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My intuition tells me that the only connected 1-dimensional topological manifolds are the real line $\mathbb{R}$ and the circle $S^1$.

Is this true?

If yes, is it possible to prove it from first principles, or is it something that needs some highly technical theorems?

If no, do we have an example of connected 1-dimensional manifold not homeomorphic to $\mathbb{R}$ or $S^1$?

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    $\begingroup$ Yes, it's true (assuming second countable and Hausdorff too, to exclude the long line and the line with a double origin, respectively). I think this has been asked before: math.stackexchange.com/questions/113705/… $\endgroup$
    – Amitesh Datta
    Commented May 29, 2015 at 15:15
  • $\begingroup$ @T-gee Oh I see, the locally homeomorphic doesn't apply at the meeting point. Cool $\endgroup$
    – Eoin
    Commented May 29, 2015 at 15:16

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By request, see here for an outline from first principles: http://www.igt.uni-stuttgart.de/eiserm/lehre/2014/Topologie/Gale%20-%201-manifolds.pdf

EDIT: this link is now dead. The reference is

Gale, D. (1987). The classification of 1-manifolds: a take-home exam. The American Mathematical Monthly, 94(2), 170-175.

This proof proves that all topological 1-manifolds (without boundary) are homeomorphic to $\mathbb{R}$ or $S^1$. From this we get the corollary that all $1$-manifolds can be endowed with $C^{\infty}$ structures. In fact, it is true that all 1-manifolds are diffeomorphic to $\mathbb{R}$ or $S^1$. For this proof, my guess is that using integral curves/orientations is probably the easiest, though check Guilliman/Pollack for sure.

Additionally, there is another proof with more of an algebraic topology flavor using CW complexes found in Lee's Topological Manifolds, Theorem 5.27.

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