Homotopy group

Definition

As homotopy classes of based maps

Suppose $(X,x_0)$ is a based topological space and $n$ is a positive integer. The homotopy group ${\pi }_n(X,x_0)$ is defined as follows:

• Consider the based $n$-sphere $(S^n,p)$ where $p$ is a chosen basepoint. As a set, ${\pi }_n(X,x_0)$ is the set of homotopy classes of all based maps from $(S^n,p)$ to $(X,x_0)$, where the homotopy classes are with respect to homotopies that preserve basepoints.
• Two maps $f_1,f_2$ are composed as follows. $S^n\setminus \{p\}$ is identified with the open northern and open southern hemisphere of a new sphere via homeomorphic identifications ${\phi }_1$ and ${\phi }_2$ from these hemispheres to $S^{\setminus }\{p\}$ (these identifications are universally fixed, independent of $X$; there's a natural choice for them). The composite map is now defined as follows: as $f_1\circ {\phi }_1$ on the northern hemisphere, as $f_2\circ {\phi }_2$ on the southern hemisphere, and as the constant map to $p$ on the equator. The basepoint is a fixed point on the equator (again, this choice is independent of $X$ and is universally fixed).

The case $n=0$

The definition of homotopy group still gives a set definition for $n=0$. $S^0$ is a two-point space, and one of these points must go to a fixed basepoint, while the other can go anywhere. Thus, the set of all based maps is the set of points in $X$, and the set of homotopy classes is the set of path components. Thus, ${\pi }_0(X,x_0)$ is the set of path components in $X$. Note that it is independent of $x_0$ because $S^0$ being discrete, the image of the basepoint does not affect where the other point goes.

However, the composition operation does not make sense for $n=0$, because $S^0$ has an empty equator. Hence, ${\pi }_0$ is only a set and has no group structure for arbitary topological spaces. (It does have a group structure when the topological space is a H-space, induced by the multiplication in the topological space).

The case $n=1$

In this case, we get the fundamental group ${\pi }_1(X,x_0)$. Recall that for $f_1,f_2$ based maps from the circle to $(X,x_0)$, we think of $f_1,f_2$ as maps from $[0,1]$ to $X$ with $f_1(0)=f_1(1)=f_2(0)=f_2(1)=x_0$. The usual way of composing is to define:

Failed to parse (unknown function "\begin{array}"): {\displaystyle (f_1 * f_2)(t) := \lbrace \begin{array} f_1(2t), & 0 \le t \le 1/2 \\ f_2(2t - 1), & 1/2 < t \le 1 \\\end{array}}

Here, the definition on $(0,1/2)$ can be viewed as the northern hemisphere definition, the definition on $(1/2,1)$ can be viewed as the southern hemisphere definition, with the equator corresponding to the two points $1/2$ and $0\sim 1$, of which we choose the latter as basepoint.

Omission of basepoint

For a path-connected space, the homotopy groups ${\pi }_n$ for all basepoints are isomorphic. In fact, any choice of path between two points can be used to define an isomorphism between the ${\pi }_n$s at these basepoints. The key fact that we need to use here is that the inclusion of a point in $S^n$ is a cofibration (which is easily seen by noting that $S^n$ is the boundary of $D^n$, or more generally from the fact that manifold implies nondegenerate).

In general, the homotopy group ${\pi }_n$ may differ for different path components. For a homogeneous space, or more generally for a space where all the path components are homeomorphic, the isomorphism class of ${\pi }_n$ does not depend upon the choice of basepoint.

Dependence on homotopy type

The homotopy groups ${\pi }_n$ depend only on the homotopy type of the based topological space. In fact, they depend only on the homotopy type of the path component of the basepoint in the topological space.

However, knowledge of the homotopy groups does not determine the homotopy type, or even the weak homotopy type, of the topological space.