**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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