It is well known that the Homoclinic Theorem, which conjugates a map near a transversal homoclinic orbit to a Bernoulli subshift, extends from invertible to specific noninvertible dynamical systems. In this paper, we provide a unifying approach, which combines such a result with a fully discrete analog of the conjugacy for finite but sufficiently long orbit segments. The underlying idea is to solve appropriate discrete boundary value problems in both cases, and to use the theory of exponential dichotomies for controlling the errors. This leads to a numerical approach which allows to compute the conjugacy to any prescribed accuracy. The method is demonstrated for several examples where invertibility of the map fails in different ways.Since B J − B ≤ 1 2C B −1 , the Banach Lemma applies and guarantees that B J is invertible and B −1 J ≤ 2C B −1 . Furthermore the following estimates hold true:This implies that for all n ∈ J,Step 3: Existence of a unique solution in B ε (ξ J (s J , N)). We apply the Lipschitz Inverse Mapping Theorem 21 with the setting18 and in case J = Z we define Z = (S Z , · ∞ ). Finally, we choose δ = ε and κ = η 2 . Assumption (63) is satisfied, since by (33) we find that N)).Assumption (64) follows from (35):Thus, Theorem 21 proves that Γ J has a unique zero x J ∈ B ε (ξ J (s J , N)).Step 4:. From (34), we conclude the existence of an ℓ ∈ [1, N ED − 1] such that Proof:Step 1: Definition of h J . Let J be a finite interval, satisfying |J| ≥ N X + 1. Due to Theorem 7, for any s J ∈ Σ J A(N X ) , there exists a unique x J (s J ) ∈ B ε X (ξ J (s J , N X )) such that Γ J (x J (s J )) = 0. Thus we conclude that x J (s J ) ∈ X J (N X , ε X ) and consequently x J (s J ) ∈ Orb(J, O) by Theorem 13. We define h J (s J ) := x J (s J ).Step 2: h J is a homeomorphism. For any x J ∈ Orb(J, O) = X J (N X , ε X ), there exists -due to Theorem 7 -a unique s J ∈ Σ J A(N X ) such that x J ∈ B ε X (ξ J (s J , N X )). Since h J (s J ) = x J , we conclude that h J is invertible. Note that Σ J A(N X ) and Orb(J, O) are finite sets with the same cardinality. Hence, h J is a homeomorphism.Step 3: Proof of conjugacy. For any symbolic sequence s J ∈ Σ J A(N X ) , we get x J = h J (s J ) ∈ B ε X (ξ J (s J , N X )). Let y J−1 = F J (x J ), then it follows that y n = x n+1 for all