# Tag Info

4

There are tons and tons of examples: $\mathbb{CP}^n$ for $n \ge 2$, $S^n \times S^m$ for $n,m \ge 2$, connected sums of these when they have the same dimensions, products of these... The list goes on and on. It might help to review why a 2- or 3-dimensional closed manifold which is simply connected is a sphere to see why it fails in higher dimensions. In ...

3

In complete generality, this is not possible. For example, consider the $1$-form $\omega = \sin x\,dx$ on $\mathbb R$. It's possible to find a countable open cover and partition of unity such that half of the terms $\int_{\phi_n(U_n)}(\phi_n^{-1})^*(\rho_n\omega)$ are equal to some constant $c$, and the other half are equal to $-c$. The limit of the sum ...

2

If the set contains three non-collinear points, then it contains a triangle, and the answer is clearly yes. If the set does not contain three non-collinear points then it is a closed segment or a point. If you are looking for closed simple curves, then in this case the answer is no.

1

Here is a topological argument: the inclusion $i:S\hookrightarrow M$ factors through $T$, hence by functoriality $i_*: H_*S \to H_*T \to H_*M$. The first map maps the fundamental class to zero. In particular the homology class which $S$ represents in $M$ is trivial. Note that the boundary of a manifold is nullhomologous, by either stratifold theory, or ...

1

Showing that the normal bundle is trivial is the same thing as demonstrating a nonvanishing global section of $\eta$ (since it's one-dimensional). At each point $n \in N$ let $v \in (TM/TN)_n$ be the unique vector with $df(v) = 1 \in T_y \Bbb R$. (Why is there a unique such vector?) This is a continuous section of the normal bundle, as you can check in local ...

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