Changes

== Group Structure ==

== Group Structure ==

The preceding definition is incomplete, because it does not specify the operation on <math>\pi_1(X)</math> which gives these classes of paths the structure of not just a set, but a group.  The definition of the group operation is simple: given two paths <math>\gamma_1</math> and <math>\gamma_2</math>, the sum <math>\gamma_1 * \gamma_2</math> is the path obtained by first following <math>\gamma_1</math>, and then following <math>\gamma_2</math>.  More precisely, set

The preceding definition is incomplete, because it does not specify the operation on <math>\pi_1(X)</math> which gives these classes of paths the structure of not just a set, but a group.  The definition of the group operation is simple: given two paths <math>\gamma_1</math> and <math>\gamma_2</math>, the sum <math>\gamma_1 * \gamma_2</math> is the path obtained by first following <math>\gamma_1</math>, and then following <math>\gamma_2</math>.  More precisely, set

:: <math>(\gamma_1 * \gamma_2)(t) = \begin{cases} \gamma_1(2t) & \textrm{if \(0 \leq t \leq 1/2\)} \\ \gamma_2(2t-1) & \textrm{if \(1/2 \leq t \leq 1\)} \end{cases}</math>

:: <math>(\gamma_1 * \gamma_2)(t) = \begin{cases} \gamma_1(2t) & \textrm{if $0 \leq t \leq 1/2$} \\ \gamma_2(2t-1) & \textrm{if $1/2 \leq t \leq 1$} \end{cases}</math>

For this to be a group, it is necessary to see that there is an identity element with respect to the operation, and that every loop has an inverse.  It is clear that the constant loop at <math>*</math> serves as an identity, and the inverse of a loop <math>\gamma</math> is obtained simply by following <math>\gamma</math> in the opposite direction:

For this to be a group, it is necessary to see that there is an identity element with respect to the operation, and that every loop has an inverse.  It is clear that the constant loop at <math>*</math> serves as an identity, and the inverse of a loop <math>\gamma</math> is obtained simply by following <math>\gamma</math> in the opposite direction:

:: <math>(\gamma^{-1})(t) = \gamma(1-t)</math>.

:: <math>(\gamma^{-1})(t) = \gamma(1-t)</math>.