1. Introduction

These definitions are needed in what follows and may or may not be familiar to everyone. A continuum X is a compact, connected metric space. A continuum X is indecomposable provided that whenever A and B are proper subcontinua of X, A ∪ B is a proper subset of X; X is hereditarily indecomposable if, and only if, every subcontinuum of X is indecomposable. A map is a continuous function. A map f from a continuum X to a continuum Y is weakly confluent provided that given any continuum M ⊆ Y there exists a continuum W ⊆ X such that f (W) = M. When X is a continuum, C(X) is the hyperspace of subcontinua of X. If a and b are points in ℝⁿ with a ≠ b, [a,b] denotes the line segment from a to b. Let Sⁿ denote the n dimensional sphere. An arc A ⊆ S³ is tame if and only if there is a homeomorphism h: S³ → S³ such that h(A) is an arc of a great circle in S³.

In ^[4] J. W. Rogers, Jr. asked whether every continuum is a continuous image of some indecomposable continuum. The author ^[1] gave an affirmative answer to this question.

Sometime later, in conversation, Rogers asked whether every continuum is a continuous image of some hereditarily indecomposable continuum. This article provides a proof that the answer to this question is also yes.

The author first announced this result in ^[1] but has not published it previously. It has come to my attention that in ^[4] this result has been extended to the non-metric case, building on the metric result.

2. Necessary Lemmas

Lemma 2.1. Let X and Y be continua. Then f: X → Y is weakly confluent if, and only if, the hyperspace map induced by f, C(f): C(X) → C(Y), is surjective.

Proof. This is just a restatement of the definition of weakly confluent.

Lemma 2.2. There exists a hereditarily indecomposable subcontinuum of ℝ⁴which separates ℝ⁴.

Remark on proof. R. H. Bing ^[2] proved this not just for n = 4, but for every n > 1.

Lemma 2.3. Each homotopically essential map from a continuum X to the three sphere, S³, is weakly confluent.

Proof. This was essentially proven, although in a different context, by S. Mazurkiewicz in [5, Theoreme I, p. 328]. This argument gives the necessary details. Let X be a continuum, and suppose g: X → S³ be a homotopically essential map. To prove that g is weakly confluent, it suffices to prove that every tame arc in S³ is equal to g(M) for some continuum M ⊆ X. This follows from Lemma 2.1 because the set of tame arcs is dense in C(S³).

First, set up some machinery and notation, as follows. Let J be a tame arc in S³; let D_n be the closed disk in the complex plane with radius (1/n) centered at 0. Let E_n be the corresponding open disk, and let T_n be the circle D_n \ E_n. Let C_n be the solid cylinder D_n x [0,1]. Since J is a tame, there exists an embedding h of C into S³ such that h({0} x [0,1]) = J. Consider C_n as a subset of S³ by identifying C₁ with h(C₁), and for each t ∈ [0,1] let t denote the point h(0, t) ∈ J.

Let F_n denote the manifold boundary of C_n , that is, F_n= (D_n x {0,1}) ∪ (T_n x [0,1]). Note that given any n and any a, b ∈ J there is an isotopy H: C_n x [0,1] → C_n satisfying the following:

1. (i) for each s ∈ [0,1], H(J x {s}) = J;

2. (ii) for each x ∈ F_n and each t ∈ [0,1], H(x,t) = x;

3. (iii) for every x ∈ C_n, H(x, 0) = x; and

4. (iv) H(b, 1) = a.

By setting H(x, t) = x for every x ∈ S³ \ Cn, and every t ∈ [0, 1], H can be considered to be a function (hence an isotopy) from S³ x [0, 1] to S³.

Now, suppose X is a continuum and let g: X → S³ be a homotopically essential map. To prove that g is weakly confluent, it suffices to prove that there exists a continum M ⊆ X such that g(M) = J.

