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The following code sets up an endomorphism of P¹_an(C₂) defined by f(z) = (z + 3)/10z²:

sage: P.<x,y> = ProjectiveSpace(Qp(2), 1)
sage: f = DynamicalSystem_Berkovich([x*y + 3*y^2, 10*x^2])
sage: B = f.domain()
sage: f
Dynamical system of Projective Berkovich line over Cp(2) of precision 20 induced by the map
  Defn: Defined on coordinates by sending (x : y) to
        (x*y + (1 + 2 + O(2^20))*y^2 : (2 + 2^3 + O(2^21))*x^2)

So far, so good. According to the official documentation "Elements of Berkovich space" (which I cannot link due to insufficient karma), the image of any Type II point ζ_a,r can be computed as f(B(a, r)). But:

sage: [f(B(1,1/2^k)) for k in range(10)]
[Type II point centered at (0 : 1 + O(2^20)) of radius 2^1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^0,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1,
 Type II point centered at (0 : 1 + O(2^20)) of radius 2^-1]

which is impossible because f is only 2-to-1. The correct answer is f(ζ_1,1/2^k) = ζ_{2/5,1/2^k−1} for all k > 0. (Proof: f(z) = Σ_n≥0 (1/10)(−1)ⁿ(3n + 4)(z − 1)ⁿ and the Weierstrass degree of f − c₀ on B(1, r) is 1 assuming r < 1.) What's going on?