Proof. [03B4]
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Proof.
By compactness of , there exists a compact strictly -analytic domain such that is a neighbourhood of and . Hence is a compact strictly -analytic domain of and we consider the piecewise linear function on defined by on and by on . Then we apply Proposition 2.6 to , in which case formal metrics correspond to piecewise linear functions (see Proposition 2.8). We deduce that there exists a piecewise linear function which agrees with on and which agrees with on . But since is a neighborhood of , we deduce that the function defined by
is still piecewise linear. Since extends and , we get the claim. ∎