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0

This response is long overdue, but I just saw this message a few months ago and I wanted to check with Birman and Hilden before responding. The lemma is actually incorrect exactly because of the line you pointed out. However, there is a certain family of cyclic branched covers of the sphere where it is true, which I describe below. Ty Ghaswala and I ...

3

Pick any non-identity involution $i$ of $\Bbb{RP}^2$. It has two lifts to a map $S^2 \to S^2$; pick the one whose square is the antipodal map. Because the antipodal map has no fixed points and is order 2, your lift must have no fixed points and be order 4. Note that necessarily the involution $i$ downstairs must have fixed points; there is no surface with ...

1

Milnor's Topology from the Differentiable Viewpoint (page 34) has a very elementary proof: Filling in the blanks: We may assume that $f(0) = 0$: Translation is an isotopy. We write $f$ in the form $f(x) = x_1g_1(x) + \cdots + x_mg_m(x)$: Hadamard's Lemma. ...the linear mapping $df_0$, which is clearly isotopic to the identity: $GL^+(m)$ is connected. ...

2

There are a couple of theorems from algebraic/geometric topology that you need to address these questions. The first theorem is that the set of isotopy classes of nontrivial simple closed curves on $\partial V$ corresponds bijectively with the set of ordered pairs $(m,n)$ of relatively prime integers, where the isotopy class of $c$ corresponds to $(m,n)$ if ...

4

Let $M$ be a connected oriented closed manifold. We know $[M,S^n]$ is isomorphic to the group of cobordism classes of framed 0-dimensional submanifolds of $M$. A framed 0-dimensional submanifold of $M$ is a set of points in $M$ with an isomorphism $T_xM \cong \Bbb R^n$ at each. A single framed point is cobordant to the same point with a different framing if ...

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In the case of the Floer homology of the cotangent bundle the answer is yes. You should have a look at this: The Viterbo transfer as a map of spectra by Thomas Kragh.

9

This follows from Gauss-Bonnet Theorem: If $f$ is the Gaussian curvature of a compact surface $S$ without boundary, then $$\int_S f=2\pi\chi(S)$$ where $\chi(S)$ is the Euler characteristics of $S$. In particular, if $S$ is $T^2$ the torus, we have $\chi(S)=\chi(T^2)=0$. Therefore, it is impossible for $f>0$ everywhere. BTW, for higher dimensional ...

7

By the Gauss–Bonnet theorem, the Euler characteristic of $T^2$ is given by $$\chi(T^2) = \frac{1}{2\pi} \int_{T^2} K dA$$ where $K$ is the curvature and $dA$ is the element area of $T^2$. If $K$ were everywhere positive, then this would be a positive number, for the same reason that the integral of a positive function is positive; but \$\chi(T^2) = \chi(...

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