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math/9908040 | Using the description of the operad MATH in terms of generators and relations, it is a straightforward exercise to check that the quadratic dual of MATH is again MATH up to a shift of grading and the change of it to the opposite. Then from the proof of REF , one can identify MATH with the cobar construction of MATH. The natural homomorphism, see REF, which easily generalizes to the graded case, from the cobar construction of an operad MATH to the quadratic dual MATH gives for MATH a morphism of operads MATH . It is also known that the operad MATH is NAME, see CITE or a purely algebraic proof by CITE. This means (by definition of CITE) that the morphism REF is a quasi-isomorphism. According to CITE, the cobar construction of the quadratic dual of a NAME operad (see CITE) is a minimal model of that operad. Applied to MATH, this implies that MATH is a minimal model of MATH. |
math/9908040 | CASE: First of all, let us prove that MATH is free. Recall that MATH . Since MATH's are made out of a stratification of MATH (see REF), we have MATH where for each MATH, MATH is the filtration from REF. Therefore, passing to cohomology, we have MATH which by definition means that MATH is a free operad generated by the collection MATH of graded vector spaces with an action of the symmetric group MATH, MATH. CASE: The next step is to show that the spectral sequence MATH collapses at the second term MATH. In order to show that the cohomology of MATH is MATH, regard MATH as a filtered complex, with the MATH-th filtration component defined by the tree degree MATH. We will compute the homology of MATH using the spectral sequence associated with this filtration. The first term of this spectral sequence is MATH with the NAME homomorphism as the differential, because of REF , that is the first term MATH of the spectral sequence associated with the filtration MATH, see REF shows that the homology of MATH is isomorphic to MATH. By construction this is the second term of the spectral sequence associated to the filtered complex MATH and the spectral sequence converges to the homology MATH of MATH. On the other hand, MATH is the MATH term MATH. Thus MATH is the second term of a spectral sequence converging to the second term of another spectral sequence converging to the same MATH. This implies that both spectral sequences collapse at the second terms. Therefore, the homology of MATH is MATH. |
math/9908040 | The operad MATH is freely generated by the spaces MATH, MATH, see REF. Thus to define a morphism MATH, it suffices to define maps MATH respecting the gradings, the symmetric group actions, and the differentials. The complements MATH and MATH are disjoint unions of NAME cells, and the spaces MATH and MATH have the NAME cells of the types MATH and MATH, respectively, see REF, as natural bases. Define the maps REF as follows: MATH permutations of the cells mapping to permutations of the generators MATH and MATH of the MATH-operad. Finally define MATH as zero. Since MATH and MATH and the action of the symmetric groups is respected by construction, the maps REF define a morphism MATH of graded operads. The only thing which remains to be checked is the compatibility of this morphism with the differentials. We will compute the boundary (differential) MATH on MATH. Each space MATH, MATH, is a disjoint union of NAME cells, which form a basis of the relative homology MATH. We will study the action of MATH on this basis. MATH. Let us start with MATH, when the points in a NAME cell group on a single vertical line. For MATH the boundary of the cell MATH in MATH may be computed as follows: where the left-hand side denotes the boundary of the cell MATH and the right-hand side denotes a linear combination of NAME cells obtained as operad compositions MATH, MATH, of two NAME cells. This equation turns into REF , where all the terms with MATH or MATH are moved to the right-hand side. The signs here and henceforth in the proof are compatible with the signs in REF - REF, if the orientations on NAME cells are chosen as in the proof of REF . MATH. Cells for MATH are configurations of points on two vertical lines. The boundary of a cell MATH may be described as follows: This equation translates under the correspondence REF into the identity REF in which all terms but those containing MATH are moved to the right-hand side. Note that the first sum on the figure corresponds to the terms MATH, where MATH is a shuffle of MATH and MATH, which show up on the left-hand side of REF for all pairs MATH being MATH or MATH. The rest of the left-hand side of REF is the last term on the figure above. MATH. Cells for MATH are configurations of points on at least three different vertical lines. The boundary of a cell with at least four vertical lines will produce the identity MATH under the morphism MATH, because of a dimension argument: The boundary of such cell has a dimension MATH, while (multiple) operad compositions of NAME cells from MATH will have dimensions MATH. Therefore, the boundary of a cell with at least four vertical lines will have no terms which are compositions of NAME cells from MATH. Thus, the only nontrivial identity to be checked in MATH comes from the lowest NAME cells in MATH, those made out of configurations of points on three vertical lines. The following figure describes the differential of such cell in MATH. Under the morphism MATH, this identity turns into REF where all terms are moved to the right-hand side. Note that the first sum on the figure corresponds to the terms in REF where all MATH or all MATH are either MATH or MATH. |
math/9908040 | In CITE NAME has constructed an operad morphism MATH, which is surjective on homology. Composing it with our morphism MATH, we get a morphism MATH. To prove the formality of MATH, it is enough to show that it is a quasi-isomorphism. Moreover, it suffices to show that MATH is surjective on homology, because of a graded dimension argument. Note that the operad MATH is generated by MATH, therefore we just need to show that the second component MATH of MATH is surjective on homology. The morphism MATH is a composition MATH. According to CITE, the induced homology morphism MATH is an isomorphism. On the other hand, notice that our morphism MATH is an isomorphism, because MATH. In particular, it induces an isomorphism MATH of the homology. Thus, the composition MATH is an isomorphism, which shows that MATH is formal. The existence of a unique up to homotopy morphism MATH follows from the fact that both operads are quasi-isomorphic to MATH and MATH is a minimal model of MATH, see REF . |
math/9908040 | Taking into account the vanishing operations MATH and MATH and rewriting the rest in terms of the dot product and braces, the identities REF through REF can be simplified as follows. The identities REF for MATH, REF, and REF, are equivalent to MATH respectively. The identities REF are nontrivial only for MATH, when they turn into the following. MATH where MATH. The identities REF are nontrivial only when MATH and REF. For MATH they rewrite as the following family of identities: MATH for each MATH. For MATH, REF turn into MATH for each MATH. All these identities for the operations on the NAME complex may be checked directly. Some of the identities are classical, see for example, CITE, the others were not noticed until more recently, see CITE. One can find a detailed verification of the identities in NAME 's paper CITE. |
math/9908042 | Let MATH be the standard square NAME curve. From REF there is a continuous decomposition MATH of MATH into acyclic continua one of which is an arc that without loss of generality we can assume is the left edge of MATH. We can also assume that MATH is the NAME curve MATH where MATH. In addition we would like to assume that the members of MATH that lie along the right edge of MATH are single points. To see that we can make this assumption without loss of generality consider the following argument. Let MATH be the natural map and consider the decomposition space MATH that results from collapsing to points all the members of MATH that intersect the right edge of MATH. We will denote by MATH this quotient map. Since the right edge of MATH is a closed set and MATH is a closed map it can be shown that MATH is closed. Also since MATH, in addition to being closed, is monotone and open, the map MATH defined by MATH is closed, monotone, and open. Finally, since MATH is a homeomorphism on the boundary points of the complementary domains of MATH it can be shown using results of CITE that MATH is homeomorphic to the NAME curve. Now MATH induces a continuous decomposition of MATH into acyclic continua so that in addition to one element of the decomposition lying along the boundary of the unbounded complementary domain there is another arc of the unbounded complementary domain that is the union of degenerate elements of the decomposition. Thus we can assume without loss of generality that the members of MATH that lie along the right edge of MATH are single points. Let MATH be the quotient map from MATH to MATH. Let MATH be a continuous decomposition of MATH into acyclic continua with the quotient map MATH so that MATH where MATH; so that MATH is the right edge of MATH; and so that MATH is the identity on the left side of MATH. See REF . Let MATH and MATH. Denote by MATH the NAME curve that results from removing the rectangles MATH and MATH from MATH and replacing them with MATH and MATH respectively. Denote by MATH the NAME curve with one pinched hole with two lobes that results from removing the rectangles MATH and MATH from MATH and replacing them with MATH and MATH respectively. We can now define the map MATH as follows: MATH . Thus MATH is a continuous function from MATH onto MATH which is monotone and open because MATH and MATH are monotone open maps. Therefore the theorem holds. |
math/9908042 | From the construction described in CITE we know that there is a monotone open map, MATH, from the NAME curve, MATH, onto MATH so that for some MATH, MATH is the boundary of a hole. Consider the identification MATH of two points of MATH. Now MATH is a closed map and it can be shown using techniques used in CITE that the image of MATH under MATH is homeomorphic to MATH. Let MATH be homeomorphism from MATH onto MATH. Since MATH is open and closed the map MATH is a monotone open map from MATH to MATH. |
