Proof.
Assume \mathcal{F} has (S1). Suppose we have ring maps f_ i : A_ i \to A in \mathcal{C}_\Lambda with f_2 surjective. Let x_ i \in \mathcal{F}(A_ i) such that the pushforwards f_{1, *}(x_1) and f_{2, *}(x_2) are isomorphic. Then we can denote x an object of \mathcal{F} over A isomorphic to both of these and we obtain a diagram as in (S1). Hence we find an object y of \mathcal{F} over A_1 \times _ A A_2 whose pushforward to A_1, resp. A_2 is isomorphic to x_1, resp. x_2. In this way we see that (S1) holds for \overline{\mathcal{F}}.
Assume \mathcal{F} has (S2). The first part of (S2) for \overline{\mathcal{F}} follows as in the argument above. The second part of (S2) for \overline{\mathcal{F}} signifies that the map
\overline{\mathcal{F}}(A \times _ k k[\epsilon ]) \to \overline{\mathcal{F}}(A) \times _{\overline{\mathcal{F}}(k)} \overline{\mathcal{F}}(k[\epsilon ])
is injective for any ring A in \mathcal{C}_\Lambda . Suppose that y, y' \in \mathcal{F}(A \times _ k k[\epsilon ]). Using the axioms of cofibred categories we can choose commutative diagrams
\vcenter { \xymatrix{ y \ar[r]_ c \ar[d]_ a & x_\epsilon \ar[d]^ e \\ x \ar[r]^ d & x_0 } } \quad \text{and}\quad \vcenter { \xymatrix{ y' \ar[r]_{c'} \ar[d]_{a'} & x'_\epsilon \ar[d]^{e'} \\ x' \ar[r]^{d'} & x'_0 } } \quad \text{lying over}\quad \vcenter { \xymatrix{ A \times _ k k[\epsilon ] \ar[d] \ar[r] & k[\epsilon ] \ar[d] \\ A \ar[r] & k } }
Assume that there exist isomorphisms \alpha : x \to x' in \mathcal{F}(A) and \beta : x_\epsilon \to x'_\epsilon in \mathcal{F}(k[\epsilon ]). This also means there exists an isomorphism \gamma : x_0 \to x'_0 compatible with \alpha . To prove (S2) for \overline{\mathcal{F}} we have to show that there exists an isomorphism y \to y' in \mathcal{F}(A \times _ k k[\epsilon ]). By (S2) for \mathcal{F} such a morphism will exist if we can choose the isomorphisms \alpha and \beta and \gamma such that
\xymatrix{ x \ar[d]^\alpha \ar[r] & x_0 \ar[d]^\gamma & x_\epsilon \ar[d]^\beta \ar[l]^ e \\ x' \ar[r] & x'_0 & x'_\epsilon \ar[l]_{e'} }
is commutative (because then we can replace x by x' and x_\epsilon by x'_\epsilon in the previous displayed diagram). The left hand square commutes by our choice of \gamma . We can factor e' \circ \beta as \gamma ' \circ e for some second map \gamma ' : x_0 \to x'_0. Now the question is whether we can arrange it so that \gamma = \gamma '? This is clear if \mathcal{F}(k) is a set, or a setoid. Moreover, if \text{Aut}_{k[\epsilon ]}(x_\epsilon ) \to \text{Aut}_ k(x_0) is surjective, then we can adjust the choice of \beta by precomposing with an automorphism of x_\epsilon whose image is \gamma ^{-1} \circ \gamma ' to make things work.
\square
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