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So I understand the mechanics of the fundamental group, but I want to gain a more natural intuition behind it. I imagine the fundamental group $\pi_{1}(X)$ to detect "holes" in a space. For example, it detects the hole in the punctured plane. However, it does not detect all type of holes, specifically the hole at $\mathbb{R}^3-0$. Presumably, $\pi_{2}(\mathbb{R}^3-0)$ could do the job in this case. I was wondering if you guys could give me some more geometrical intuition behind the fundamental group. For example, what types of "holes" can $\pi_{1}(X)$ detect? Let's restrict the discussion to subspaces of $\mathbb{R}^{n}$.

Sorry if this is a repost.

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The topological spaces $\mathbb{R}^n - \textbf{0}$ and $S^{n-1}$ (the unit $(n-1)$-sphere in $\mathbb{R}^n$) are homotopy equivalent because there is a deformation retraction of $\mathbb{R}^n - \textbf{0}$ onto $S^{n-1}$. In particular, you can also consider the $2$-sphere $S^2$ for the purposes of your question. In some sense, the fundamental group "detects" "two-dimensional holes". – Amitesh Datta Dec 23 '11 at 1:19

Recall the definition of the fundamental group: $\pi_1(X)$ is the set of homotopy classes of maps from $S^1$ into $X$, made into a group by concatenation.

So imagine a particular map $f:S^1 \to X$, I visualize this as taking a (one-dimensional) rubber band, and putting it somewhere in $X$ (I'm allowed to twist it, and have it intersect itself). If $f$ is homotopy equivalent to a constant map -- that is, if you can deform your rubber band within $X$ to a point in $X$ -- then $f$ is a representative of the identity in $\pi_1(X)$. A representative of a non-constant class in $\pi_1$ is one where your rubber band is somehow looped around something, in the sense that you can't shrink it to a point within $X$.

In your example, a rubber band in the punctured plane which loops around $0$ can't be deformed to a point within the punctured plane, so a map such as the inclusion of the unit circle into the punctured plane represents a non-trivial element of $\pi_1$ -- so $\pi_1$ does detect this hole.

In $\mathbb{R}^3 - 0$, suppose your rubber band is sitting on the unit circle in the $z=0$ plane. You can lift the band straight up into, say, the $z=1$ plane, and then within the $z=1$ plane shrink it to a point. You should be able to convince yourself that wherever in $\mathbb{R}^3 - 0$ you put the rubber band, you'll be able to move it within $\mathbb{R}^3 - 0$ enough to shrink it to a point. So $\pi_1$ doesn't detect this hole.

As you say, this hole would be picked up by $\pi_2$. The group $\pi_2(X)$ is defined as the group of homotopy classes of maps from $S^2$ to $X$. I imagine maps from $S^2$ to $X$ as placing a stretchy $2$-sphere in $X$ (again, I'm allowed to stretch, twist, squish, and have it intersect itself). A $2$-sphere sitting around $0$ in $\mathbb{R}^3 - 0$ can't be deformed to a point within $\mathbb{R}^3 - 0$, so it represents a non-trivial element of $\pi_2$ -- that is, $\pi_2$ does detect this hole.

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