math/9908042 | Since MATH to the left of MATH it can be shown that MATH because of the definition of MATH. Since MATH must be between MATH and MATH, we know that for any MATH there some point of MATH that lies between MATH and MATH. Thus for any MATH we have that MATH is strictly to the right of MATH. (See, for example, the proof of REF). Applying Claim C of CITE we have that for all MATH . First we assume that MATH. If MATH lies between MATH and MATH, then we are done because it can be shown that MATH. See, for example, the proof of REF. So assume that there is a point MATH that is below MATH. The situation when there is a point MATH that is above MATH is handled similarly. CASE: Assume that MATH is odd in addition to assuming MATH. Since MATH there is a MATH so that MATH. Note that MATH is a horizontal cell since MATH is even. Because of the way polygonal lines were defined, specifically that MATH and that MATH, we know that MATH is the identity below MATH and above MATH between MATH and MATH and because of the way stretching is controlled near boundaries we know that MATH is the identity below MATH. Note that MATH is the identity to the left of MATH. Thus since MATH is below MATH, we have that MATH and that MATH. Let MATH be a cell-piece of the cell MATH that lies between MATH and MATH. Such a cell-piece exists since MATH and the width of a cell-piece is MATH. Let MATH and MATH be the horizontal lines that run along the top and bottom boundaries of MATH respectively. Define MATH to be the part of MATH that lies between MATH and MATH. Notice that since MATH, MATH is below MATH. See REF . Now MATH, and MATH. Consider MATH. There is MATH so that MATH crosses some cell-point of MATH. But MATH so some point of MATH lies above MATH and since MATH we can conclude from REF that some point of MATH lies above MATH. Therefore MATH must lie along MATH and ``cross" one of the ``cell points" of MATH. Now MATH so there exists a trapezoid MATH that satisfies REF - REF Assume MATH is even in addition to assuming MATH. Consider MATH so that MATH. Note that MATH is a horizontal cell. If MATH then denote MATH by MATH. Since MATH is the identity to the left of MATH we have that MATH. Now MATH is the identity below MATH. Also we know that MATH is the identity below MATH and to the left of MATH. Recall that MATH. Finally, because MATH maps horizontal lines onto horizontal lines and because the image under MATH of any point of MATH is to the left of MATH, we can conclude that MATH and that MATH which is below MATH. Let MATH be a cell-piece of the cell MATH that lies between MATH and MATH. Let MATH and MATH be the horizontal lines that run along the top and bottom boundaries of MATH respectively. Define MATH to be the part of MATH that lies between MATH and MATH. Notice that since MATH, MATH is below MATH and that MATH. Again from REF and the fact that MATH we can conclude that some point of MATH lies above MATH and that MATH is forced to lie along the image under MATH of a ``cell-piece" of MATH. Because of the behavior of MATH to the left of MATH we can find a trapezoid MATH that satisfies REF - REF. Now we will assume that MATH. We also assume that there is a point MATH that is below MATH. The situation when MATH is close to the top boundary of MATH is handled similarly. CASE: Assume that MATH is odd in addition to assuming MATH. Consider MATH so that MATH. Note that MATH is a horizontal cell. If MATH then denote MATH by MATH. Since MATH is the identity to the left of MATH we have that MATH. As in REF above we see that MATH is the identity below MATH. Also we know that MATH is the identity below MATH and to the left of MATH. Finally because MATH maps horizontal lines onto horizontal lines and because the image under MATH of any point of MATH is to the left of MATH we can conclude that MATH and that MATH which is below MATH. Let MATH be a cell-piece of the cell MATH that lies between MATH and MATH. Let MATH and MATH be the horizontal lines that run along the top and bottom boundaries of MATH respectively. Define MATH to be the part of MATH that lies between MATH and MATH. Notice that since MATH, MATH is below MATH, and that MATH. From REF and the fact that MATH we can conclude that some point of MATH lies above MATH and that MATH is forced to lie along the image under MATH of a ``cell-piece" of MATH. Because of the behavior of MATH near MATH we can find a rectangle MATH that satisfies conditions MATH - MATH of the lemma. CASE: Assume MATH is odd in addition to assuming MATH. Exactly as in REF above. |
math/9908042 | Since we have that MATH is a continuous decomposition of MATH and that MATH is a homeomorphism, then MATH must be continuous at every MATH except possibly when MATH. But REF guarantees that the elements of MATH get bent along MATH so MATH is lower semicontinuous at MATH and by the argument preceding this lemma we have that MATH is upper semicontinuous at MATH. Thus MATH is continuous at MATH also. |
math/9908042 | Let MATH be the natural projection. We know that MATH is locally connected because MATH is. Since MATH is upper semicontinuous, we know that by adding the points of the compliment of MATH, we get an upper semicontinuous decomposition of MATH which we will call MATH. Thus we extend MATH to MATH. No member of MATH separates the plane so by NAME 's Theorem CITE we know that MATH is homeomorphic to the plane. Thus MATH is planar. By REF, we get that MATH is a homeomorphism for any MATH a bounded component of MATH. Finally, if MATH is the unbounded component of MATH, then MATH is also a simple closed curve. Thus the images under MATH of the boundaries of the components of MATH are simple closed curves, are dense in MATH, and have diameters that go to zero. Therefore MATH is homeomorphic to the NAME curve by CITE. |
math/9908045 | CASE: By the equality MATH and NAME 's theorem, the equatity follows. CASE: By using the differential equation for MATH, we have MATH . By the commutativity condition of MATH, we have MATH for MATH. Using the relation in REF., MATH, and we obtain the statement of REF. |
math/9908045 | If MATH, then it is nothing but the first statement of REF . Suppose MATH. Then MATH-part of MATH is a linear combination of MATH . Since the set MATH is stable under the action of MATH, REF satisfies the relation MATH. |
math/9908045 | For a vector MATH, we set MATH. CASE: If MATH, then MATH and MATH-th component of MATH is equal to MATH for MATH. CASE: If MATH, then MATH and MATH-th component and MATH-th component of MATH is equal to MATH and MATH respectively. Therefore if we define MATH inductively by MATH and MATH. Then MATH. If MATH, then MATH. This proves REF . For REF , we can prove MATH if MATH by induction on MATH using the infinitesimal pure braind relation REF . (In case MATH and MATH, MATH and MATH respectively.) This complete REF. |
math/9908045 | We write MATH for short. First we prove that MATH by induction on the cardinarity of MATH. By the assumption of induction for MATH, we have MATH and the last expression gives the expression of MATH. Therefore we have MATH . This implies the proposition. |
math/9908045 | Let MATH and suppose MATH. As in REF , we make the numbering of the path in MATH from MATH to MATH as MATH and MATH. First we claim the set MATH contains MATH. Since MATH is not zero, MATH contains MATH, that is, MATH. If the path connecting MATH and MATH in the corresponding graph MATH is MATH, then MATH. Therefore MATH. If MATH and MATH is a minimal element satisfying MATH, then MATH contins an edge MATH by the definition of MATH, that is, MATH, where MATH . Therefore MATH. Conversely, for any subset MATH of MATH satisfying CASE: MATH is contained in the same connected component of MATH, CASE: MATH, and CASE: MATH, there exists unique MATH satisfying CASE: MATH, CASE: MATH, and CASE: MATH. Therefore MATH . |
math/9908045 | Since MATH, using identity MATH, we get the statements. |
math/9908045 | The first equality follows from the expression MATH . The second equality is obtined directly by the equality MATH . |
math/9908045 | We prove the proposition by induction. By two lemmata, we have MATH using homomorphism MATH and assumption of independence of MATH. |
math/9908045 | By REF , we have MATH. To prove the statement, it is enough to prove that MATH satisfies the estmation of REF for an ordered rooted tree MATH with the root set MATH. We change variable by MATH for MATH. Then MATH and MATH . Here we put MATH. In particular, MATH if MATH contains an edge adjacent to REF. The signature in REF arise from the subtutition for separating edges of MATH adjacent to REF and those does not adjacent to REF. If MATH contains no edges adjacent to REF, we get the second statement. |
math/9908045 | By the definition of MATH, if MATH for some MATH, then any MATH has an edge adjacent to REF. If MATH for all MATH, then any MATH contains no edges adjacent to REF. Therefore the statement follows from REF . |
math/9908045 | This is the fundamental property of MATH-function MATH. |
math/9908045 | Suppose that MATH contains an edge adjacent to REF. Let MATH be the minimal number such that MATH is an edge of MATH. Set MATH where MATH if MATH is an edge of MATH and MATH otherwise. Then MATH . Therefore if MATH or there exist at least tow p such that MATH is an edge of MATH, then MATH. If MATH and there is no edge adjacent to REF other than MATH, then MATH. |
math/9908045 | (Proof of the Main REF ) We can proceed by the induction on MATH. We consider the limit of REF for MATH. Then all the entries of MATH are contined in MATH by REF . By REF , all the entries of MATH are contained in MATH. Therefore all the entries of MATH are also contained in MATH. Therefore MATH is an element of MATH for MATH under the restriction REF : MATH for all MATH. On the other hand, by the relation REF , the restriction REF is not necessary. This completes the main therem. |
math/9908050 | Suppose the answer to Question B is yes and the NAME representation of MATH has kernel for some MATH. As the NAME representation is known to be faithful for MATH, assume MATH is at least MATH. Then there is a pseudo-Anosov element MATH in the kernel CITE. Replacing MATH with a power of MATH if necessary, we can assume MATH is a pure braid, that is, fixes each puncture. Let MATH be the braid MATH where MATH is the MATH standard generator of MATH (see REF ). Taking a power of MATH if necessary, we can assume that MATH is pseudo-Anosov. Now MATH induces a MATH-cycle on the punctures because MATH was a pure braid and MATH induces a MATH-cycle. Since MATH, the NAME polynomials of MATH and MATH are the same. The manifold MATH is NAME fibered, and it's easy to see that the NAME norm ball is as shown in REF , where the two infinite faces are fibered faces. Thus by REF , the NAME norm ball of MATH has exactly the same shape as the NAME norm ball. Since MATH and MATH have the same NAME polynomials, the NAME norm ball of MATH is as shown. But MATH is hyperbolic, and hence the NAME norm is non-degenerate. So any face of the NAME norm ball is bounded. Thus a fibered face of the NAME norm ball of MATH is properly contained in the corresponding face of the NAME norm ball. This contradicts the assumption that the answer to Question B is yes. |
math/9908050 | By REF the short exact sequence MATH implies that there are integers MATH such that MATH . Combining and multiplying by MATH gives MATH . Taking radicals of the above and using that MATH gives MATH . Now MATH is radical since it is the kernel of the ring homomorphism MATH which sends every MATH to REF. By REF we have MATH. Combining, we get MATH as desired. |
