So, if I take one torus and take off one single point, what will be its fundamental group? I think that one single point will not change the topology in this sense. Or will it? If so, how?
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asked Jun 30 '15 at 1:06
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1$\begingroup$ In fact, if I think well it will be like an eight, right? Then, the correct value of the fundamental group for this figure is $\mathbb{Z}*\mathbb{Z}$, is this? $\endgroup$ – L.F. Cavenaghi Jun 30 '15 at 1:10
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2$\begingroup$ removing a point will often change the topology drastically. For example, $\Bbb R^2$ minus a point gives you $\Bbb Z$ $\endgroup$ – Ben Grossmann Jun 30 '15 at 1:10
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3$\begingroup$ It's easiest to think about the fundamental domain of a square, remove the center of the square, and then retract the remainder onto the boundary. Then, perform the typical identifications of the boundary and you should get a figure 8. $\endgroup$ – Michael Burr Jun 30 '15 at 1:17
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3$\begingroup$ The reason that the fundamental group changes is that the fundamental group measures how to "go around" holes in the object. When you remove a point, you have a new hole to "go around" $\endgroup$ – Michael Burr Jun 30 '15 at 1:19
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1$\begingroup$ @Leonardo Just for extra information: If the dimension of the manifold is greater or equal to $3$, then removing a point does not change the fundamental group. You can use Seifert-Van Kampen to prove that, but I guess this is done somewhere in Lee's Topological Manifolds book. $\endgroup$ – Ivo Terek Mar 16 '16 at 4:54
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If we remove a point the topology will change and the fundamental group will change drastically! The fundamental group of the torus is $$ \pi_1(T)=Z\oplus Z $$ but if you remove a point the fundamental will be even a non abelian group: $$ \pi_1(T-\{p\})= F(\alpha, \beta) $$ where $\alpha$ and $\beta$ are the two cycles of the figure 8, and $F(\alpha, \beta)$ is the free group generated by $\alpha$ and $\beta$.
All theses problems appear because of the existence of the so called "Peano curves" https://en.wikipedia.org/wiki/Space-filling_curve (which are really complicated curves that are surjective). For example following this naive idea that removing a point does not change the topology we could end up with the following WRONG proof for the sphere:
1-"We remove a point to $S^2$ and does not change anything", then $S^2-{p} \simeq R^2$, but $R^2$ is contractible and we get $\pi_1(S^2) = \pi_1(S^2-{p}) = \pi_1(R^2)=0$.
Although the achieved result is true, we have to be careful because this proof is COMPLETELY WRONG! And the reason is the subtle idea that there exist paths filling $S^2$ which clearly will not have a homotopy equivalence class in $\pi_1(S^2-{p})$.
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5$\begingroup$ It's not clear at all why you think the existence of space filling curves is what's leads to a change in the fundamental group... $\endgroup$ – Najib Idrissi Feb 26 '16 at 9:17
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$\begingroup$ You are right. Certainly is not a proof, but gives you a pretty simple idea why it should not work. I am not saying is the REASON that changes the fundamental group, it is just a counter example curve that does not fit in the fundamental curve of the space minus a point. $\endgroup$ – ycidruiz Feb 26 '16 at 9:30
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$\begingroup$ I don't believe your comment about space filling curves is correct: a consequence of the simplicial approximation theorem is that every path is homotopic to a non-space fillling path, so you can fix the wrong proof that $S^2$ is simply connected this way. The reason this doesn't work for $T$ isn't due to the existence of space filling curves, but rather you can't make the same conclusion since $\pi_1(T-\{p\})$ is non-trivial. $\endgroup$ – ktoi Mar 16 '17 at 17:01
