On hyperbolicity and tautness modulo an analytic subset of complex spaces

JOURNAL OF SCIENCE OF HNUE Mathematical and Physical Sci., 2014, Vol. 59, No. 7, pp. 34-43 This paper is available online at ON HYPERBOLICITY AND TAUTNESS MODULO AN ANALYTIC SUBSET OF COMPLEX SPACES Mai Anh Duc Faculty of Mathematics - Physics and Informatics, Tay Bac University Abstract. The main goal of this article is to give necessary and sufficient conditions on the hyperbolicity or tautness modulo an analytic subset of complex spaces. Keywords: Hyperbolicity modulo, analytic subset, tautness modulo. 1. Introduction The notions of hyperbolicity and tautness modulo an analytic subset of complex spaces are due to S. Kobayashi (see [2, p. 68]). Much attention has been given to these notions, and the results on this problem can be applied to many areas of mathematics, in particular to the extensions of holomorphic mappings. For details see [2] and [3]. The main goal of this article is to give necessary and sufficient conditions on the hyperbolicity or tautness modulo an analytic subset of complex spaces. 2. Basic Notions First of all, we recall the definitions of hyperbolicity and tautness modulo an analytic subset of complex spaces. Definition 2.1. (see [3, p. 68]) Let X be a complex space and S be an analytic subset of X. We say that X is hyperbolic modulo S if for every pair of distinct points p, q of X we have dX(p, q) > 0 unless both are contained in S, where dX is the Kobayashi pseudodistance of X. If S = ∅, then X is said to be hyperbolic. Received July 28, 2014. Accepted August 30, 2014. Contact Mai Anh Duc, e-mail address: ducphuongma@gmail.com 34 On hyperbolicity and tautness modulo an analytic subset of complex spaces Definition 2.2. (see [3, p.240]) LetX be a complex space and S be an analytic subset in X . We say thatX is taut modulo S if it is normal modulo S, i.e., for every sequence {fn} in Hol(D, X) one of the following holds: i. There exists a subsequence of {fn} which converges uniformly to f ∈ Hol(D, X) in Hol(D, X); ii. The sequence {fn} is compactly divergent modulo S in Hol(D, X), i.e., for each compact set K ⊂ D and each compact set L ⊂ X \ S, there exists an integer N such that fn(K) ∩ L = ∅ for all n ≥ N. If S = ∅, then X is said to be taut. It is immediate from the definition that if S ⊂ S ′ ⊂ X and X is taut modulo S, then it is taut modulo S ′, so in particular if X is taut, it is taut modulo S for any analytic subset S. Of course, the converse does not hold. Definition 2.3. (see [5]) Let X be a complex space. X is said to be weakly disc-convex if every sequence {fn} ⊂ Hol(∆, X) converges in Hol(∆, X) whenever the sequence {fn ∆∗} ⊂ Hol(∆∗, X) converges in Hol(∆∗, X). Definition 2.4. (see [1]) Let X be a complex space. A (strictly) plurisubharmonic function on X is a function φ : X → [−∞,∞) having the following property. For every x ∈ X there exists an open neighbourhood U of x in X with a biholomorphic map h : U → V into a closed complex subspace V of some domain G ⊂ Cm and a (strictly) plurisubharmonic function eφ : G→ [−∞,∞) such that φ|U = eφ ◦ h. 3. Main results We now prove Eastwood’s theorem for hyperbolicity and tautness modulo of an analytic subset. First of all, we need the following: Lemma 