math/9908050 | Let MATH. I will show: Let MATH be as above. Then MATH. Let me now deduce the theorem assuming the lemma. By REF , the components of MATH correspond to the vertices of MATH whose coefficients are MATH. Such a vertex MATH corresponds to: MATH where MATH is the hemisphere MATH . To prove the theorem it suffices to show MATH is the same as the projection into MATH of the interior of the face MATH of MATH corresponding to MATH. Translate MATH so it is balanced about MATH - this doesn't change MATH or MATH. Now note that the cone over the interior of MATH is MATH . It's easy to see that this cone projects to MATH in MATH. This proves the theorem modulo the lemma. Let's go back and prove the lemma. The idea of the proof is that REF says that MATH is close, in some sense, to MATH. Using the properties in REF, we have (notation changed for clarity): MATH . For any finitely generated module MATH we have MATH, and so MATH . Let MATH be the augmentation ideal of MATH. Since MATH, the invariant MATH is all of MATH. So for any ideal MATH, we have MATH. Thus MATH . By REF , MATH, so MATH as required. This completes the proof of the lemma and thus the theorem. |
math/9908050 | Let MATH be the kernel of the map from MATH to its free abelianization. It is clear that MATH as the NAME invariant MATH is a quotient of MATH. By REF, MATH and we are done. |
math/9908056 | Fix MATH; we have to prove that MATH. We claim that the following equality holds: MATH . It follows easily by applying each element MATH to both sides of REF and using the separating property of MATH. Denoting by MATH the norm of MATH, it follows: MATH the continuity of MATH concludes the argument. |
math/9908056 | By possibly passing to a smaller MATH, we can assume the existence of a MATH-curve MATH of isomorphisms of MATH such that MATH carries MATH to a fixed subspace MATH of MATH. We can now replace each MATH by the push-forward MATH, and each MATH by MATH. Such replacements will not affect the hypotheses of the Proposition, nor the quantities involved in the equality REF. For instance, thanks to REF , the index of the restriction of MATH to MATH does not change; namely, for MATH, it is: MATH . We can therefore assume without loss of generality that MATH and MATH for all MATH. Moreover, we observe here that, by a convenient choice of the NAME space inner product on MATH, we can assume that MATH is represented by a compact perturbation of the identity of MATH, MATH. Now, the subspace MATH is the eigenspace of MATH corresponding to the eigenvalue MATH, hence it is finite dimensional. We start considering the case that MATH is positive semi-definite on MATH and that MATH is positive definite on MATH. In this case, the thesis means that MATH is positive definite on MATH for MATH sufficiently close to MATH. Let MATH be any closed complementary subspace of MATH in MATH; clearly MATH is positive definite on MATH. We claim that there exists a positive constant MATH such that, for MATH sufficiently close to MATH, it is: MATH . Namely, for MATH, REF follows from the fact that the restriction of MATH to MATH is of the form MATH for some compact operator MATH. In this case, MATH may be chosen to be the least eigenvalue of MATH. The continuity of MATH concludes the proof of the claim. We set: MATH . Since MATH is MATH, it is easy to see that, for MATH sufficiently close to MATH, it is: MATH so that MATH is positive definite on both MATH and MATH for MATH sufficiently close to MATH. We want to show that, if MATH is sufficiently close to MATH, then for all MATH and MATH, MATH is positive definite on the two dimensional subspace of MATH generated by MATH and MATH. By the positivity on MATH and MATH, it suffices to prove that, for MATH is sufficiently close to MATH, the following inequality holds: MATH for all MATH, MATH, MATH. Obviously, we can assume MATH. As MATH vanishes on MATH and MATH is of class MATH, there exists MATH such that, for all MATH is sufficiently close to MATH, we have: MATH for all MATH, MATH with MATH. By REF, for all MATH is sufficiently close to MATH we get: MATH for all MATH, MATH with MATH. This yields REF and concludes the first part of the proof. For the general case, we use the spectral decomposition of MATH to write an orthogonal decomposition MATH, where MATH is positive definite on MATH and negative definite on MATH; observe that MATH is finite dimensional, and MATH. Moreover, we write MATH, where MATH is positive definite on MATH and negative definite on MATH. We then apply the result proven in the first part of the proof to the restriction of MATH to MATH once, and again to the restriction of MATH to MATH. The conclusion follows by observing that MATH is positive definite on MATH and negative definite on MATH, which implies that MATH for MATH sufficiently close to MATH. Clearly, this also implies that MATH is non degenerate. |
math/9908056 | Use REF twice, once to MATH and once to a backwards reparameterization of MATH (see REF ). |
math/9908056 | We exhibit local trivializations for the family MATH. For MATH, the map MATH maps the orthogonal complement MATH isomorphically onto MATH; by continuity, this also holds for MATH sufficiently close to MATH. This implies that we have a direct sum decomposition MATH and the projection MATH onto MATH is given by: MATH . Obviously, MATH is MATH. For MATH sufficiently close to MATH, we define MATH to be the inverse of the isomorphism: MATH . Such a map MATH gives the required local trivialization for the family MATH. |
math/9908056 | Substituting MATH in REF, we get the following expression for MATH: MATH . Differentiating REF with respect to MATH we get: MATH . We now apply REF to MATH, MATH is the right hand side of equality REF, MATH, MATH and MATH, where MATH . It is easy to check that MATH is continuous, by the continuity of MATH and MATH, and clearly MATH is separating for MATH, which concludes the first part of the proof. From REF we compute easily: MATH for all MATH. Its regularity is established analogously applying REF . |
math/9908056 | For MATH, MATH is strongly non degenerate (positive) if and only if MATH is strongly non degenerate (positive). From REF, MATH is strongly non degenerate because MATH is non degenerate; by continuity, MATH is also strongly non degenerate for MATH small enough. If MATH is positive definite, then MATH is a NAME space inner product, and therefore it is positive definite and away from MATH. By continuity, MATH is positive definite for MATH small enough. |
math/9908056 | Let MATH and MATH be fixed; observe that MATH and MATH are maps of class MATH, because they are affine reparameterizations of solutions to REF; they satisfy the following differential equations: MATH . We differentiate REF with respect to MATH and, observing that MATH, we obtain: MATH . Using REF, we eliminate from REF the terms involving the operator MATH, and we get: MATH . This concludes the proof. |
math/9908056 | Let MATH be fixed. By REF , MATH is represented by a compact perturbation of the identity map with respect to some suitably chosen NAME space inner product on MATH. By REF , MATH is of class MATH, and we are under the hypotheses of REF . If MATH, applying REF , we obtain that the integer valued function MATH is constant around MATH if MATH is not a MATH-focal instant, whereas it has a jump of exactly MATH at MATH if MATH is a MATH-focal. If MATH is small enough, by REF , it is MATH, and this concludes the proof in the case that MATH is not MATH-focal. Applying REF to backwards reparameterizations of MATH (see REF ), we see that the map MATH is indeed a left-continuous function on MATH, and therefore MATH for MATH small enough. With this observation the proof is concluded. |
math/9908056 | As in the proof of REF , we study the evolution of the function MATH when MATH runs from MATH to MATH; observe that MATH. Observe that, by REF , the MATH-focal instants coincide with the instants MATH where MATH is degenerate on MATH. By REF (see also REF ), there is only a finite number of MATH-focal instants, hence, by REF , MATH is piecewise constant on MATH. Namely, MATH is constant on any interval that does not contain MATH-focal instants. Since MATH for MATH, by the non degeneracy of MATH on MATH and REF , MATH for MATH sufficiently small. When MATH passes through a MATH-focal instant MATH, by REF and by REF the jump of the function MATH is equal to the signature MATH. Finally, applying REF to a backwards reparameterization of MATH around MATH (see REF ), by REF for MATH sufficiently close to MATH we have MATH, which concludes the proof. |
math/9908056 | We formally differentiate REF, obtaining: MATH . Using the fact that MATH is of class MATH, it is easily seen that REF defines a continuous curve in MATH. We now use REF by considering MATH to be the set of evaluations at fixed vectors MATH; the conclusion will follow once we prove that the map MATH is of class MATH for all MATH, and that its derivative is given by REF. Let MATH be the NAME space of MATH-valued MATH-maps on MATH; we define the following bounded linear operator MATH by: MATH . We observe that the map MATH is given by the composition of MATH and the map MATH . It remains to show that the map REF is of class MATH. This is again an easy consequence of REF , where MATH is the set of evaluations at fixed instants MATH. |
math/9908056 | For MATH, we compute : MATH . Hence, for the proof we need to show that, given MATH there exists MATH and MATH such that the following differential equation is satisfied: MATH . It suffices to take: MATH . Observe that the above formulas make sense because MATH. |
math/9908056 | It follows directly from REF . |
math/9908056 | By REF , an easy linear algebra argument shows that, for MATH, MATH. For MATH we apply the isomorphism MATH and we get the conclusion. |
math/9908056 | Let MATH and MATH be fixed, with MATH. From REF, we compute using integration by parts as follows: MATH . Similarly, if MATH and MATH are fixed, MATH, since MATH is constant, from REF we have: MATH which concludes the proof. |
math/9908056 | Let MATH be fixed. From REF it follows immediately that MATH, hence MATH. For the opposite inclusion, observe that, if MATH, then MATH for all MATH, and, by REF , also MATH for all MATH. By REF it then follows that MATH for all MATH, proving that MATH. Similarly, MATH. Since MATH is non degenerate, from REF it is easy to see that MATH is non degenerate in MATH, which proves that MATH is non degenerate in MATH. |
math/9908056 | Let MATH be fixed; from REF we write MATH on MATH as follows: MATH . Now, the bilinear form on MATH given by the first integral in REF is a NAME space inner product on MATH, and therefore it is represented by the identity operator on MATH. We now observe that the bounded linear operator MATH from MATH to MATH has a continuous extension to MATH. Namely: MATH and the latter expression is clearly continuous with respect to the uniform topology. It follows that the bilinear form on MATH given by the second integral of REF has a continuous extension to MATH, and we have observed that this implies that it is represented by a compact operator on MATH. The terms in the last line of REF are also continuous in the MATH-topology, and again the corresponding bilinear form is represented by a compact operator on MATH, which proves the first part of the Proposition. As to the bilinear form MATH on MATH, observe that, by definition of MATH (see REF ), if MATH then the quantity MATH is constant, and thus: MATH . Then, for MATH, it is: MATH . Again, the integral in the above formula is a NAME space inner product in MATH, and the last term is continuous in the MATH-topology, which proves that MATH is represented by a compact perturbation of a positive isomorphism of MATH. |
