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

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 ?

• In munkres topology 2 edition which is online and free, you can find on chapter four a complete answer to the imbedding of manifold question. Jul 21, 2016 at 12:00
• Every connected one-manifold without boundary is homeomorphic to a circle or open interval. There's an elementary proof in Differential Topology by Guillemin and Pollack. The intuitive idea is easy enough: Fix a point arbitrarily, and exhaust by compact intervals. If the ends join up in the limit, you're on a circle. If not, you're on an open interval. Jul 21, 2016 at 12:00
• Reference : Appendix in J. Milnor's beautiful book Topology from the Differential Viewpoint Jul 25, 2016 at 2:54
• Interestingly, you need to use both the Hausdorff condition and second countability for this result to be true. The line with two origins and the long line give counterexamples otherwise. Jul 26, 2016 at 17:53

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.

• This assumes the manifold is smooth. Jul 26, 2016 at 17:56
• I know man, life is a sad story. I'm sorry for the inconvenience. :) Jul 27, 2016 at 7:59
• How does the existence of fixed point give a contradiction? Nov 20, 2021 at 13:53