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4

You are using that $\partial B^2 = S^1$, but by doing so you assumed already that your $S^1$ is homotopically trivial (it bounds a disk). So you have used the condition (simply connectedness) already. Your first answer is correct (module the fact that you can find a smooth disk $B^2$) For your second question, as $g$ is a constant map, then $g^*\omega = ...


3

The first Laplacian you mention (sometimes called the Laplace-Beltrami operator) acts on scalar functions, that is, functions $S^2 \to \mathbb{R}$. The de Rham (a.k.a. Hodge) Laplacian acts on differential forms. In particular, the de Rham Laplacian acts on zero-forms, which are precisely scalar functions $S^2 \to \mathbb{R}$, on which it agrees with the ...


3

Answer to the first one is technical. If we would take $U_i$ too big, then the map defined as it has been done $$(\mathscr{T}f)_{ij}: \mathscr{T}U_i \to \mathscr{T}V_j$$ $$[x,i,a] \mapsto [f(x),j,\partial(\theta_jf\psi^{-1}_i)(\psi_i(x))a],$$ wouldn't have much sense, because for $x\in U_i\setminus f^{-1}(V_j)$ point $y=f(x)$ wouldn't lie in $V_j.$ However ...


3

This is not quite true, since, for example, the open unit disk is embedded in $\mathbb{R}^2$ by the inclusion, and the inverse image of the closed unit disk is the open unit disk. What we will show is that embeddings with closed image are the same thing as proper injective immersions. We note that the property of being closed in a topological space is a ...


2

First, manifolds "exist in their own". However, every closed iriented 3-manifold does bound a compact 4-manifold (the latter is far from being unique). My suggestion is to pick up a copy of Munkres' "Topology" book and read first few chapters. This will help to clear many issues that you currently have.


2

This is Hopf's Umlaufsatz. There are proofs in most undergraduate differential geometry books (including my own text, on my webpage, pp. 28 ff.). The technical details are at the level of establishing path lifting for $\gamma'\colon S^1\to S^1$ and an extension of this lifting to the chord map of the curve.


2

A connected smooth $1$-manifold is either a line or a circle. The above Helix is a line. The Trefoil is a circle. Both can be smoothly embedded in the plane.


1

Sometimes you want to study the properties of the embedding in the sense that you want to study the embedded manifold together with its surrounding. Notice that knot theory is not equivalent to the classification of closed 1-manifolds! Every connected 1-manifold (being either an interval or $S^1$) embeds in a standard way into $\mathbb R^2$.


1

The following is a heuristic argument in terms of complex analysis: The given curve $\gamma$ bounds a simply connected domain $\Omega$ in the complex $w$-plane. By Riemann's mapping theorem there is a conformal map $f$ of the unit disk $D$ onto $\Omega$. We assume that $f$ is in fact analytic in some larger disk $D_\rho$, and that $f'(z)\ne0$ in all of ...


1

It doesn't make sense to integrate $d(\gamma^*\omega)$ on $B_2$ since $\gamma^*\omega$ is not defined on $B_2$ and consequently neither is $d(\gamma^*\omega)$. Put differently an interior point of $B_2$ is not related in any way to any point of $X$. Thus your use of Stokes is indeed unwarranted.


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As far as I concerned ,this property of embedding requires that $f(X)$ is closed in Y. There is a counter-example .(When $f(X)$ is not closed in Y) Let Y = $\{k_\lambda : \lambda \in A\} $ where A is a uncountable set. Let X be $\{k_1,\cdots,k_n,\cdots\}$ , an infinite countable subset of A. Topology of X is discrete topology , hence X is not compact in ...


1

Closed embeddings (equivalently, embeddings with closed image) are precisely proper injective immersions. $(\Rightarrow)$ If $f : M \to N$ is an embedding with $f(M)$ closed, then it is obviously proper (since the intersection of a compact set with $f(M)$ is compact). $(\Leftarrow)$ On the other hand, if $f : M \to N$ is a proper, injective immersion, it ...


1

Since $X=S^n\times\mathbb{R}\cong S^n\times (0,1)$, you can think of $X$ as an n-dimensional annulus or shell, without boundary. Usually, $X$ retracts onto $S^n$. But after removing a point, $X\setminus\{*\}$ retracts onto $S^n$, with a bump. This "bump" is caused by the removed point and looks like a copy of $S^n$ (you can easily visualize this for ...



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