math/9908056 | Let MATH be a direct sum decomposition of MATH, with MATH positive definite and MATH negative definite (recall that MATH is non degenerate on MATH). Then, it is easy to see that we have a direct sum decomposition MATH, where: MATH and MATH . Clearly, MATH; to conclude the proof, it suffices to show that MATH is positive semi-definite on MATH and negative definite in MATH. If MATH, MATH, then MATH for some MATH, MATH, and for all MATH; then, from REF, we have: MATH . If MATH, then, by REF, we have: MATH . Since MATH ad the function MATH is convex in MATH, we use the NAME 's inequality to prove the following: MATH . Finally, from REF we obtain: MATH which concludes the proof. |
math/9908056 | The curve MATH is clearly a timelike geodesic in the opposite Lorentzian manifold MATH with the same conjugate points. We know that all the conjugate points along a causal geodesic are positive, hence MATH has only negative conjugate points in MATH. Then, there cannot be any conjugate point, because the sum of their signatures must be non negative integer. |
math/9908056 | Let MATH denote the restriction of the action functional MATH to the manifold MATH; as we have observed, MATH is a smooth submanifold of MATH and the critical points of MATH on MATH are precisely the geodesics joining MATH and MATH in MATH. We endow MATH with the following Riemannian structure. We consider an auxiliary Riemannian metric MATH on MATH, and for all MATH we define a NAME space inner product MATH in MATH by: MATH . Using the MATH-precompactness assumption, as well as the density of MATH in MATH, the following facts are proven in CITE: CASE: MATH is bounded from below, that is, there exists MATH such that MATH for all MATH; CASE: for all MATH, the sublevel MATH is a complete metric subspace of MATH; CASE: for all MATH, MATH satisfies the NAME - NAME condition at the level MATH when MATH is endowed with the NAME structure given by REF. Finally, the condition that MATH and MATH be non conjugate in MATH implies that MATH is a NAME functional, that is, all its critical points in MATH are non degenerate. Namely, as we have already observed, the second variation of MATH at any geodesic MATH is given by the restriction of the index form MATH, and its kernel in MATH coincides with the set of NAME fields along MATH vanishing at the endpoints. If MATH and MATH are non conjugate in MATH, then MATH has trivial kernel, and MATH is a NAME functional. Then, by standard results of Global Analysis on Manifolds (see for instance CITE), denoting by MATH the NAME index of the critical point MATH of MATH, we have the following NAME relations. For all field MATH there exists a formal power series MATH in the variable MATH, with coefficients in MATH such that the following identity between formal power series is satisfied: MATH . By REF , for all MATH we have MATH; moreover, since MATH is complete, it is proven in CITE that the spaces MATH and MATH are homotopically equivalent, which implies that MATH for all field MATH. Finally, also the spaces MATH and MATH have the same homotopy type (see CITE), and so the NAME relations REF are easily obtained from REF. |
math/9908057 | The Schwarzian derivative of MATH is given by MATH where MATH . Let MATH and consider MATH for MATH. We have MATH . Moreover, its derivative with respect to MATH satisfies MATH . We see that, inside MATH, the cubic polynomial MATH can only have a single critical point and it is a minimum. Hence MATH. |
math/9908057 | To find fixed points of MATH, one tries to solve the equation MATH which can be reduced to MATH . Since MATH, for MATH, the equation has solution if and only if MATH . The fixed point of MATH at the boundary parameter values MATH occurs at MATH. It is easy to see that for MATH and any MATH, MATH and MATH, so as MATH crosses this boundary, a saddle node bifurcation occurs. |
math/9908057 | If MATH corresponds to an attracting fixed point of MATH, we have MATH and MATH. From the expression of MATH, this is equivalent to MATH . Then, MATH. Since MATH is increasing on both MATH and MATH, MATH lies in the intervals defined by the image under MATH, namely, MATH. To see the period doubling bifurcation, it is sufficient to show MATH . Here we need some calculations which will also be useful in the future. We have MATH . Furthermore, MATH . In particular, if MATH or MATH, one has MATH . The first two results only require simple calculus. The third one is a repeated application of the chain rule. Then, using that both MATH and MATH are even functions in MATH, the last result follows. At MATH, we have MATH and MATH, therefore MATH . Moreover, for MATH, MATH except at MATH. Thus, generically, period-doubling bifurcation occurs at MATH. |
math/9908057 | We mainly use two properties of MATH to conclude this. First, each MATH has negative Schwarzian derivative, the technique of CITE or CITE is applicable. Therefore, once the map MATH has an attracting cycle, at least one critical point of it must be attracted to this attracting orbit. Since MATH has only two critical points, there are at most two attracting orbits. If the attracting orbit is a reflection symmetric one, then this orbit attracts both critical points. If the orbit is an asymmetric one, its twin orbit attracts another critical point. Multiplicity occurs if there is a self-twin orbit. |
math/9908057 | The orbit for a single asymmetric REF-cycle can be obtained by solving the equation MATH . The only simultaneous solution exists when MATH and MATH. Then, whether the orbit is attracting can be easily decided by calculating MATH. |
math/9908057 | Let MATH be a symmetric REF-cycle, that is, MATH. Equivalently, MATH . If we subtract the second equation from the first one, we obtained that MATH and MATH. Clearly, MATH. For MATH, the trivial solutions MATH corresponds to fixed points of MATH. Further calculation on MATH shows that the equation MATH has only a nontrivial solution for MATH. In fact, let MATH such that MATH, then MATH form a symmetric REF-cycle. Since MATH, it follows that the cycle is attracting for MATH. Moreover, since MATH, one can only obtain trivial solution MATH from MATH for MATH. For MATH, the above equations always has solution for all MATH. In fact, its solution within MATH is given by the following. Let MATH satisfy MATH. We claim that MATH and MATH are solutions to the equation MATH and MATH respectively. Taking tangent to both sides and using that MATH, we have MATH . It follows that MATH. Hence MATH is a symmetric REF-cycle. It can be easily checked that MATH . Therefore, the symmetric REF-cycle is repelling. Again, by that MATH, we do not have other solutions to MATH for MATH. |
math/9908057 | For REF-cycle MATH of MATH, the derivatives are given by MATH . Thus, REF-cycle undergoes a period-doubling bifurcation when MATH, which is exactly MATH. We claim that the new attracting REF-cycle is reflection symmetric. In fact, the equation for a reflection symmetric REF-cycle of MATH is MATH . Let MATH. Then we have to solve for MATH. Clearly, MATH is an odd function with MATH, MATH, and MATH. If MATH is decreasing at MATH, equivalently, it is so at MATH, then MATH must have a zero modulo MATH in neighborhoods of MATH. It is easy to compute that MATH . Thus, MATH if and only if MATH. This shows that for MATH, MATH has a symmetric REF-cycle. Moreover, it must be attracting when MATH by continuity. We have shown in the previous section that an attracting symmetric REF-cycle exists for MATH with MATH. It is given by MATH . One may further calculate according to the lemma in REF to obtain that, at MATH and MATH, MATH . Then, by an argument making use of the Inverse Function Theorem, one concludes that MATH has a REF-period preserving pitch fork bifurcation at MATH. |
math/9908057 | The key is to consider the zero set of MATH in the MATH-plane. In the above, we have shown that for MATH slightly larger than MATH, this can be obtained from the solution set of a symmetric REF-cycle, MATH. We then solve for the specific values of MATH such that MATH and MATH. Our calculation involves the expression of MATH as a NAME product given in CITE, MATH . Then, a symmetric REF-cycle can be solved from MATH and MATH . After factoring out the obvious factors MATH, and letting MATH, we have the polynomial equation MATH . For MATH, all the four roots of this equation are real. Two of them are always MATH, one lies within MATH and the other within MATH. After taking arccosines, the four solutions form a symmetric REF-cycle. Let MATH be a solution, then we solve for MATH in MATH, we obtain a numerical value of MATH approximately equal to REF. We further verify that MATH . Many of the above calculations are lengthy and we indeed make use of computation software to help us. Finally, we apply the Inverse Function Theorem to conclude that there is a REF-period preserving pitch fork bifurcation. |
math/9908068 | CASE: MATH. CASE: MATH. CASE: This follows from the NAME formula and the fact that MATH vanishes on commutators. CASE: This follows from MATH . See CITE for details. |
math/9908068 | MATH . The sum converges uniformly since by REF, MATH, and since MATH. Now, recall that if MATH is invertible, the NAME determinant of MATH is a positive real defined by MATH where MATH is defined using the functional calculus for self-adjoint operators. Now we have MATH . |
math/9908068 | Fix MATH, and recall that MATH denotes the stabilizer in MATH of MATH. For MATH, there is a function MATH supported only on the orbit of MATH, MATH . It is easy to check that the correspondence MATH induces an isomorphism of MATH-Hilbert modules MATH . This proves the Proposition, because the trace on MATH is normal. |
math/9908068 | Stabilizers in MATH of MATH must fix MATH and must have translation length MATH to fix MATH. That is, MATH. Using REF, MATH . |
math/9908068 | MATH is naturally a disjoint union over MATH of subsets MATH. Then MATH splits as a finite direct sum over MATH, and the dimension of the MATH summand is exactly MATH times the left hand side of REF. |
math/9908068 | Convergence for small MATH will follow from REF . So, compute formally: MATH from REF . The coefficient of MATH converges absolutely for each MATH. Rearranging terms and substituting MATH, REF becomes MATH . |
math/9908068 | First we prove that the right hand side of REF is absolutely convergent. Notice that MATH is a projection to MATH followed by a unitary operator, hence by REF MATH . Using REF MATH which is finite by assumption. As in REF, fix a lift MATH of each edge MATH. We then have a MATH-equivariant isometric embedding MATH. To compute the trace of MATH, we calculate the trace of MATH acting on the MATH summand for each MATH. More precisely, for MATH let MATH denote the composition MATH . The first map in REF is the projection given by averaging over MATH. The second map is MATH described in REF. The final map to MATH is evaluation at MATH. Thus MATH where MATH . Using the definition of MATH, for any MATH we have MATH . We can now take the trace, to obtain MATH . CITE shows that all hyperbolic elements associated to an end have the same axis. We can therefore change REF to a sum over ends. We have MATH . The trace of MATH is the sum over MATH of MATH. That is, we have expressed the trace as a sum over orbit representatives of edges and then a sum over all ends. The key step of this computation is to switch to a sum over orbit representatives of ends and then a sum over all edges. Note that each MATH orbit of MATH has MATH elements. Also, the final summand in REF is MATH-invariant as a function of MATH. Thus, MATH . The interchange of summations is allowed because the last sum converges absolutely. |
