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11

No $\mathbb{R}$-vector bundle (of dimension $\geq 1$) is ever compact. Proof: Let $S$ be an arbitrary topological space and let $T$ be a vector bundle of dimension $n\geq 1$, and let $p:T\rightarrow S$ be the bundle projection. If $S$ a manifold, or is even $T_1$, or even more generally if any point of $S$ is closed, then the proof is particularly easy: ...


8

Yes, the obvious vector space inclusion $\iota: \mathbb{R}^3 \hookrightarrow \mathbb{R}^4$, or its restriction $\iota\vert_{\mathbb{S}^2} : \mathbb{S}^2 \hookrightarrow \mathbb{S}^3$, induces an embedding of $\mathbb{RP}^2$ into $\mathbb{RP}^3$, and the latter is orientable.


4

Hints: It suffices to find a closed, non-compact subset of the total space. In an arbitrary vector bundle (of positive rank), there's a particularly natural way to do this.


3

The usual Euler characteristic is a homotopy invariant so you can calculate it for $U$ by replacing $U$ with something homotopy equivalent. I believe $U$ deformation retracts to a wedge sum of $2g + r - 1$ circles. This computes the Euler characteristic as $1 - (2g + r -1) = 2 - 2g - r$ using the fact that the homology of a wedge sum is the direct sum of ...


2

As stated it is false. Let $$ U = (-1,0) \cup (10,11) \subset \mathbb{R}$$ and $$ V = (1,2) \cup (15,16) \subset \mathbb{R} $$ Take $$ K = \{ -2/3, -1/3 , 10.5\} $$ Let $\gamma$ be the map $$ \gamma(x) = \begin{cases} x + 16 & x \in (-1,0) \\ x - 9 & x \in (10,11)\end{cases} $$ which is clearly a diffeomorphism of $U$ and $V$. But ...


2

The Mayer-Vietoris sequence goes $\cdots \rightarrow H^{n - 1}(U) \oplus H^{n - 1}(V) \rightarrow H^{n-1}(U \cap V) \rightarrow H^{n}(M) \rightarrow H^{n}(U) \oplus H^{n}(V) \rightarrow 0$. Note that since $V \cong \mathbb{R}^{n}$ we have $H^{n}(V) = 0$. We also already know that $H^{n-1}(U \cap V)$ and $H^{n}(M)$ are both $\mathbb{R}$, so in particular they ...


1

$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Basis}{\mathbf{e}}$Let $V$ be a $k$-dimensional subspace of $\Reals^{N}$, let $(v_{i})_{i=1}^{k}$ be an ordered basis of $V$, and let $(\Basis_{i})_{i=1}^{k}$ denote the standard basis of $\Reals^{k}$. The mapping $\varphi:V \to \Reals^{k}$ defined by $$ \varphi(x_{1} v_{1} + \dots + x_{k} v_{k}) = (x_{1}, \dots, ...


1

If you restrict the inner product on $\mathbb{R}^3$ to the sphere $\mathbb{S}^2$, the result is a Riemannian metric. Every Riemannian metric comes with a canonical top-dimensional form, the volume form, which tells you which bases for a tangent space have unit volume. This 2-form is the volume form for the metric on the sphere. It's called an "area form" ...



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