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8

That $f$ is a local diffeomorphism implies that it is an open map. Hence $f(M)$ is an open subset of $S^n$. But $M$ is compact, so $f(M)$ is also closed. Since $S^n$ is connected, it follows that $f$ is surjective. Further, since $df_x$ has full rank everywhere, the fibres $f^{-1}(p)$ are discrete. Hence the fibres are finite. Use that to find that $f$ is a ...


6

Tensor products are taken with respect to $C^{\infty}(M)$; (multi)linear means $C^{\infty}(M)$-(multi)linear; $\Gamma(E)$ denotes the space of sections of a vector bundle $E\xrightarrow{\pi}M$; an open set $U\subset M$ is said to trivialize $E$ if the restriction $E\big|_{U}\to U$ of $E$ to $U$ is trivial. There is a first general fact: given vector ...


5

Once you choose local coordinates, a basis for the derivations is $\left\{\dfrac{\partial}{\partial x^1}\Big|_p,\dots, \dfrac{\partial}{\partial x^n}\Big|_p\right\}$. Then there's an obvious curve with tangent vector $\sum\limits_{i=1}^n a_i \dfrac{\partial}{\partial x^i}\Big|_p$.


5

Read about both simultaneously in the same book: Differential forms in Algebraic Topology ! (Bott was one of the best twentieth century geometers, and it shows in this extraordinary book)


5

This is certainly a valid proof, and whilst the method behind it is very intuitive, actually writing out what's being doing is very messy, especially considering the problem feels like it should be "obvious". A much "nicer" proof comes from using the following equivalent definition of derivative, if $f:X\rightarrow Y$ is a smooth map of manifolds embedded ...


5

Yes. One of many ways to see this is to fit $\mathbb{CP}^n$ into a fiber sequence $$S^1 \to S^{2n+1} \to \mathbb{CP}^n$$ (since $S^1 \cong \text{U}(1)$ acts by scalars on the unit sphere $S^{2n+1} \subset \mathbb{C}^{n+1}$ with quotient $\mathbb{CP}^n$) and apply the long exact sequence in homotopy. The long exact sequence also shows that ...


4

It is quite amazing but such maps do exist! Call a map between two metric spaces (say, Riemannian manifolds) $f: M\to N$ a path-isometry if it preserves lengths of all paths. Recall that if $p: [a,b]\to M$ is a path (say, Lipschitz-continuous), then its length is defined as $$ \int_a^b |p'(t)|dt. $$ Example. $f: R\to R$, $f(x)=|x|$. This map is not a ...


4

The result is: The product of spheres $X=S^{n_1}\times S^{n_2}\times \ldots \times S^{n_k}$ (containing at least two factors) is parallelizable iff at least one of the $n_i$ is odd. First, if all the $n_i$ are even, the $\chi(S^{n_1}) = 2$ for all $i$, and hence $\chi(X) = 2^k \neq 0$. This implies $X$ is not parallizable. In fact, it implies that ...


4

Theorem Every closed embedded hypersurface of $S\subset \mathbb R^{k+1}$ is orientable. Proof The hypersurface $S$ is defined by family $(U_i,f_i)$ where the $U_i$'s are an open covering of $\mathbb R^{k+1}$ and the $f_i:U_i\to \mathbb R$ are $C^\infty$ functions subject the condition that there exist $C^\infty$ functions $g_{ij}:U_i\cap U_j\to \mathbb ...


4

Consider this part of the long exact sequence for $\partial M\subset M$: $$H_{m+1}(M,\partial M;\mathbb R)\xrightarrow{a} H_m(\partial M;\mathbb R)\xrightarrow{b} H_m(M;\mathbb R).$$ This sequence is isomorphic to its dual $$H_m(M;\mathbb R)^*\xrightarrow{b^t} H_m(\partial M;\mathbb R)^*\xrightarrow{a^t} H_{m+1}(M,\partial M;\mathbb R)^*$$ (dual spaces, ...


4

I think @squirrel's approach can probably be made to work. But here's another approach that doesn't require dealing with a finite cover. Suppose $K\subseteq P\times P$ is a compact set. Let $\pi\colon P\to P/G$ denote the bundle projection. To show that the preimage of $K$ is compact, we'll show that it's sequentially compact (which is equivalent for ...


3

If a smooth $n$-manifold $M \subset \mathbb{R}^m$ is globally defined as the preimage of a regular value of a smooth function $f:\mathbb{R}^n \to \mathbb{R}^{m-n}$, then we have the desired setup $$TM=\{(x,v) \in \mathbb{R}^{m} \times \mathbb{R}^{m} \mid f(x)=0 \text{ and } df_{x}(v)=0\}.$$ If you read his answer more carefully, @jef808 was saying that we ...


3

Of course this is not true as stated because there exist nonorientable closed $3$-manifolds, for example $S^1 \times \mathbb{RP}^2$. Once you have orientability $w_1 = v_1 = 0$.