math/9908068 | The proof is very similar to that of CITE. Define the operators: MATH . Notice that MATH and MATH. Let MATH and define the following two operators on MATH: MATH A simple computation shows that MATH and MATH . MATH is invertible CITE so MATH by REF . Using REF , compute MATH . The last equality is the computation MATH in CITE. The theorem follows easily. |
math/9908068 | Fix MATH, with lift MATH and MATH. From REF, we need to compute MATH for each MATH. First notice that MATH. The action of MATH on MATH factors through MATH, and so MATH . Orthogonal projection MATH onto MATH is therefore given by averaging over MATH, and so MATH. |
math/9908068 | From REF we have MATH so that MATH . Now fix MATH. Fix MATH. Choose any MATH with MATH. We have MATH . If MATH, no hyperbolic element of MATH lies in MATH. So, for such MATH, MATH. Define MATH. Then for MATH, MATH . Now, putting MATH, REF becomes MATH which proves the theorem. |
math/9908068 | Define MATH . Then we have a commutative diagram: MATH . It is not difficult to check that all rows and columns are exact, and the lemma follows. |
math/9908068 | CASE: For any MATH, let MATH be a lift of MATH to MATH, and let MATH be the edge in MATH covered by MATH. We have MATH because MATH is free and MATH is torsion. Then by REF , MATH is the edge-volume of MATH REF , which is finite. CASE: By combining REF with the computation in REF , MATH . Let MATH be the bound on the degree of vertices of MATH. We claim that MATH, where MATH. To see the claim, let MATH. Then the orbit of MATH under MATH has at most MATH elements. The group MATH is the stabilizer in MATH of MATH, and this proves the claim. Choose a lift MATH for each edge MATH. Each class MATH has a representative which contains at least one MATH. On the other hand, each edge MATH is contained in only finitely many MATH (a loose bound is MATH). Then continuing REF, MATH which is finite. |
math/9908071 | If MATH, then consider the direct MATH-system MATH, where MATH for each MATH and MATH for each MATH with MATH. If MATH, then consider any subset MATH of MATH such that MATH and MATH. Without loss of generality we may assume that MATH contains the unit of MATH. Each MATH can obviously be identified with a subset (denoted by the same letter MATH) of MATH of cardinality MATH. Let MATH be the smallest MATH-subalgebra of MATH containing MATH. If MATH and MATH, then MATH (as subsets of MATH) and consequently MATH. This inclusion map is denoted by MATH. It is easy to verify that the collection MATH is indeed a direct MATH-system such that MATH. |
math/9908071 | Clearly MATH is dense in MATH (this fact remains true for arbitrary direct systems of MATH-algebras). Consequently, for any point MATH there exists a sequence MATH, consisting of elements from MATH, such that MATH. For each MATH choose an index MATH such that MATH. By REF , there exists an index MATH such that MATH for each MATH. Since MATH, it follows that MATH . Finally, since MATH is closed in MATH, it follows that MATH . |
math/9908071 | Since MATH, there exists a dense subset MATH of MATH such that MATH. For each MATH there exists, by REF , an index MATH such that MATH. Since MATH is a MATH-complete REF , there exists, by REF , an index MATH such that MATH for each MATH. As in the proof of REF we can conclude that MATH . Since MATH is dense in MATH and since MATH is closed in MATH it follows that MATH . By REF , the MATH-th limit homomorphism MATH of the direct MATH-system MATH is an injective unital MATH-homomorphism. Thus the composition MATH is a well defined unital MATH-homomorphism. It only remains to note that MATH, as required. |
math/9908071 | We perform the spectral search (see REF ) with respect to the relation MATH which is defined as follows. An ordered pair MATH of indeces is an element of MATH if and only if MATH and there exists a unital MATH-homomorphism MATH such that MATH, that is, if the diagram MATH commutes. Let us verify conditions of REF . Existence. For each MATH we need to find an index MATH such that MATH. Indeed, according to REF , MATH. Consider the unital MATH-homomorphism MATH. By REF , there exist an index MATH (which, without loss of generality, may be assumed to be greater than MATH) and a unital MATH-homomorphism MATH such that MATH. This obviously means that MATH. NAME. Let MATH and MATH. In order to show that MATH, consider the composition MATH, where the unital MATH-homomorphism MATH is supplied by the condition MATH. Clearly MATH. This shows that MATH. MATH-closeness. Let MATH be a chain of elements in MATH with MATH. Suppose that MATH for some MATH and each MATH. We need to show that MATH, where MATH. First observe that if MATH for MATH, then MATH . Since, by REF , the MATH-th limit MATH-homomorphism MATH of the direct system MATH is injective, it follows that MATH. This means that the collection MATH forms a morphism of the subsystem MATH of the direct MATH-system MATH into the MATH-algebra MATH. Consider (see Subsection REF) the unital MATH-homomorphism MATH . Finally, applying REF , we define the unital MATH-homomorphism MATH as the composition MATH . The straightforward verification shows that MATH indeed satisfies the required equality MATH and, consequently, MATH. Now, by applying REF , we conclude that the set MATH of MATH-reflexive elements is cofinal and MATH-closed in MATH. Observe that an element MATH is MATH-reflexive if and only if there exists a unital MATH-homomorphism MATH such that MATH. It follows from the above construction that the collection MATH is indeed a morphism between the systems MATH and MATH such that MATH. |
math/9908071 | By REF , applied to the unital MATH-homomorphism MATH, there exist a cofinal and MATH-closed subset MATH and a morphism MATH such that MATH. Similarly, by REF , applied to the unital MATH-homomorphism MATH (recall that MATH is a unital MATH-isomorphism), there exist a cofinal and MATH-closed subset MATH and a morphism MATH such that MATH. By REF , the intersection MATH is still cofinal and MATH-closed subset of MATH. Note that for each MATH we have two unital MATH-homomorphisms MATH and MATH satisfying the equalities MATH and MATH. Consequently, having also in mind REF , we have MATH . Similarly, MATH. This obviously means that both MATH and MATH are MATH-isomorphisms (inverses of each other). |
math/9908071 | Let MATH and MATH denote the canonical inclusions (see CITE). Let MATH and MATH also denote canonical inclusions into the corresponding unital free products. Now consider the homomorphisms MATH and MATH . These two homomorphisms define the unique unital MATH-homomorphism MATH such that MATH and MATH . Here MATH. Similarly consider the unique unital MATH-homomorphism MATH satisfying the equalities MATH . Next observe that if MATH, then MATH . Similarly, if MATH, then MATH . Now observe that REF guarantee the validity of the equality MATH . In order to prove the equality MATH it suffices to show that MATH and MATH . Note that REF follows from the following observation CASE : MATH . Similarly REF follows from the following observation CASE : MATH . This finishes proof of REF . It only remains to note that, by REF , both MATH and MATH are isomorphisms as required. |
math/9908071 | It can be shown, by applying the argument similar to the one used in the proof of REF , that the homomorphism MATH coincides with the homomorphism MATH . It only remains to note that MATH is an inclusion by CITE. |
math/9908071 | For each MATH consider the unital MATH-homomorphism MATH defined in REF . Clearly MATH for each MATH. Consider the unique unital MATH-homomorphism (see Subsection REF) MATH such that MATH for each MATH (here MATH denotes the MATH-th limit injection of the above direct system). Applying property MATH it is easy to see that MATH is an isomorphism. |
math/9908071 | The first part of this statement follows from the above given definition of unital free products. In order to prove the second part we need to show that MATH is a direct MATH-system associated with the unital free product MATH. Let us verify REF - REF is obvious since the set MATH is MATH-complete. REF follow from REF . Finally REF , that is, the fact that MATH is separable for a countable subset MATH, follows from CITE. |
math/9908071 | The first part is trivial. If MATH is a projective unital MATH-algebra, then the projection MATH of the direct product MATH onto the first coordinate has the inverse, that is, there exists a unital MATH-homomorphism MATH such that MATH. Clearly the projection MATH onto the second coordinate is a unital MATH-homomorphism. It only remains to note that the composition MATH is a unital MATH-homomorphism. |
math/9908071 | Let MATH be a doubly projective homomorphism. Consider the following commutative diagram MATH . Since MATH is doubly projective, there exists a unital MATH-homomorphism MATH such that MATH. |
math/9908071 | First suppose that MATH is projective. In order to show that MATH is projective, consider a surjective unital MATH-homomorphism MATH and a unital MATH-homomorphism MATH. Our goal is to find a unital MATH-homomorphism MATH such that MATH. Since MATH is projective, there exists a unital MATH-homomorphism MATH such that MATH. Since MATH is doubly projective there exists a unital MATH-homomorphism MATH such that MATH (and MATH). Obviously MATH is a required lift of MATH and, consequently, MATH is projective. Now assume that MATH is projective. In order to show that MATH is projective, consider a surjective unital MATH-homomorphism MATH and a unital MATH-homomorphism MATH. Our goal is to find a unital MATH-homomorphism MATH such that MATH. By REF , there exists a unital MATH-homomorphism MATH such that MATH. Consider the composition MATH. Since MATH is projective, there exists a unital MATH-homomorphism MATH such that MATH. Let MATH. It only remains to note that MATH. |
math/9908071 | Let MATH and MATH be doubly projective homomorphisms of unital MATH-algebras. We need to show that the composition MATH is also doubly projective. Consider a surjective unital MATH-homomorphism MATH and two unital MATH-homomorphisms MATH and MATH such that MATH. Consider the following commutative diagram MATH . Since MATH is doubly projective and since MATH, there exists a unital MATH-homomorphism MATH such that MATH and MATH. Next consider the commutative diagram MATH . Since MATH is doubly projective and since MATH, there exists a unital MATH-homomorphism MATH such that MATH and MATH. It only remains to note that MATH . |
math/9908071 | Let MATH be a surjective homomorphism of unital MATH-algebras. Let also MATH and MATH be unital MATH-homomorphisms such that MATH. We need to find a unital MATH-homomorphism MATH such that MATH and MATH. Note that MATH . Since MATH is doubly projective, there exists a unital MATH-homomorphism MATH such that MATH and MATH. Finally note that the composition MATH has all the required properties. Indeed, MATH and MATH . |
