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Does a bounded helix; for instance $\{(\cos 2\pi t, \sin 2\pi t, t); -5\leq t\leq5\}$ in $\mathbb R^3$ with the projection map $(x,y,z)\mapsto (x,y)$ form a covering space for the unit circle $\mathbb S^1$?

If yes, does it disprove: There is a bijection between the fundamental group of a space and the fibers of the base point into a simply connected covering space? enter image description here

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    $\begingroup$ This re-write seems wrong, since the points $(\cos(x),\sin(y),z)$ do not lie on a helix, but instead since $x,y,z$ are unrelated form a box in 3-space. $\endgroup$ – coffeemath Jul 21 '13 at 4:06
  • $\begingroup$ I changed it again! $\endgroup$ – Ronald Jul 21 '13 at 4:19
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Yes it's a covering map. Since the fundamental group of $S^1$ is $\mathbb{Z}$, and since each fiber is the set of $(\cos(t_0),\sin(t_0),t_0+2\pi k)$, $k \in \mathbb{Z}$, there is no conflict to the bijection you mention.

Note the question has been reformulated and this answer is no longer relevant; I'm awaiting further clarification from OP as to what the rephrased bijection assertion means.

The re-edit now has the parameter $t$ bounded, i.e. $t \in [-5,5]$, in the helix parametrization $(\cos 2\pi t,\sin 2\pi t, 2\pi t).$ This is not a covering space, since for $t$ near an end, say near 5, the neighborhood of $(0,0)$ in the base is not covered by an open interval along the helix. In a simple covering space over the circle, if you are on a sheet of the cover over any point in the base, you should be able to move back or forth a bit and remain on that sheet, locally.

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  • $\begingroup$ sorry, my symbols were misleading I edited the question. So the answer will be changed or no? $\endgroup$ – Ronald Jul 21 '13 at 3:13
  • $\begingroup$ In this particular case the codimension is 1 and there is only one fiber of the base point, which is what I have above with $t_0$ replaced by $0$. By your phrase here "fundamental group of the base point and the fibers of the base point" do you mean the Cartesian product of these? If so then it is $\mathbb{Z} \times S^1$ which is trivially a simply connected covering space of $S^1$ $\endgroup$ – coffeemath Jul 21 '13 at 3:37
  • $\begingroup$ I've read in some book a fact that there is a bijection between the fundamental group of a space and the fibers of the base point into a simply connected covering space. But if the semi-helix in the question is covering space of the circle then $p^{-1}(x_0)$ is finite whereas $\pi_1(S^1)=\mathbb Z$. $\endgroup$ – Ronald Jul 21 '13 at 3:41
  • $\begingroup$ My comment above shows I'm confused. I would appreciate it if you included in your question an exact quote of the theorem concerning the bijection you mention. "A bijection between the fundamental group of a space and the fibers of the base point into a simply connected covering space" leaves one wondering what the and means, maybe Cartesian product, while the into suggests the bijection is into a covering space. By the way what is a simply connected covering space, is that just that the covering manifold is simply connected? $\endgroup$ – coffeemath Jul 21 '13 at 3:55
  • $\begingroup$ i attached the theorem $\endgroup$ – Ronald Jul 21 '13 at 4:28

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