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=. Exch and *. Epi, From now on, assume that ||f || > #A and assume f ? P. By Lemma 6.12, all images of letters begin with the same letter, or, all images of letters end with the same letter. Assume that the first case holds (the second case is symmetric) and let ? be the first letter of images of letters. Let w ? E. By hypothesis f (w) ? E. As E ? Stab(L ? R) by Theorem 6.3, f (w) = L ? (w ? ) or f (w) = R ? (w ? ) for some letter ? and some word w ? . When f (w) = L ? (w ? ), we have ? = ? and we can find a morphism g such that f = L ? g. By Lemma 6.11, g preserves E. As f ? P, ||g|| < ||f || and by induction g ? S epi . So f ? S epi . Assume now that f (w) = R ? (w ? ). If all images of letters by f end with ?, then f = R ? g and, as in the case f = L ? g above, f ? S epi . If some image of a letter by f does not end with ?, as the image of letters is followed by ? in f (w) and by, Let ||f || = ??A |f (?)|. We prove by induction on ||f || that f ? S epi . Assume first that ||f || = #A. If f (A)

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