math/9908071 | Consider a surjective unital MATH-homomorphism MATH and two unital MATH-homomorphisms MATH and MATH such that MATH. Here is the corresponding diagram MATH . Let MATH and MATH. Note that MATH . Since MATH is doubly projective, there exists a unital MATH-homomorphism MATH such that MATH and MATH. Now consider the composition MATH and observe that MATH and MATH . This shows that MATH is doubly projective. |
math/9908071 | Consider a surjective unital MATH-homomorphism MATH and two unital MATH-homomorphisms MATH and MATH such that MATH. Our goal is to construct a unital MATH-homomorphism MATH such that MATH and MATH. Let MATH denote the canonical embedding of MATH into MATH. Since MATH is projective, there exists a unital MATH-homomorphism MATH such that MATH. The two unital MATH-homomorphisms MATH and MATH define the unique unital MATH-homomorphism MATH such that MATH and MATH. Finally, observe that MATH and MATH. This shows that MATH. |
math/9908071 | Let MATH be a doubly projective homomorphism and MATH and MATH be separable projective unital MATH-algebras. Consider the unital free product MATH and note that the diagram MATH commutes. Also observe that MATH is a projective unital MATH-algebra. Clearly MATH is surjective, because MATH is surjective. |
math/9908071 | Let MATH be a surjective unital MATH-homomorphism of unital MATH-algebras. Let also MATH and MATH be unital MATH-homomorphisms such that MATH. By induction we construct a well ordered collection MATH of unital MATH-homomorphisms. Let MATH and suppose that we have already constructed MATH-homomorphisms MATH for each MATH, where MATH, in such a way that the following conditions are satisfied: CASE: MATH for each MATH. CASE: MATH for each MATH. CASE: MATH whenever MATH is a limit ordinal number with MATH. Let us construct a MATH-homomorphism MATH. If MATH is a limit ordinal number, then let MATH. If MATH, then consider the following commutative diagram MATH . Since MATH is doubly projective there exists a unital MATH-homomorphism MATH such that MATH and MATH. Thus, the homomorphisms MATH are constructed for each MATH and satisfy the above stated properties for each MATH. It only remains to note that for the unital MATH-homomorphism MATH we have MATH and MATH as required. |
math/9908071 | CASE: Let MATH and MATH be objects of the category MATH and MATH be an epimorphism of the same category. Our goal is to show that for any morphism MATH of MATH there exists a morphism MATH of MATH such that MATH. MATH . Since MATH is an epimorphism in MATH it follows that each of the homomorphisms MATH and MATH is surjective. Since MATH is projective, there exists a unital MATH-homomorphism MATH such that MATH. Clearly MATH. Consequently, since MATH is doubly projective, there exists a unital MATH-homomorphism MATH such that MATH and MATH. In other words the following diagram MATH commutes. The straitforward verification shows that MATH as required. CASE: Now suppose that MATH is a projective object of the category MATH. In order to show that MATH is doubly projective, consider a surjective unital MATH-homomorphism MATH and two unital MATH-homomorphisms MATH and MATH such that MATH. Clearly the pair MATH forms a morphism MATH in the category MATH. Consider also the epimorphism (in the category MATH) MATH sending the left vertical arrow in the following diagram onto the middle one. MATH . Since MATH is a projective object in the category MATH, it follows that there exists a morphism MATH, consisting of the unital MATH-homomorphisms MATH and MATH, such that MATH. This implies that MATH, that is, MATH, and MATH. In order to prove the equality MATH, simply note that MATH is a morphism in the category MATH. Thus MATH is doubly projective. |
math/9908071 | Let MATH be a dense subset of MATH such that MATH. Let also MATH. Since MATH, it follows that MATH. As in the proof of REF , we can conclude that MATH is the limit of the direct system MATH, consisting of separable unital MATH-subalgebras of MATH (generated by countable subsets of MATH) and associated inclusion maps. Next consider the unital MATH-homomorphism MATH, generated by the homomorphisms MATH. This means that MATH for each MATH (here MATH denotes the canonical inclusion). Note that MATH is a surjective unital MATH-homomorphism. This follows from REF . Recall that by REF , the collection MATH is a direct MATH-system such that MATH. For each MATH let MATH. Also by MATH we denote the restriction of the homomorphism MATH onto the unital free product MATH. We have the following commutative diagram MATH where MATH denotes the inclusion. It is obvious that the system MATH, consisting of MATH-subalgebras MATH of MATH and their natural inclusions MATH, forms a direct MATH-system such that MATH. Also note that MATH is a morphism between the indicated direct systems such that MATH. Since MATH is a projective MATH-algebra, there exists a unital MATH-homomorphism MATH such that MATH. According to REF , applied to the homomorphism MATH, there exist a cofinal and MATH-closed subset MATH of MATH and a morphism MATH such that MATH. In particular, the square diagram MATH commutes for each MATH. Note also that MATH for each MATH. According to REF , the MATH-algebra MATH, and hence each MATH, MATH, admits a unital MATH-homomorphism into MATH. Consequently, by REF , the inclusion MATH is a coretraction with the associated retraction MATH. Consider the unital MATH-homomorphism MATH, defined as the composition MATH. Note that MATH which shows that MATH is a retraction. It only remains to note that MATH, as a retract of MATH, is projective. |
math/9908071 | First we show that if MATH is a projective MATH-algebra, then there exists a well ordered continuous direct system MATH, satisfying REF . While proving this we will show the existence of a direct MATH-system MATH, satisfying REF . According to REF there exists a collection MATH, consisting of separable unital projective MATH-subalgebra of MATH, such that MATH and MATH. Below we follow the proof of REF . The fact that each MATH, MATH, is projective becomes crucial later in this proof. As in the proof of REF , the homomorphisms MATH, MATH, generate the surjective unital MATH-homomorphism MATH such that MATH for each MATH (here MATH denotes the canonical inclusion). Recall that by REF , the collection MATH is a direct MATH-system such that MATH. For each MATH let MATH. Also by MATH we denote the restriction of the homomorphism MATH onto the unital free product MATH. We have the following commutative diagram MATH where MATH denotes the inclusion. It is obvious that the system MATH, consisting of MATH-subalgebras MATH of MATH and their natural inclusions MATH, forms a direct MATH-system such that MATH. Since, by REF , MATH is a projective MATH-algebra, there exists a unital MATH-homomorphism MATH such that MATH. Let us say that a subset MATH is admissible if MATH. This clearly means that the diagram MATH where MATH, commutes. We need to state some of the properties of admissible subsets. CASE: If MATH is admissible, then MATH. Proof of REF . Follows form the above constructions and the equality MATH (see the proof of REF ). CASE: If MATH is an admissible subset of MATH, then MATH. Proof of REF . Follows from REF (see the proof of REF ). CASE: The union of an arbitrary collection of admissible subsets of MATH is admissible. Proof of REF . Let MATH, MATH be an admissible subset of MATH and let MATH. First observe that MATH . Consequently MATH . CASE: If MATH is an admissible subset of MATH, then MATH is a projective MATH-subalgebra of MATH. Proof of REF . See the proof of REF . Every countable subset of MATH is contained in a countable admissible subset of MATH. Proof of REF . According to REF , applied to the homomorphism MATH, there exist a cofinal and MATH-closed subset MATH of MATH and a morphism MATH such that MATH. Clearly each MATH is admissible. CASE: If MATH is an admissible subset of MATH, then the inclusion MATH is doubly projective. Proof of REF . Recall that the following diagram MATH commutes and that MATH and MATH. Since each MATH, MATH, is projective (this is where REF is actually being used) we easily conclude that MATH is also projective (compare to CITE). By REF , the inclusion MATH is doubly projective. Finally, REF guarantees that the inclusion MATH is also doubly projective. This completes proof of REF . Now consider the direct system MATH. Clearly MATH is a direct MATH-system such that MATH (see REF ). By REF , each MATH, MATH, is a separable unital projective subalgebra of MATH and, by REF , each limit inclusion MATH, MATH, is doubly projective. This finishes the proof of the implication MATH . Next we prove the implication MATH . Since MATH, we can write MATH. By REF , for each MATH there exists a countable admissible subset MATH such that MATH. Let MATH and MATH. Also let MATH denote the inclusion. Thus we have the well ordered continuous direct system MATH. It follows from the above constructions that MATH. According to REF , each MATH, MATH, is a unital projective MATH-subalgebra of MATH. Since MATH is countable, we conclude that MATH is separable. REF guarantees that for each MATH, both limit inclusions MATH and MATH are doubly projective. Note that MATH. By REF , MATH is a coretraction. Consequently, by REF , MATH is also doubly projective. Finally, in order to see that MATH has a separable type, note that according to the above constructions and REF , we have the following commuting diagram MATH with surjective MATH and MATH. Clearly both MATH are projective (as unital free products of projective MATH-algebras). It only remains to note that since MATH is countable and since each MATH is separable, the unital free product MATH is also separable CITE. This completes the proof of the implication MATH . In order to prove the implication MATH observe that if MATH is a direct MATH-system satisfying properties indicated in REF , then for any MATH the MATH-th limit inclusion MATH is doubly projective and the MATH-algebra MATH is projective. Consequently, by REF , MATH is also projective. Finally, the implication MATH follows from REF . |
math/9908071 | Let MATH be a surjective unital MATH-homomorphism of unital MATH-algebras. Consider also two unital MATH-homomorphisms MATH and MATH such that MATH. Clearly MATH . Since MATH is doubly projective, there exists a unital MATH-homomorphism MATH such that MATH and MATH. Since the given diagram is a pushout, we have a unital MATH-homomorphism MATH. Recall that MATH and MATH. Consequently it only remains to show that MATH. In order to prove this equality note that MATH . Again, since the given diagram is a pushout, the above equalities imply that MATH as required. |