3.1. Let X,Y be complex spaces and dX , dY be Kobayashi pseudodistances on X,Y, respectively. Let π : X → Y be a holomorphic mapping and Y be a hyperbolic complex space. Let y = π(x) ∈ Y , B(y, s) = {y′ ∈ Y | dY (y, y′) < s}, and V = π−1(B(y, 2s)). Then, for each x ∈ X, there exists a positive constant C > 0 such that dX(x, x ′) ≥ min{s, C.dV (x, x′)} for every x′ ∈ V, where C is dependent of s. Proof. Assume that x′ ∈ V with x′ ̸= x. Let γ be a chain of holomorphic disks joining x to x′ in X : x = x0, x1, ..., xm = x′; q1, q2, ..., qm ∈ ∆; f1, f2, ..., fm ∈ Hol(∆, X) : f1(0) = x0 = x; fi(qi) = fi+1(0), (i = 1, 2, ...,m− 1), fm(qm) = xm = x′. Put Γ = m∪ i=1 fi([0, qi]). It is easy to see that Γ is a compact subset in X . There are two following cases: 35 Mai Anh Duc Case 1. Γ ̸⊂ π−1(B(y, s)). Then there exists j ∈ {1, · · · ,m} such that π(fj(qj)) ̸∈ B(y, s). Hence, we have l(γ) = m∑ i=1 d∆(0, qi) ≥ m∑ i=1 dX(fi(0), fi(qi)) ≥ m∑ i=1 dY (π(fi(0)), π(fi(qi))) ≥ j∑ i=1 dY (π(fi(0)), π(fi(qi))) ≥ dY (y, π(fj(qj))) > s. Thus, we get dX(x, x′) = inf γ l(γ) ≥ s. Case 2. Γ ⊂ π−1(B(y, s)). Let r (depends on s) be a constant such that 0 < r < 1 and d∆(0, z) < s for all z ∈ ∆r := {z ∈ C : |z| < r}. Then there exists a constant C (depends on r) such that C.d∆r(0, z) ≤ d∆(0, z) for any z ∈ ∆ r2 . Without loss of generality we may assume that qi ∈ ∆ r 2 for 1 ≤ i ≤ m. Since π(Γ) ⊂ B(y, s), it follows that π(fi(qi)) ∈ B(y, s), π(fi(0)) ∈ B(y, s) ∀1 ≤ i ≤ m. It is easy to see, for each z ∈ ∆r and for 1 ≤ i ≤ m, that dY (π(fi(0)), π(fi(z))) ≤ dX(fi(0), fi(z)) ≤ d∆(0, z) < s. Since dY (y, π(fi(z)) ≤ dY (y, π(fi(0))) + dY (π(fi(0)), π(fi(z))) < s+ s = 2s, π(fi(∆r)) ⊂ B(y, 2s). Thus fi(∆r) ⊂ V for z ∈ ∆r and for 1 ≤ i ≤ m. It implies that l(γ) = m∑ i=1 d∆(0, qi) ≥ C m∑ i=1 d∆ r 2 (0, qi) ≥ C m∑ i=1 dV (fi(0), fi(qi)) ≥ C.dV (x, x′). Therefore, dX(x, x′) = inf γ l(γ) ≥ C.dV (x, x′). Theorem 3.1. (Eastwood theorem for hyperbolicity modulo of an analytic subset) Let π : X → Y be a holomorphic map between complex spaces and SY be an analytic subset in Y . Assume that Y is hyperbolic modulo SY and for each y ∈ Y \ SY , there is an open neighbourhood U of y in Y \ SY such that π−1(U) is hyperbolic. Then X is hyperbolic modulo SX := π−1(SY ). 36 On hyperbolicity and tautness modulo an analytic subset of complex spaces Proof. Assume that x, x′ ∈ X such that x ̸= x′ and x ̸∈ SX . We now show that dX(x, x ′) > 0. Indeed, consider two following cases. Case 1. π(x) ̸= π(x′). Since x ̸∈ SX , π(x) ̸∈ π(SX) = SY , Y is hyperbolic modulo SY and π is a holomorphic map, it implies that dX(x, x′) ≥ dY (π(x), π(x′)) > 0. Case 2. π(x) = π(x′) = y ∈ Y. Since x ̸∈ SX and y ̸∈ SY , we get y ∈ Y \ SY . Therefore, there exists an open neighbourhood U of y in Y \ SY such that π−1(U) is hyperbolic. Let s be a small enough positive constant such that B(y, 2s) ⊂ U . Let V = π−1(B(y, 2s)) ⊂ π−1(U). Since π−1(U) is hyperbolic, V is hyperbolic, and hence dV (x, x′) > 0. Using Lemma 3.1, there exists a constant C > 0 such that dX(x, x′) ≥ min{s, C.dV (x, x′)} > 0. Thus X is hyperbolic modulo SX . Theorem 3.2. (Eastwood theorem for tautness modulo of an analytic subset) Let X, eX be complex spaces and S be an analytic subset in X . Assume that eX is weakly