For any smooth manifold, is it true that for any two points on the manifold, there exists a chart that covers the two points?

Question

Some say that we can connect two points with a continuous curve and a small contractible neighborhood of the path together with the charts of the points can be regarded as the chart. But I don't know how to verify this statement.

Answer 1

A connected smooth manifold $M$ has "$n$-transitive automorphism group" - this means that, for any set of $n$ distinct points $x_i$, and any set of $n$ distinct points $y_i$, there is a diffeomorphism $\varphi$ with $\varphi(x_i) = y_i$.

If $x_1$ and $x_2$ are your two points, pick a small chart around $x_1$; pick $y_1 = x_1$ and $y_2$ to be some point in the chart around $x_1$. If $(U,\psi)$ is the chart around $x_1$, then your desired chart is $(\varphi^{-1}(U),\psi \circ\varphi)$.

An alternate approach finds a smooth embedded path from $x_1$ to $x_2$ and takes a tubular neighborhood. To be precise: find such an embedded path and slightly extend it to a path $(-\varepsilon, 1+\varepsilon) \to M$ with $\gamma(0) = x_1$, $\gamma(1) = x_2$. The tubular neighborhood theorem says that there is an open neighborhood of this path diffeomorphic to the total space of its normal bundle; the only vector bundle over an interval is trivial, so the tubular neighborhood theorem says that a neighborhood of this path is diffeomorphic to $(-\varepsilon, 1+\varepsilon) \times \Bbb R^{n-1}$, giving us the desired chart.