3

Your definition of a differential $p$-form on an $n$-dimensional manifold $M$ is slightly different from the one I learned. The definition I'm used to is: given an $n$-dimensional manifold $M$, a differential $p$-form $\omega$ is a function which assigns to each point of $M$ an alternating $p$-tensor whose domain is the tangent space at that point. That is, ...


2

The same happens with usual derivatives. If you work on a closed interval, say $[-1,1],$ then the function $f(x)=\sqrt{1-x^2}$ is $C^{\infty}$ in $(-1,1).$ However, it is not $\require{cancel} \cancel{\textrm{derivable}}$ differentiable at $x=-1,x=1.$ So, if you want to have derivatives (partial derivatives in higher dimensions) you have to consider the ...


2

The definition of the $i$th partial derivative is really as follows: at a point $x = (x_1,\dots,x_n)$, we say $$ \left.\frac{\partial f}{\partial x_i}\right|_{x} = \lim_{t \to 0} \frac{f(x_1,\dots,x_{i-1},x_i+t,x_{i+1},\dots,x_n) - f(x_1,\dots,x_{i-1},x_i,x_{i+1},\dots,x_n)}{t} $$ However, in order for this limit to make sense, $f$ needs to be defined ...


2

The two are fairly closely linked, and it wouldn't be a bad idea to study them concurrently. That said, I think if you have to pick one, go with algebraic topology first. The first reason is historical. Before the modern theory of differential topology was developed, algebraic (and combinatorial) methods were employed. It is usually much easier to calculate ...


2

There's probably a more complete and efficient proof out there, but here's a partial answer. Let's show it for trivial bundles $P \cong N \times G$ for some smooth manifold $N$ with trivial action $\theta:G \times( G \times N) \to G \times N$ given by $\theta(g,(h,n))=(gh,n)$. Then we'll wave our hands with some local trivializations for the general case. ...


2

When you ask that $G \times G \to G$ is a smooth map, it means "with respect to the smooth structure you put on $G$ and the product differentiable structure on $G \times G$", so that asking this map to be smooth makes sense since it becomes a map between smooth manifolds and you can ask yourself this question. So somewhere you already assumed there was a ...


2

If you take the case where $a$ is negative, just say $-1$ for ease, you get, $$z_0-z_3+z_1=0$$ or $$z_0-z_3-z_1=0$$ the union of two planes, so their points of intersection are not smooth. If $a$ is positive, say $1$, then you have $$z_0=z_3$$and $$z_1=0$$ this seems to be the intersection and its a line, so it seems to me to be smooth ? But its dimension ...


2

Here a counterexample to 2) (without "properness" assumptions) Let $B$ the ball of center $0$ and radius $1$. Let $g:B\to\mathbb R$ be a function with property 2) (positive with two minima where the function is zero and no other critical point). This is easily constructed. Now consider any diffeomorphism $\phi:\mathbb R^2\to B$. Now the function ...


2

A topological manifold is by definition locally homeomorphic to $\mathbb{R}^n$. Locally meaning there is an open set containing each point that is homeomorphic. If you're talking about topology you would expect open sets to come up since the collection of open sets is by definition the topology on a set.


2

According to the definition I have seen $\Omega^n_{fr}(X)$ is the set of cobordism classes of continuous functions $f\colon M^n\to X$. The map $f_1\colon M\to X$ is said to be cobordant to $f_2\colon N\to X$ if there is a cobounding $(n+1)$-manifold $W$ so that the maps $f_1$ and $f_2$ can be extended over all of $W$ to a map to $X$. I'm spelling this out ...


1

I think I have it. There is only one covering map $S^n\rightarrow S^n$, up to isomorphism, for $n\geq 2$. If I can show that $f$ is a covering map I will be done. Since every value $y\in S^n$ is regular, the set $f^{-1}(y)$ is finite, say $\{x_1\ldots,x_k\}$. By the Inverse Function Theorem and the fact that $S^n$ is Hausdorff, I can find pairwise ...


1

Informally said, open subsets are those which represent “full pieces” of the space. They are “saturated” in the sense that they contain all the local information for each point they contain. Every local property for every point holds in open subset if and only if it holds in the whole space.


1

You're probably looking at the first edition of Introduction to Topological Manifolds. I realized that my hint to "use stereographic projection" was not very helpful, so in the second edition I expanded on it: Consider the map $\pi\circ\sigma^{-1}\colon\mathbb R^n\to \mathbb R^n$, where $\sigma$ is the stereographic projection and $\pi$ is a projection ...


1

You're right, this is a mistake. I've added a correction to my online errata list. Thanks for pointing it out.


1

I agree that this is a mistake in the text (that doesn't affect the truth of the lemma, as you noted). For instance, if you start with the constant function $f=0$ and apply the construction, you get $\operatorname{supp}\widetilde{f}=\emptyset$, which can't be equal to the right side unless $A =\emptyset$ as well.


1

Is it true for compact simply-connected oriented smooth 4-manifolds that a real 4-dimensional vector bundle is determined by its Stiefel-Whitney and Pontrjagin classes? No. For example, the tangent bundle of $S^4$ is stably trivial (because the outward pointing normal gives a trivialization of its normal bundle in $\mathbb{R}^5$; true more generally of ...



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