disc-convex and π : eX → X is a holomorphic mapping such that for each x ∈ X \ S, there exists an open neighbourhood U of x inX \ S such that π−1(U) is taut. Then eX is taut modulo eS := π−1(S) if X is taut modulo S. Proof. Suppose that X is taut modulo S and { efn} ⊂ Hol(∆, eX) is not compactly divergent modulo S inHol(∆, X). Hence, without loss of generality there exist a compact set K ⊂ ∆ and a compact set L ⊂ eX \ eS such that efn(K) ∩ L ̸= ∅ for n ≥ 1. For each n ≥ 1, there exists zn ∈ K ⊂ ∆ such that efn(zn) ∈ L. Since K and L are compact sets, by taking subsequences if necessary, we may assume that {zn} ⊂ K ⊂ ∆ such that zn → z ∈ K ⊂ ∆ and efn(zn) → ep ∈ L ⊂ eX \ eS. Since π : eX → X is holomorphic, π(L) is a compact set in X \ S. Therefore, {fn := πo efn} ⊂ Hol(∆, X) is not compactly divergent modulo S in Hol(∆, X). Since X is taut modulo S, we may assume that {fn} converges uniformly to a mapping F ∈ Hol(∆, X). Obviously, π( efn(zn)) → π(ep) and π( efn(zn)) = πo efn(zn) = fn(zn) → F (z) as n → ∞. Therefore, we can let p = π(ep) = F (z). Since ep ∈ eX \ eS, p = π(ep) ̸∈ S. According to the hypothesis of this theorem, there exists an open neighbourhood U of p in X \ S such that π−1(U) is taut. Taking an open neighbourhood V ⊂ F−1(U) of z in ∆ \ F−1(S) and since the sequence {fn} converges uniformly to a mapping F , we may assume that fn(V ) ⊂ U . This implies thatefn(V ) ⊂ π−1(U) for every n ≥ 1. Note that efn(zn) converges to ep, so put K = {zn} ∪ {z} and L = { efn(zn)} ∪ {ep}. Then efn(K) ∩ L ̸= ∅ for all n. Thus, the sequence { efnV } is not compactly divergent modulo eS in Hol(∆, eX). On the other hand, since π−1(U) is taut and efn(V ) ⊂ π−1(U), { efnV } converges uniformly to mapping eF in Hol(V, π−1(U)). 37 Mai Anh Duc Consider a family Γ consisting of pairs (W,Φ), where W is open in ∆ \ F−1(S) and Φ ∈ Hol(W, eX \ eS) such that there exists a subsequence { efnkW} of { efnW} which converges uniformly to mapping Φ in Hol(W, eX \ eS). According to the proof above, we have Γ ̸= ∅. In the family Γ, consider the following order relation: (W1,Φ1) ≤ (W2,Φ2) if i)W1 ⊂ W2 and ii) for any subsequence { efnkW1} of { efnW1} that converges uniformly to mapping Φ1 in Hol(W1, eX \ eS), there exists a subsequence of { efnk} that converges uniformly to mapping Φ2 in Hol(W2, eX \ eS). Assume that {(Wα,Φα)}α∈Λ is a complete order subset of Γ. Let W0 = ∪ α∈Λ Wα and define a mapping Φ0 ∈ Hol(W0, eX \ eS) such that Φ0W = Φα for α ∈ Λ. Take a sequence {(Wi,Φi)}∞i=1 ⊂ {(Wα,Φα)}α∈Λ such that (W1,Φ1) ≤ (W2,Φ2) ≤ · · · , W0 = ∞∪ i=1 Wi. As the definition of Γ, the subsequence { ef 1nW1} of { efnW1} converges uniformly to the mapping Φ1 in Hol(W1, eX \ eS). As the definition of the order relation on Γ, there exists a subsequence { ef 2n} of { ef 1n} which converges uniformly to Φ2 in Hol(W2, eX \ eS). By continuing this process, we get the sequence { efkn} such that { efkn} ⊂ { efk−1n } for all k ≥ 2 and { efkn Wk} converges uniformly to Φk in Hol(Wk, eX \ eS). Thus, a diagonal sequence { efkk } converges uniformly to Φ0 in Hol(W0, eX \ eS). Hence (W0,Φ0) ∈ Γ and the subset {(Wα,Φα)}α∈Λ of Γ has a supremum. By Zorn lemma, there exists a maximum element (W,Φ) of the family Γ. Assume that { efnkW} is a subsequence