math/9908071 | Consider the pushout MATH generated by the homomorphisms MATH and MATH. Since, by the commutativity of the first diagram, MATH it follows that there exists unique unital MATH-homomorphism MATH such that MATH and MATH . Let MATH denote the canonical injection of MATH into the unital free product MATH. Consider the homomorphisms MATH and MATH. Since MATH is the unital free product, there exists unique unital MATH-homomorphism MATH such that MATH and MATH where MATH denotes the canonical injection (not to be confused with MATH). Note that MATH . In order to prove our statement we need to show that MATH is an isomorphism. We accomplish this by proving that MATH and MATH. The following diagram helps to visualize the situation. MATH . First let us show that MATH . Since both MATH and MATH are defined on the unital free product MATH, REF will be proved by examining compositions of the above homomorphisms with MATH and MATH. Observe that MATH and MATH . Note that REF imply REF . Next note that MATH and MATH . Clearly REF imply the equality MATH. In order to establish the second equality MATH we proceed in a similar way. Observe that MATH and MATH . As above, REF imply the required equality MATH. This shows that MATH is an isomorphism and completes the proof. |
math/9908071 | Since MATH has a separable type, we have the following commutative diagram MATH where MATH and MATH are projective unital MATH-algebras, MATH in addition is separable and the unital MATH-homomorphisms MATH and MATH are surjective. By REF , the diagram MATH is a pushout. Consequently there exists a unital MATH-homomorphism MATH such that MATH. Since MATH is surjective, the latter equality guarantees that MATH is also surjective. Thus we have the commutative diagram MATH . Next consider the following diagram MATH in which, according to REF , the subdiagram, represented by the back face of the above diagram, is a pushout. Since MATH is surjective and since MATH is doubly projective, there exists a unital MATH-homomorphism MATH such that MATH and MATH. Now consider the unital MATH-homomorphisms MATH and MATH. Note that MATH. Since, as was indicated, the back face is a pushout, there exists the unique unital MATH-homomorphism MATH such that MATH . It only remains to show that MATH is surjective. To see this consider the homomorphisms MATH and MATH. Clearly MATH . Since the originally given diagram is a pushout, there exists a unital MATH-homomorphism MATH such that MATH and MATH. Straightforward verification (based on the universality properties of the two pushout diagrams involved) shows that MATH. This suffices to conclude that MATH is surjective. Consequently the homomorphism MATH has a separable type. |
math/9908071 | Consider the pushout diagram MATH generated by the MATH-homomorphisms MATH and MATH. Since MATH is doubly projective, it follows, by REF , that MATH is also doubly projective. Since the characteristic MATH-homomorphism MATH of the originally given diagram is doubly projective, it follows, by REF , that the composition MATH is doubly projective. This proves the first part of our statement. In order to prove the second part of Lemma consider the following diagram in which all objects satisfy the above formulated assumptions: MATH . Since MATH is a pushout diagram and since the MATH-homomorphisms MATH and MATH satisfy the equality MATH, there exists unique MATH-homomorphism MATH such that MATH and MATH . In order to prove that MATH, first observe that MATH . Secondly, MATH . Since MATH is a pushout diagram, REF , imply the required equality MATH. Since MATH is doubly projective the latter equality guarantees the existence of a unital MATH-homomorphism MATH such that MATH and MATH. The straitforward verification shows that MATH and MATH and MATH . This completes the proof of REF . |
math/9908071 | Let MATH be a unital surjective MATH-homomorphism of unital MATH-algebras. Consider two unital MATH-homomorphisms MATH such that MATH. Our goal is to construct a unital MATH-homomorphism MATH such that MATH and MATH. Let MATH . We now construct (by induction) a collection of unital MATH-homomorphisms MATH so that the following conditions are satisfied: CASE: MATH, MATH. CASE: MATH, MATH. CASE: MATH, MATH. CASE: MATH, whenever MATH is a limit ordinal number with MATH. By our assumption, the MATH-homomorphism MATH is doubly projective. Consequently there exists a unital MATH-homomorphism MATH such that MATH and MATH. Suppose that for each MATH, where MATH, we have already constructed unital MATH-homomorphisms MATH satisfying REF - REF for appropriate indices. Let us construct a unital MATH-homomorphism MATH. If MATH is a limit ordinal number, then let (consult with Subsection REF) MATH . The continuity of the direct systems MATH and MATH guarantees that MATH, MATH and MATH for each MATH. If MATH, then, by the assumption, the diagram MATH is doubly projective. Therefore, by REF , there exists a unital MATH-homomorphism MATH such that MATH, MATH and MATH. Thus, the MATH-homomorphisms MATH are now constructed for each MATH. It only remains to note that the MATH-homomorphism MATH satisfies all the required properties. |
math/9908071 | Let MATH be a doubly projective homomorphism of separable type. We will show the existence of the above indicated direct MATH-systems and of a morphism, satisfying the required properties. If MATH is separable, then the statement is trivial. Indeed, by REF , MATH is injective and consequently MATH is also separable. Let MATH, MATH, MATH, MATH and MATH. Obviously the diagram MATH is a pushout. Now consider the case MATH. By our assumption, the homomorphism MATH has a separable type. This means (see REF ) that there exist a projective unital MATH-algebra MATH such that MATH, a separable projective unital MATH-algebra MATH and two surjective unital MATH-homomorphisms MATH and MATH such that MATH, where MATH denotes the natural inclusion. In other words, the following diagram MATH commutes. Since MATH is projective and MATH is surjective, there exists a unital MATH-homomorphism MATH such that MATH. Now consider the square diagram MATH which obviously commutes. To see this note that MATH . Since MATH is surjective and since MATH is doubly projective, there exists a unital MATH-homomorphism MATH (indicated in the above diagram as the diagonal arrow) such that MATH and MATH. Thus we have the commutative diagram MATH . Next observe that the MATH-algebras MATH, MATH, MATH and MATH all have density MATH. Consequently, by REF , MATH, MATH and MATH, where MATH, MATH and MATH are direct MATH-systems consisting of separable unital projective MATH-algebras and doubly projective limit inclusions MATH, MATH, MATH, MATH, and MATH, MATH. Also note that all three indexing sets MATH, MATH and MATH are cofinal and MATH-closed subsets of MATH. Next observe that the unital free product MATH is also the limit of the direct system MATH (straightforward verification using the universality properties of unital free products and limits of direct systems; see also REF). An important consequence of the fact that MATH has a separable type is that MATH is a separable MATH-algebra. This guarantees, according to CITE, that each MATH-algebra MATH, MATH, is separable and, as a result, MATH is actually a direct MATH-system. For each MATH let MATH. Let also MATH, MATH, MATH denote the corresponding inclusion. Similarly, for each MATH let MATH and MATH, MATH, MATH denote the corresponding inclusion. It is easy to see that the systems MATH and MATH are direct MATH-systems such that MATH and MATH. Since the indexing sets MATH, MATH and MATH are cofinal and MATH-closed in MATH, we can conclude, by REF , that the intersection MATH is still cofinal and MATH-closed in MATH. Next we consider six homomorphisms MATH . NAME of these homomorphisms are, by construction, the limits of associated morphisms MATH and MATH . We apply REF to the remaining three homomorphisms MATH, MATH and MATH and conclude that there exist cofinal and MATH-complete subsets MATH, MATH and MATH of MATH and morphisms MATH and MATH such that MATH . Note that, by REF , the intersection MATH is cofinal and MATH-closed in MATH (and consequently in MATH). For each MATH we have the following commutative diagram: MATH . Note that, by REF , we may without loss of generality assume that the limit inclusions MATH and MATH, MATH, are doubly projective. This observation coupled with REF guarantees that the homomorphism MATH, MATH, is also doubly projective. It is now clear that in order to complete the proof it suffices to show that the diagram (the front face of the above cubic diagram) MATH is a pushout for an arbitrary index MATH. Let MATH and MATH be unital MATH-homomorphisms into a unital MATH-algebra MATH such that MATH. Consider the homomorphisms MATH and MATH. Note that MATH . Now let MATH. Since, by REF , the diagram (the back face of the above cubic diagram) MATH is a pushout, it follows that there exists a unique unital MATH-homomorphism MATH such that MATH and MATH. Now let MATH. We have MATH and MATH . This simply means that the diagram under consideration has the corresponding universality property. Finally the uniqueness of MATH guarantees that MATH is the only unital MATH-homomorphism with the just indicated properties. This shows that our diagram is pushout and completes the proof of REF. Suppose that we are given direct MATH-systems MATH, MATH and a morphism MATH, satisfying the above indicated properties. Let MATH. By REF , the composition MATH is doubly projective. Since, by REF , the inclusion MATH is doubly projective, it follows from REF , that MATH is also doubly projective. By REF , the homomorphism MATH has a separable type. Finally, by REF , MATH also has a separable type. |
math/9908072 | Let MATH be a continuous one-to-one NAME isomorphism, where MATH is a zero-dimensional Polish space CITE. Next consider the pullback square MATH where MATH and the maps MATH and MATH are the restrictions of the natural projections MATH and MATH onto MATH respectively. Since the above diagram is a pullback, it follows that MATH is an open surjection and MATH is a contiunuous one-to-one NAME isomorphism. Since MATH and since MATH is open, there exists a continuous map MATH such that MATH. Clearly the NAME map MATH satisfies the equality MATH. This proves the first part of the Lemma. Obviously, MATH is a closed subset of MATH. Now consider a Polish space MATH. Clearly the map MATH is open and surjective (since MATH for each MATH). As above, there exists a continuous map MATH such that MATH. Note that MATH is also closed in MATH and MATH. Finally let MATH, MATH. |
math/9908072 | Identify MATH with the graph of the map MATH, that is, MATH. Clearly MATH is a closed subset of the product MATH and the map MATH coincides with the restriction of the projection MATH onto MATH. By our assumption, there exists a NAME subset MATH of MATH such that MATH is a copy of the NAME set. By the main result of CITE, MATH has a NAME parametrization. Since any two uncountable Polish spaces are NAME isomorphic, the latter means that there exists a NAME isomorphism MATH such that MATH. Since MATH, it follows that MATH as required. |
