A separable and metrizable space X is a matchbox manifold if each point x of X has an open neighborhood which is homeomorphic to 4 x 1 for some zero-dimensional space S x . Each arc component of a matchbox manifold admits a parameterization by the reals 1 in a natural way. This is the main tool in defining the orientability of matchbox manifolds. The orientable matchbox manifolds are precisely the phase spaces of one-dimensional flows without rest points. We show in this paper that a compact homogeneous matchbox manifold is orientable.As an application a new proof is given of Hagopian's theorem that a homogeneous metrizable continuum whose only proper nondegenerate subcontinua are arcs must be a solenoid. This is achieved by combining our work on matchbox manifolds with Whitney's theory of regular curves.

ABSTRACT. In this paper it is proved that every bounded arcwise connected plane continuum which does not separate the plane has the fixed point property.A set X is said to have the fixed point property if each continuous function ƒ on X into itself leaves some point fixed (that is, there is a point x belonging to X such that f(x)=x).The problem "Must a bounded plane continuum which does not separate the plane have the fixed point property?" has motivated a great deal of research in plane topology. K. Borsuk in 1932 proved that every Peano continuum which lies in the plane and does not separate the plane has the fixed point property [2]. Since that time, other general conditions have been found which insure that a plane continuum has this property. In 1967, H. Bell proved that every bounded plane continuum which does not separate the plane and has a hereditarily decomposable boundary has the fixed point property [l]. The following question is still outstanding. If a bounded plane continuum is arcwise connected and does not separate the plane, then must it have the fixed point property? Here an affirmative answer is given to this question by proving the following theorem. If M is a bounded arcwise connected plane continuum which does not have infinitely many complementary domains, then the boundary of M does not contain an indecomposable continuum.Throughout this paper 5 is the set of points of a simple closed surface (that is, a 2-sphere).The proof of the following theorem is based on techniques which are closely related to the folded complementary domain concept defined by F. Burton Jones [3, p. 173]. Key words and phrases. The fixed point property, arcwise connected continua, folded complementary domain, plane continua which do not separate the plane.

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ABSTRACT. F. Burton Jones [7] proved that every decomposable homogeneous plane continuum is either a simple closed curve or a circle of homogeneous nonseparating plane continua. Recently the author [5] showed that no subcontinuum of an indecomposable homogeneous plane continuum is hereditarily decomposable. It follows from these results that every homogeneous plane continuum that has a hereditarily decomposable subcontinuum is a simple closed curve. In this paper we prove that no subcontinuum of an indecomposable homogeneous plane continuum is decomposable. Consequently every homogeneous nonseparating plane continuum is hereditarily indecomposable. 1. Definitions. A space is homogeneous if for each pair p, q of its points there exists a homeomorphism of the space onto itself that takes p to q.A continuum is a nondegenerate compact connected metric space. A continuum is of type A' provided it is irreducible between a pair of its points and admits a monotone upper semicontinuous decomposition, each of whose elements has void interior, whose quotient space is an arc [14].A finite collection {L¡: 1 < z < zz} of open sets is a chain provided that L¡ nLj-^0 if and only if \i-j\ < 1.Throughout this paper R2 is the Cartesian plane with metric p. The closed interval with end points p and q in R2 is denoted by