How to prove that there are only two kinds of 1-dim manifolds without boundary

Question

I just know a conclusion that all 1-dim manifolds without boundary is homomorphism to $S^1$ or $\mathbb{R}$ , but I don't know how to prove it . Why is so ?

Answer 1

Here a sketch of the proof.

Suppose that $M$ is a smooth 1-dimensional manifold.

Let's first suppose that $M$ is orientable. This means that we can find a global top-dimensional never-vanishing differential form over $M$, i.e. (up to fixing a Riemannian metric over $M$) a never-vanishing global vector field $X \in \Gamma(TM)$.

It is easy to check (the details are up to you) that a flow line $\gamma: \mathbb{R} \to M$ of $X$ gives either a diffeomorphism $M \simeq \mathbb{R}$ , or a periodic map (with say period $T$) that descends to a diffeomorphism $M \simeq \mathbb{R}/T \mathbb{Z} \simeq S^1$.

Now we have just to rule out the evenience that there are non-orientable 1-dimensional manifolds. In order to do this suppose that $M$ is non-orientable and consider its universal cover $\widetilde{M}$. This is an orientable 1-diemnsional manifold (in fact, it is simply connected), and you can check (using what we proved so far) that $\widetilde{M}\simeq \mathbb{R}$.

Now, if $M$ is non-orientable, there is $\gamma \in \pi_1(M)$ acting on the universal covering $\widetilde{M}\simeq \mathbb{R}$ as an orientation-reversing diffeomorphism. In order to obtain the contradiction, notice (this is elementary analysis) that an orientation reversing diffeomorphism of the real line always has a fixed point.