Daniele Zuddas
• Member for 9 years, 9 months
• Last seen this week
• Trieste, Italia

The answer is no in infinite dimension (and yes in the finite dimensional case, of course). Because the sphere in a Banach space is Baire. But an infinite dimensional CW-complex is not, being a ...

For TOP Morse function a reference is the classical book of Kirby and Siebenmann "Foundational essays on topological manifolds, smoothings, and triangulations" (1977). The key point is to consider the ...

This is the special case announced in the comment. We assume $M$ to be an oriented 4-manifold with the homotopy type of a 2-complex. We prove that $M$ immerses in $\mathbb{C}P^2$. In the case of ...

The answer is not. Calabi and Eckmann ("A class of compact, complex manifolds which are not algebraic", Ann. of Math. 58 (1953) 494-500) proved that $\Bbb R^{2n}$, $n > 1$, has a complex structure ...

A theorem of Eliashberg implies that an open subset of $\Bbb C^n$, $n \neq 2$, is isotopic to a Stein domain (hence to a domain of holomorphy) if and only if it admits a handlebody structure with all ...

I think that a complete set of moves for cobordisms are the following: Kirby moves (that preserve the 4-manifold), handle trading (dotted circles become 0-framed 2-handles), addition/deletion of pairs ...

There are even infinitely many inequivalent complex structures without non-constant holomorphic functions on $\Bbb R^4$ download.

First, your assumption imply that $\bar D$ is a compact smooth manifold with boundary a topological sphere (because is a simply connected homology sphere). So, $\bar D$ is a topological $n$-ball, by ...

For $A$ compact the answer is that given by Joseph Van Name. If $A$ is not compact then the answer is negative. For example take the standard $\Bbb R \subset \Bbb R^2$. Then the quotient is not II-...

As Paul Siegel pointed out, the fundamental group of a smooth closed orientable 4-manifold can be an arbitrary finitely presented group, and for this reason a general classification is not possible, ...

The Schoenflies theorem: if $\Sigma^n \subset \Bbb R^{n+1}$ is homeomorphic to $S^n$, then there is a homeomorphism $h \colon \Bbb R^{n+1} \to \Bbb R^{n+1}$ such that $h(\Sigma^n) = S^n$ (the round $n$...

In this G&T paper there are examples of complex structures on $\Bbb R^4$ withouth compatible Kähler metrics. The point is that in this $(\Bbb R^4, J)$ there are holomorphic elliptic compact curves,...

In my understanding, I guess that the following strategy could be be attempted. Let $M$ be a closed connected orientable 4-manifold (while nonorientable 4-manifolds are doubly covered by orientable ...

May be a possible way to solve the problem is to set up it in the context of closed relations instead of functions, by putting $R = g^{-1} \circ f : B^n \to B^n$, and assuming for instance $f(B^n) \... View answer Accepted answer 4 votes This question is badly posed. If you have a not simply connected manifold (of any dimension$> 3$) then two connected closed curves are isotopic iff they are homotopic, hence you have as many ... View answer 3 votes This is a remarkable application in topology. It has been given by Mike Freedman in his work about the classification of simply connected closed topological 4-manifolds, which had as a main ... View answer Accepted answer 3 votes The answer is yes, as it has been remarked in the previous answers. Here is a very explicit construction. Let$T$be the sphere$S^2 \subset \Bbb R^3$with three disks removed. Take these three disks ... View answer 3 votes The videos of Mike Freedman lectures on the topology of 4-manifolds, broadcasted from UC Santa Barbara: Freedman's Lectures Also other videos on 4-manifolds and related topics given at MPIM during ... View answer 2 votes This is a well-known basic fact, so it might be not suitable for MO, but let me sketch an answer. You are looking for a local extension, and everything can be arranged to happen in a local chart. So, ... View answer 2 votes This is too long for a comment, it's just an idea for this computation. Start with a Seifert surface$F \subset \Bbb R^3$for the knot$K$. Up to isotopy, we can assume that$F$projects regularly to ... View answer 2 votes Another approach (for any base surface$S$): take the disk bundle over$S$with Euler number$e$(any integer). The total space is a smooth 4-manifold with boundary with the required property (the 0-... View answer 2 votes The most explicit and geometric generator for$\pi_4(S^2)$is probably the achiral genus-1 Lefschetz fibration$f : S^4 \to S^2$with two Lefschetz singularies (one positive and one negative, with ... View answer Accepted answer 2 votes There are some related results about compact Stein 4-manifolds with boundary as Lefschetz fibrations over the disk (whose fiber has non-empty boundary). Corners in this case arise naturally on the ... View answer 2 votes This is a simple exercise and the answer is no (cf. comments of Lee Mosher and Misha). Consider the two degree-4 coverings of the torus to itself with monodromies$\omega_1, \omega_2 : \pi_1(T^2) \to ...

No. Under your assumptions the map can have fold singularities (and also other kinds of singularities), namely those of the form $f(x,y) = (x, y^2)$ in local coordinates. In order to have a local ...

The answer is in Wajnryb's paper from which I attach Figure 12.

I show that $\partial D$ is a Jordan curve in the plane, may be this will be of some help. To do this, define a map $\phi : S^1 \to \Bbb R^2$ such that $\phi(v) = (\Bbb R_+ v) \cap \partial D$. Your ...

Well, it is well-known that both of $X$ and $Y$ are contractible (hence they are homotopy equivalent, this should answer your question). Notice that $Y$ is exactly the set of unit vectors of $E$. To ...

As the question is posed, the answer is not. If you take connected sum with $\Bbb{CP}^2$ you preserve the determinant, but change the intersection form.
The question should be posed in this way: given a real analytic $l$-form $\alpha$ on $M$, is there a real analytic extension $\widetilde \alpha$ defined in an open neighborhood of $M$ in $X$? The ...