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 ?

  • $\begingroup$ 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. $\endgroup$ Jul 21, 2016 at 12:00
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    $\begingroup$ 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. $\endgroup$ Jul 21, 2016 at 12:00
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    $\begingroup$ Reference : Appendix in J. Milnor's beautiful book Topology from the Differential Viewpoint $\endgroup$ Jul 25, 2016 at 2:54
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    $\begingroup$ 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. $\endgroup$ Jul 26, 2016 at 17:53

1 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.

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

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