Proof.
Proof of (1). Say y \to x lies over f : B \to A. Let y' \to y lying over B' \subset B be a minimal object of \mathcal{S}_ y. Let
\vcenter { \xymatrix{ y' \ar[d] \ar[r] & x' \ar[d] \\ y \ar[r] & x } } \quad \text{lying over}\quad \vcenter { \xymatrix{ B' \ar[d] \ar[r] & f(B') \ar[d] \\ B \ar[r] & A } }
be as in the construction of f_* above. Suppose that (x'' \to x) \to (x' \to x) is a morphism of \mathcal{S}_ x with x'' \to x' lying over A'' \subset f(B'). By (S1) there exists y'' \to y' lying over B' \times _{f(B')} A'' \to B'. Since y' \to y is minimal we conclude that B' \times _{f(B')} A'' \to B' is an isomorphism, which implies that A'' = f(B'), i.e., x' \to x is minimal.
Proof of (2). Suppose f : B \to A is surjective and y \to x lies over f. Let x' \to x be a minimal object of \mathcal{S}_ x lying over A' \subset A. By (S1) there exists y' \to y lying over B' = f^{-1}(A') = B \times _ A A' \to B whose image in \mathcal{S}_ x is x' \to x. So f_*(y' \to y) = x' \to x. Choose a morphism (y'' \to y) \to (y' \to y) in \mathcal{S}_ y with y'' \to y a minimal object (this is possible by the remark on lengths above the lemma). Then f_*(y'' \to y) is an object of \mathcal{S}_ x which maps to x' \to x (by functoriality of f_*) hence is isomorphic to x' \to x by minimality of x' \to x.
\square
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