{Spec}}(K) \to X is given by a pair (x, \kappa (x) \to K). Property (3) says exactly that associating to the pair (x, \kappa (x) \to K) the pair (s, \kappa (s) \to \kappa (x) \to K) is injective. In other words (1) holds. At this point we know that (1), (2) and (3) are all equivalent.
Finally, we prove the equivalence of (4) with (1), (2) and (3). A point of X \times _ S X is given by a quadruple (x_1, x_2, s, \mathfrak p), where x_1, x_2 \in X, f(x_1) = f(x_2) = s and \mathfrak p \subset \kappa (x_1) \otimes _{\kappa (s)} \kappa (x_2) is a prime ideal, see Schemes, Lemma 26.17.5. If f is universally injective, then by taking S'=X in the definition of universally injective, \Delta _{X/S} must be surjective since it is a section of the injective morphism X \times _ S X \longrightarrow X. Conversely, if \Delta _{X/S} is surjective, then always x_1 = x_2 = x and there is exactly one such prime ideal \mathfrak p, which means that \kappa (s) \subset \kappa (x) is purely inseparable. Hence f is radicial. Alternatively, if \Delta _{X/S} is surjective, then for any S' \to S the base change \Delta _{X_{S'}/S'} is surjective which implies that f is universally injective. This finishes the proof of the lemma.
\square
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