math/9908072 | According to REF , for each MATH there exist NAME maps MATH such that MATH, MATH, MATH and MATH is a NAME subset of MATH, MATH. Let MATH and suppose that for each MATH we have already constructed a subset MATH so that the following conditions are satisfied: CASE: MATH is a NAME subset of MATH whenever MATH. CASE: MATH whenever MATH. CASE: MATH whenever MATH and MATH. We let MATH. Since MATH and MATH are NAME maps, it follows that MATH is a NAME subset of MATH. CASE: MATH and REF MATH are also satisfied by the construction. This completes the inductive step and consequently we may assume that the NAME sets MATH, satisfying REF MATH - REF MATH, have been constructed for each MATH. Next consider the subset MATH. CASE: MATH, MATH, imply that MATH is a NAME subset of MATH. CASE: MATH and REF MATH, MATH, ensure that MATH is a topological copy of the NAME cube for each MATH. Finally, by REF (with MATH, MATH and MATH), we conclude the existence of the required NAME isomorphism MATH. |
math/9908072 | If MATH, then all spaces MATH, MATH, are Polish and the statement follows from REF . Thus we may assume that MATH. Since each short projection of the spectrum MATH has a Polish kernel, we conclude that MATH for each MATH. Represent the space MATH, MATH, as the limit space of a factorizing MATH-spectrum MATH, consisting of Polish spaces and MATH-soft limit projections CITE. Observe, in the meantime, that the indexing sets of all these spectra coincide with MATH (which has cardinality MATH). Since all short projections MATH of the spectrum MATH are MATH-soft and have Polish kernels, we see, by CITE, that MATH, MATH, is the limit of some Cartesian morphism MATH consisting of open maps between Polish spaces, where MATH is a cofinal and MATH-closed subset of the indexing set MATH. Let MATH and note that MATH is still a cofinal and MATH-closed subset of MATH CITE. In particular, MATH. For each MATH consider the inverse sequence MATH and let MATH. If MATH, MATH, then there is a Cartesian morphism MATH consisting of open surjective maps MATH, MATH. Denote by MATH the limit map of the morphism MATH. Thus, the following infinite commutative diagram arises: Straightforward verification shows that the limit space of the spectrum MATH coincides with the space MATH, and that all newly arising square diagrams are also Cartesian squares. Now take an index MATH. Since the above mentioned diagrams are Cartesian (pullback) squares, we conclude that MATH for each MATH and MATH. By REF , the limit projection MATH of the spectrum MATH is NAME isomorphic to the projection MATH, that is, there exists a NAME isomorphism MATH such that MATH. The required NAME isomorphism MATH can now be defined by letting MATH . |
math/9908072 | First let us assume that MATH has a Polish kernel. Then, by CITE, there exists pullback diagram MATH where MATH and MATH are Polish spaces and the map MATH is an open surjection. By REF , there exists a NAME map MATH such that MATH. Then the diagonal product MATH is a NAME map satisfying the equality MATH. Now consider the general case. It follows from consideration in CITE, that there exists a well ordered continuous inverse spectrum MATH consisting of MATH-spaces and MATH-soft short projections with Polish kernels such that MATH, MATH and the first limit projection MATH of the spectrum MATH coincides with the given map MATH. Let MATH and suppose that NAME for each ordinal number MATH, where MATH, we have already constructed a NAME map MATH so that the following conditions are satisfied: CASE: MATH whenever MATH. CASE: MATH whenever MATH. CASE: MATH whenever MATH is a limit ordinal number. Let us construct a NAME map MATH. If MATH is a limit ordinal number, then let MATH. Since the spectrum MATH is continuous, MATH is a well defined NAME map. If MATH, then according to the first part of the proof there exists a NAME map MATH (recall that the short projection MATH has a Polish kernel) such that MATH. Then the NAME map MATH satisfies all the required properties. This completes the inductive step and hence we may assume that NAME maps MATH with the above properties are constructed for each MATH. Let finally MATH. Then MATH is a NAME map and MATH. |
math/9908072 | CASE: Let MATH be a NAME isomorphism and consider a NAME map MATH. Take a neighbourhood MATH of MATH in the space MATH. By the definition of the topology MATH there exists a collection MATH such that MATH for each MATH, MATH and MATH . Let MATH and note that MATH. Since MATH is a MATH-space it can be represented as the limit space of a (factorizing) MATH-spectrum MATH consisting of MATH-spaces of weight MATH and MATH-soft limit projections (see CITE). Let also MATH be the standard MATH spectrum consisting of MATH subproducts of MATH and natural projections whose limit coincides with MATH. By the Spectral Theorem for NAME maps (see CITE), we may assume without loss of generality that MATH is the limit of a morphism MATH consisting of NAME isomorphisms (that is, MATH for each MATH). Note that if MATH is a NAME subset of MATH, then there exists an index MATH such that MATH (recall that MATH is a factorizing MATH-spectrum) - we say in such a case that MATH is MATH-cylindrical. This implies that for each MATH there exists an index MATH such that every element of MATH is MATH-cylindrical (recall that collection MATH is countable). Finally, since MATH and since MATH is a MATH-complete set, it follows that there exists an index MATH such that for each MATH every element of MATH is MATH-cylindrical. In this situation we have MATH . NAME observation here is that the set MATH is a neighbourhood of the point MATH in MATH (see CITE). Now consider the NAME map MATH and represent it as the diagonal product MATH . Since the MATH-weight of the product MATH does not exceed MATH and since MATH, there exists a MATH-embedding MATH. Then the diagonal product MATH is a NAME map such that CASE: MATH. CASE: The sets MATH and MATH are NAME separated in MATH. Let now MATH. It follows that the sets MATH and MATH are NAME separated in MATH. Finally observe that MATH which shows that MATH. CASE: Trivial. CASE: Embed MATH as a MATH-embedded subspace into MATH, where MATH. By repeating the argument presented in the proof of CITE we construct a collection MATH of countable subsets of the indexing set MATH satisfying the following properties: CASE: The collection MATH is cofinal and MATH-closed in MATH. CASE: If MATH is an (arbitrary) union of elements of MATH, then MATH is a closed and MATH-embedded MATH-subspace of the product MATH. CASE: If MATH and MATH are (arbitrary) unions of elements of MATH and MATH, then the restrictions MATH are MATH-soft. By using the above collection we construct a well ordered continuous inverse spectrum MATH consisting of MATH spaces MATH and MATH-soft short projections MATH with Polish kernels so that MATH. Since MATH, we can write MATH. By REF , there exists an element MATH such that MATH. Let MATH. Without loss of generality we may assume that MATH. By REF , MATH is a closed subspace of MATH and the map MATH is MATH-soft. Suppose that for each ordinal MATH, where MATH, we have already constructed a subset MATH as an union of less than MATH elements of the collection MATH so that MATH whenever MATH and all point inverses of the map MATH contain at least two points. Here MATH and MATH. Suppose also that the construction has been carried out in such a way that MATH for each MATH with MATH. This ensures that the MATH-soft map MATH has a Polish kernel. If MATH is a limit ordinal number we let MATH. If MATH, then consider the MATH-soft map MATH. By REF , there exists a NAME map MATH such that MATH. Next consider the composition MATH where MATH denotes the natural projection onto the first coordinate. According to CITE, the collection of maps MATH satisfying the equality MATH is a neighbourhood of the composition MATH in MATH. Consequently, by REF , for at least one map MATH with MATH the sets MATH and MATH are NAME separated. Since MATH is MATH-embeddwed in MATH, it follows that there exists a NAME subset MATH of MATH such that MATH . Choose a countable subset MATH such that MATH. This obviously implies that MATH . Since MATH, it follows that MATH. The cofinality of the collection MATH in MATH allows us to find an element MATH such that MATH. Clearly MATH . Finally let MATH. It then follows that the MATH-soft map MATH has a Polish kernel and all its fibers contain at least two points. Thus the required well ordered continuous inverse spectrum MATH has been constructed so that all pont inverses of all short projections MATH (which have Polish kernels) contain at least two points. A straightforward transfinite induction, based on REF , shows that MATH is NAME isomorphic to the product MATH. Since MATH is an uncountable Polish space, it is NAME isomorphic to MATH. Thus MATH is NAME isomorphic to MATH. |
math/9908073 | Implications REF are trivial. Proof of implication MATH follows the proof of CITE. Let MATH denote the set of all maps MATH such that domain MATH is a Polish subspace of MATH, MATH and MATH. Let MATH. Clearly, MATH. Consider also the map MATH which coincides with MATH on MATH for each MATH. Let MATH be the unique continuous extension of MATH to the NAME compactification MATH of MATH. By REF , MATH. By CITE and by the compactness of MATH, the latter is the limit space of a Polish spectrum MATH consisting of metrizable compacta MATH (compactness of MATH follows from the fact that MATH is dense in MATH, according to assmption made in CITE) with MATH. Write MATH where MATH, MATH, denotes a copy of MATH. Let also MATH, MATH, denote the corresponding projection. Since the spectrum MATH is factorizing, for each MATH there exist an index MATH and a map MATH such that MATH, where MATH is the MATH-th limit projection of the spectrum MATH. Since MATH is a Polish spectrum (see CITE) there exists an index MATH such that MATH for each MATH. Next consider the map MATH where MATH, MATH, denotes the corresponding projection of the spectrum MATH. It is easy to see that MATH, where MATH is the MATH-th limit projection of the spectrum MATH. It now suffices to let MATH and MATH. Let us show that MATH is indeed MATH-invertible. Since the spaces MATH and MATH are Polish (even compact and metrizable), it suffices (according to CITE) to consider only Polish spaces MATH in the definition of MATH-invertibility given above. Indeed, let MATH be a map defined on a Polish space MATH with MATH. We may as well assume that MATH. By the definition of MATH, there is an index MATH such that MATH. Let MATH denote the corresponding embedding. Clearly, MATH. Then the composition MATH lifts the map MATH, that is, MATH. CASE: As in the proof of CITE (see also CITE where the case MATH is considered) one shows that for any uncountable cardinal number MATH there exists a MATH-invertible map MATH, where MATH is a compactum of weight MATH such that MATH and MATH denotes the NAME cube of weight MATH (for MATH the existence of such a map is guaranteed by REF ; note also that a MATH-space of countable weight is Polish CITE). Consider now a space MATH with MATH and choose MATH large enough so that MATH can be identified with a subspace of MATH. Since the map MATH is MATH-invertible there exists a map MATH such that MATH. Since MATH is compact, the map MATH admits a continuous extension MATH. Since MATH and since MATH it follows that MATH. In this situation it can easily be seen that MATH is an embedding. In other words, MATH is a topological copy of MATH. Finally, since MATH it follows that MATH. |
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