fundamental group intuition 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.
 A: since the question is about gaining intuition of fundamental groups,i would like to explain through basic examples.for eg,say you have a unit circle and a torus and one asks you to distinguish between these two surfaces.A very obvious fact distinguishing these surfaces is that a circle has presence of one hole while torus has two holes.But,this has no meaning until we define a way to mathematically represent holes and show that it is topologically invariant. 
Now,how can one mathematically represent these holes?Consider a surface with a hole in it.now,consider a point $x_0$ in it and a loop based at $x_0$ going around the hole and one not enclosing the hole as we can see in the diagrams 1 and 2 below.Now try to continuously deform this loop to the point $x_0$ in figure 1 and 2.As we can clearly see in figure 1,inability to deform the loop continuously to the point $x_0$ indicates a hole.

now,we generalize above method even more.We know that the relation of homotopy splits set of all loops based at a point,say $x_0$ into disjoint equivalence classes,where a homotopy class represents set of loops continuously deformable to one another.Moreover, we see that a loop going twice around a hole is not in the homotopy class of loop going once aroud the hole.so,we can associate integers n$\gt0$ ,n =0,n$\lt0$ to the homotopy class of loops according the no of times loop goes aound the hole in postive direction,dooes not wind around the loop and winds around the loop in negative direction.Moreover,the operation n+m has a geometrical meaning :n+m corresponds to going round the hole first n times and then m times.Thus,the set of homotopy classes is endowed with a group structure called the fundamental group which we can define as:
([f]*[g])(t) =
\begin{cases}
[f(2t)],  & \text{if $0\lt t \lt \frac{1}{2}$} \\
[g(2t-1)], & \text{if $\frac{1}{2} \lt t \lt 0$ }
\end{cases}
Now,we can  rigorously prove that homotopy classes of loops based at a point forms group under the above defined operation (as it was intuitvely clear).Moreover,we show that fundamental groups,which gives measure of no of holes in the space is a topological invariant. Fundamental group as outlined above,detects one dimensional holes in a space.
A: 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. 
