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How i see this the homotopy group of order 1 is giving information about the "holes" in a topological space. In that way what kind of information is giving the homotopy group of order 2 or n in general?

I can't think pretty clear about it, but in the first case it's seems clear that a hole is defining different equivalent classes of homotopic loops in the space.

But the FG of order 2 is like homotopic classes of loops in the loop space of the space X, so is basically holes in the loop space, and this actually is like: "The actual points for the loop are in the set but you can't walk through the loop in some cases", a loop passing in some points can be absent but other loop passing thru the same points with different parametrization can be there, so, is not about "holes" in the real space, but like "phantom holes" that are just holes in the loop space. Is hard to imagine this.

So, how do you imaging the info giving by the homotopy group of order 2 of X (or n in general) in the space X?

Thanks.

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    Yes, thank you, I changed in the post.2012-11-26

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In regards to holes, the first fundamental group only tells you about holes that can be enclosed by a loop -- if you remove a point from Euclidean 3-space, it turns out the first fundamental group is trivial. (but if you remove a line, or a loop, or a knot, the first fundamental group can tell you about that)

To detect holes that are enclosed by spheres, such as removing a single point from Euclidean 3-space, you need the second fundamental group. And so forth.

The higher homotopy groups are usually defined as homotopy classes of $n$-spheres.

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    The connection between higher homotopy groups of a space, and the iterated loop spaces of a space is given by the adjunction between the loop space functor and the suspension functor. Let $[A,B]$ be the set of classes of homotopic maps $A\rightarrow B$ which preserve basepoint. Then there is a natural isomorphism $[\Sigma S^n,Y]\cong [S^n,\Omega Y]$. The suspension of the $n$-sphere is the $(n-1)$-sphere and so this says that $\pi_n(Y)\cong\pi_{n-1}(\Omega Y)$. You can obviously continue this to get $\pi_n(Y)\cong\pi_1(\Omega^{n-1}Y)$.2013-01-13