Proceed by contradiction; assume there is no such M. Then no component of g^-1(J) intersects both g^-1(0) and g^-1(1). By compactness, there is a separation, R₀ ∪ R₁ of g^-1(J) satisfying g^-1(0) ⊆ R₀ and g^-1(1) ⊆ R₁. Since R₀ and R₁ are disjoint closed sets in X, there exist open subsets S₀ and S₁ of X such that R₀ ⊆ S₀ and R₁ ⊆ S₁ and Cl(S₀) ∩ Cl(S₁) = ∅. There exists n such that g^-1(Cn) ⊆ S₀ ∪ S₁. Let p = inf g(R₁) and let q = sup g(R₀), and let a, b ∈ J be such that 0 < a < p and q < b < 1. If p > q, then g is not surjective and hence not essential, so 0 <a<p ≤ q<b< 1. ∪sing the number n and the points a and b just chosen, let H: S³ x [0, 1] → S³ be the isotopy described above. Define a homotopy G: X x [0,1] → S³ by G(x, t) = g(x) if x ∈ X \ S₀ and G(x,t) = H(g(x),t) if x ∈ Cl(S₀). Define f: X → S³ by f (x) = G(x, 1).

Then, note that if y ∈ J and a < y < p, then there does not exist z ∈ X such that f (z) = y, so f is nonsurjective. Hence, f is inessential. Since g is homotopic to f, g is inessential also, a contradiction, which completes the proof.

Lemma 2.4. A continuum X ⊆ R⁴admits a homotopically essential map onto S³if, and only if, R⁴ X is not connected S³.

Remark on Proof. This is a special case of the Borsuk separation theorem. I do not have a reference to the original proof, but a proof can be found in almost any advanced topology or algebraic topology book.

Lemma 2.5. Given any continuum Y, there is a continuum X ⊆ S³that admits a continuous surjection f : X → Y.

Proof. Let Y be a continuum and let C and D be Cantor sets in R³ such that C and D lie on lines skew to each other. Then, whenever a, p ∈ C and b, q ∈ D, and a, p, b, and q are all different, the line segments [a, b] and [p, q] are disjoint. Let g: C ∪ D → Y be a map such that g|C: C → Y and g|D: D → Y are both onto. Such a g exists since a Cantor set can be mapped onto every compact metric space. Define and g(a) = g(b)}. Then X is a continuum in R³. For each x ∈ X, let [a(x), b(x)] be a segment in X satisfying a(x) ∈ C; b(x) ∈ D, and x ∈ [a(x), b(x)]. (This segment is unique unless x = a(x) or x = b(x).) Define f: X → Y by f (x) = g(a(x)) = g(b(x)). It is straightforward to verify that f : X → Y is continuous and onto. Since for any point p ∈ S³, S³ \ {p} is a copy of R³, X can be treated as a subcontinuum of S³.

3. Main Result

Theorem 3.1. Let Y be an arbitrary continuum. There exists a hereditarily indecomposable continuum K that admits a surjective map f: K → Y.

Proof. Let Y be a continuum. By Lemma 2.5, there is a continuum T ⊆ S³ and an onto map g : T → Y. By Lemma 2.2, there exists a hereditarily indecomposable continuum L ⊆ R⁴ that separates R⁴. Thus by Lemma 2.4, there is a homotopically essential map h: L → S³. By Lemma 2.3, h is weakly confluent, so there exists a continuum K C L such that h(K) = T. Let f = g o (h|K). Then f: K → Y is the desired map; K is hereditarily indecomposable since it is a subcontinuum of L.

References

[1] Bellamy D.P., “Continuous mappings between continua", in Topology Conference Guilford College, 1979, Guilford College (1980), 101-112.

[2] Bellamy D.P., “Mappings of indecomposable continua”, Proc. Amer. Math. Soc. 30 (1971), 179-180.

[3] Bing R.H., “Higher-dimensional hereditarily indecomposable continua”, Trans. Amer. Math. Soc. 71 (1951), 267-273.

[4] Hart K.P., van Mill J. and Pol R., “Remarks on hereditarily indecomposable continua”, https://arXiv.org/pdf/math/0010234.pdf

[5] Mazurkiewicz S., “Sur l’existence des continus indécomposables”, Fund. Math. 25 (1935), No. 1, 327-328.

[6] Rogers J.W.Jr., “Continuous mappings on continua”, Proc. Auburn Topology Conference, Auburn University, Auburn, USA, 1969, 94-97.

Notes

Author notes

^* E-mail: bellamy@udel.edu