of { efnW} such that { efnkW} converges uniformly to Φ in Hol(W, eX \ eS). We now show that W = ∆ \ F−1(S). Suppose that there exists z0 ∈ W ∩ (∆ \ F−1(S)). Take an open neighbourhood U of F (z0) in X \ S such that π−1(U) is taut. Since {fn} converges uniformly to mapping F in Hol(∆, X), there exists an open neighbourhood W0 of z0 in ∆ \ F−1(S) such that π ◦ efn(W0) ⊂ U . Henceefn(W0) ⊂ π−1(U), for all n ≥ 1. Fix z1 ∈ W0 ∩ W . Then the sequence { efnk(z1)} is convergent. Since Hol(W0, π −1(U)) is a normal family, { efnkW0} converges uniformly to Φ0 in Hol(W0, π −1(U)). Thus (W0,Φ0) ∈ Γ. It implies thatW0 ⊂ W . HenceW = ∆\F−1(S). Since F−1(S) is an analytic subset in the open unit disc∆, F−1(S) is a discrete set and F−1(S) does not have any accumulation point. Therefore, for each z ∈ F−1(S), there exists a number 0 < r < 1 such that B(z, r) ∩ F−1(S) = {z}. Thus, ∆ \ F−1(S) is ∆∗. 38 On hyperbolicity and tautness modulo an analytic subset of complex spaces Hence the sequence { efnk∆∗} is a subsequence of { efn∆∗}, where { efnk∆∗} converges uniformly in Hol(∆∗, eX). Since eX is weakly disc-convex, { efnk} converges uniformly in Hol(∆, eX). Thus eX is taut modulo eS. We now prove the invariance of hyperbolicity and tautness modulo an analytic subset under holomorphic covering mappings and holomorphic fiber mappings. Theorem 3.3. Let π : eX → X be a covering space of a complex space X and S be an analytic subset in X . Then eX is taut modulo π−1(S) if and only if X is taut modulo S. Proof. (⇒) Assume that eX is taut modulo eS := π−1(S) and {fn} ⊂ Hol(∆, X) is not compactly divergent modulo S inHol(∆, X). Then there exists {zn} ⊂ ∆ such that {zn} converges to z ∈ ∆ and {yn := fn(zn)} converges to p ∈ X \ S. Take ep ∈ π−1(p). Since dX(yn, p) = infeyn∈π−1(yn) d eX( eyn, ep), for each n ≥ 1 there exists eyn ∈ π−1(yn) such that d eX(eyn, ep) < dX(yn, p) + 1n . Since {yn} converges to p ∈ X \ S as n→∞, d eX(eyn, ep)→ 0 as n→∞. Moreover, since p ∈ X \ S, ep ∈ eX \ eS and since eX is taut modulo eS, eX is hyperbolic modulo eS. Thus eyn converges to ep as n→∞. Lifting fn to efn such that πo efn = fn and efn(zn) = eyn. Since eyn converges toep, { efn} ⊂ Hol(∆, eX) is not compactly divergent modulo eS in Hol(∆, eX). Since eX is taut modulo eS, { efn} converges uniformly to a mapping ef in Hol(∆, eX). Hence {fn} converges uniformly to a mapping f := π ◦ ef in Hol(∆, X). So X is taut modulo S. (⇐) Assume that X is taut modulo S. For each x ∈ X \ S, take an open neighbourhood U of x in X \ S such that U ∩ S = ∅ and U is taut. Since π : eX → X is covering holomorphic, π−1(U) = ∪ i∈I Ui, where Ui are open subsets in eX such that Ui ∩ Uj = ∅ (i, j ∈ I, i ̸= j) and π Ui : Ui → U is biholomorphic for all i. This implies that Ui is taut. So π−1(U) = ∪ i∈I Ui is taut. Assume that { efn} ⊂ Hol(∆, eX) is not compactly divergent modulo eS in Hol(∆, eX). Repeating the proof of Theorem 3.2, it follows that { efnk∆∗} is a subsequence of { efn∆∗} such that { efnk∆∗} converges uniformly to eΦ inHol(∆∗, eX) and π◦eΦ = F ∆∗ . Fix z0 ∈ ∆∗. Since F ∈ Hol(∆, X) and π : eX → X is a holomorphic covering mapping, we can lift F to a mapping Φ ∈ Hol(∆, eX) such that π ◦ Φ = F and eΦ(z0) = Φ(z0). Since eΦ, Φ ∆∗ are lifting mappings of F ∆∗ with eΦ(z0) = Φ(z0) and by the unique property of lifting mapping, it implies that eΦ = Φ ∆∗ . Thus Φ is a holomorphic extension of eΦ from ∆∗ to ∆. Hence eX is taut modulo eS. 39 Mai Anh Duc Theorem 3.4. Let X be a complex space and S be an analytic subset in X . Assume that X = ∪ i∈I Xi, where {Xi} are irreducible components of X. Then X is taut modulo S if and only if Xi is taut modulo Si := Xi ∩ S for all i ∈ I . Proof. Assume that {Xi}i∈I are irreducible components of X. Then {Xi}i∈I is a locally finite family and Hol(∆, X) = ∪ i∈I Hol(∆, Xi). (⇒) Assume thatX is taut modulo S. Fix i ∈ I and assume that {fn} ⊂ Hol(∆, Xi) is not compactly divergent modulo Si inHol(∆, Xi). Then there exists {zn} ⊂ ∆ such that {zn} converges to z ∈ ∆ and {fn(zn)} converges to a point p ∈ Xi\Si. SinceX is taut modulo S, {fn} ⊂ Hol(∆, X) converges uniformly to f in Hol(∆, X) = ∪ i≥1 Hol(∆, Xi). Thus Xi is taut modulo Xi ∩ S for all i. (⇐) Assume that Xi is taut modulo Si for all i and {fn} ⊂ Hol(∆, X) is not compactly divergent modulo S in Hol(∆, X). It implies that there exist a compact subset K ⊂ ∆, a compact subset K ′ ⊂ X \ S, and a subsequence {fnk} of {fn} such that fnk(K) ∩ K ′ ̸= ∅ for all k ≥ 1. We may assume that fn(K) ∩ K ′ ̸= ∅ for all n ≥ 1. By the locally finite property of the family {Xi}i≥1, there exists n0 such thatK ′ ⊂ n0∪ i=1 Xi and K ′ ∩ Xi = ∅ for all i > n0. Since fn(K) ∩ K ′ ̸= ∅ and fn(K) ⊂ fn(∆) for all n ≥ 1, fn(∆) ̸⊂ Xi for all i > n0. Since fn(∆) is a subset of some irreducible component of X, it follows that fn(∆) ⊂ n0∪ i=1 Xi for all n ≥ 1, and hence, there exist an index i0 ∈ {1, 2, ..., n0} and a subsequence {fnk} of {fn} such that fnk(∆) ⊂ Xi for all k ≥ 1. SinceXi is taut modulo Si for all i and {fnk} is not compactly divergent modulo Si inHol(∆, Xi), {fnk} has a subsequence which is convergent uniformly inHol(∆, Xi) ⊂ Hol(∆, X). Hence X is taut modulo S. Theorem 3.5. Let X be a complex space and S be an analytic subset in X . Let D be a Cartier divisor in X . Then X \D is taut modulo S \D if X is taut modulo S. Proof. Assume that X is taut modulo S. We now show that X \D is taut modulo S \D. Assume that {fn} ⊂ Hol(∆, X \ D) is not compactly divergent modulo S \ D in Hol(∆, X \ D). Then there exist a compact subset K ⊂ ∆, a compact subset L ⊂ (X \ D) \ (S \ D) = X \ (S ∪ D) such that fn(K) ∩ L ̸= ∅ for all n ≥ 1. Thus {fn} is not compactly divergent modulo S in Hol(∆, X). Hence {fn} converges uniformly on compact subsets of ∆ to a mapping f in Hol(∆, X). We now show that f(∆) ∩D = ∅. Suppose on the contrary that f(∆)∩D ̸= ∅. Let A = f−1(D). It is easy to see that A ̸= ∅ and A is closed in ∆. We have to prove that A is open in ∆. Take z0 ∈ A and U is a neighbourhood of f(z0) ∈ D inX such that U ∩D = {x ∈ X : φ(x) = 0}, where φ ∈ Hol(∆,C). Choose relatively compact complete hyperbolic 40 On hyperbolicity and tautness modulo an analytic subset of complex spaces neighbourhoodsW1,W2 of f(z0) such thatW1 ⊂ W2 ⊂ W2 ⊂ U . Since {fn} converges uniformly to f in Hol(∆, X), there exists an open neighbourhood V of z0 in ∆ such that fn(V ) ⊂ W1 \ D for all n ≥ 1. Since W2 is complete hyperbolic, W2 \ D is complete hyperbolic too and thusW2 \D is taut i.e., the family Hol(V,W2 \D) is normal. Assume that there exists z ∈ V such that f(z) ̸∈ D. Since {fn(z)} converges uniformly to f(z) ∈ W2 \D, {fn V } is not compactly divergent modulo (W2 ∩ S) \D in Hol(V,W2 \D). SinceHol(V,W2 \D) is normal, we can assume that {fn V } converges uniformly to ef ∈ Hol(V,W2 \ D). It implies that ef(z0) = f(z0) ∈ D. This contradics the hypothesis that ef ∈ Hol(V,W2 \D). Thus f(V ) ⊂ D i.e.,V ⊂ f−1(D) = A. Hence A is an open subset ∆. Thus A = ∆ i.e., f(∆) ⊂ D. This is a contradition. We get f(∆) ∩D = ∅ and hence X \D is taut modulo S \D. Lemma 3.2. Let E be a holomorphic bundle on M with a hyperbolic bundle F and the projection π : E →M , where E,F,M are complex manifold. Then dM(x, y) = dE(ex, π−1(y)) = infey∈π−1(y) dE(ex, ey) for x, y ∈M , and ex ∈ π−1(x). Proof. Obviously, dM(x, y) ≤ dE(ex, π−1(y)) for ex ∈ π−1(x). Now we proof dM(x, y) ≥ dE(ex, π−1(y)) for ex ∈ π−1(x). Indeed, take two arbitrary points x, y ∈ M and ex ∈ π−1(x). Assume that a1, a2, ..., ak ∈ ∆ and f1, f2, ..., fk ∈ Hol(∆,M) such that f1(0) = a1, fi(ai) = fi+1(0) for 1 ≤ i ≤ k − 1, and fk(ak) = y. Consider a pullback bundle ∆×M E σ1−−−→ E θ1 y yπ ∆ −−−→ f1 M. Then there exists an isomorphimsΦ1 : ∆×F → ∆×ME between two holomorphic bundles on ∆. Thus, there exists a point c1 ∈ F such that σ1 ◦ Φ1(0, c1) = ex. Let φ1 : ∆→ E be a holomorphic map defined by φ1(z) = σ1 ◦ Φ1(z, c1) for z ∈ ∆. Consider a pullback bundle ∆×M E σ2−−−→ E θ2 y yπ ∆ −−−→ f2 M. 41 Mai Anh Duc Then there exists an isomorphimsΦ2 : ∆×F → ∆×ME between two holomorphic bundles on ∆. Thus, there exists a point c2 ∈ F such that σ2 ◦ Φ2(0, c2) = φ1(a1). Let φ2 : ∆→ E be a holomorphic map defined by φ2(z) = σ2 ◦ Φ2(z, c2) for z ∈ ∆. Continuing this process, it implies that φ1, φ2, ..., φk ∈ Hol(∆, E) satisfying φ1(0) = ex, φi(ai) = φi+1(0) for 1 ≤ i ≤ k − 1 and φk(ak) ∈ π−1(y). Therefore, dM(x, y) ≥ dE(ex, π−1(y)) for ex ∈ π−1(x). Theorem 3.6. Let E be a holomorphic bundle on M with a taut bundle F and the projection π : E → M , where E,F,M are complex manifolds. Let S be an analytic subset inM . Then E is taut modulo eS := π−1(S) if and only ifM is taut modulo S. Proof. (⇒) Assume that E is taut modulo eS and {fn} ⊂ Hol(∆,M) is not compactly divergent modulo S in Hol(∆,M). Then there exists {zn} ⊂ ∆ such that {zn} → z ∈ ∆ and {yn := fn(zn)} → p ∈ M \ S. Take ep ∈ π−1(p) ⊂ E \ eS. By Lemma 3.2, it implies thatdM(p, yn) = infeyn∈π−1(yn) dE(ep, eyn). Hence, for each n ≥ 1, there exists eyn ∈ E such that dE(ep, eyn) < dM(p, yn) + 1 n . Since dM(p, yn) + 1n → 0 as n → ∞, dE(ep, eyn) = 0. On the other hand, E is hyperbolic modulo eS since E is taut modulo eS. Hence eyn → ep as n→∞. For each n ≥ 1, consider the pullback bundle θn : ∆ ×M E → ∆ of the bundle π : E →M . We have the following diagram. ∆×M E σn−−−→ E θn y yπ ∆ −−−→ fn M. By the proof of Lemma 3.2, there exist an isomorphism Φn : ∆ × F → ∆ ×M E between two holomorphic bundles on∆ and cn ∈ F such that σn◦Φn(0, cn) = eyn. Define the holomorphic map φn : ∆→ E by φn(z) = σn ◦ Φn(z, cn) for all z ∈ ∆. It is easy to see that φn(zn)→ ep ∈ E. SinceE is taut modulo eS, the sequence φn converges uniformly to a φ inHol(∆, E) as n→∞. Thus, {fn} cenverges uniformly to a mapping f := π ◦φ in Hol(∆,M). Hence,M is taut modulo S. (⇐) Assume thatM is taut mo