# Mathematics Prelims

## March 17, 2009

### The Fundamental Group, Part I – Connecting Paths

Filed under: Topology — cjohnson @ 4:27 pm

Imagine that you have two paths, $f : [0, 1] \to X$ and $g : [0, 1] \to X$.  If it should happen that the terminal point of $f$ is the initial point of $g$ (that is, if $f(1) = g(0)$), then we can basically concatenate $f$ and $g$ into a new path $h$.  To do this formally we need to define $h : [0, 1] \to X$, which means we have to reparameterize $f$ and $g$ a little bit.  We’ll define $h$ so that it behaves as $f$ on [0, 1/2] and as $g$ on [1/2, 1] (notice this will be well-defined since we’re assuming the terminal point of $f$ is the initial point of $g$).

$\displaystyle h(t) = \left\{ \begin{array}{ll} f(2t) &: t \in [0, 1/2] \\ g(2t - 1) &: t \in [1/2, 1] \end{array} \right.$

We will let the asterisk denote this concatenation operation: $f * g = h$.

In the case that $f$ and $g$ are loops with the same basepoint, we’re just gluing the two loops together.  In the image below the basepoint is represented by the red dot.  We’re taking these two loops and gluing them together so that we first wind around the first loop (the one on the left) and then around the second loop.

Note that the red dot is the same basepoint for both loops, even though in our picture it doesn’t stay aligned (just for graphical convenience).

If $f$, $g$ and $h$ are loops (with the same basepoint) and $f \simeq g$, we can show that $f * h \simeq g * h$.  Letting $\{ f_t : [0, 1] \to X \}$ be the homotopy taking $f$ to $g$, we construct the homotopy $\{ h_t : [0, 1] \to X \}$ by defining $h_t = (f_t * h)$.  This always behaves as $h$ on the “second half” of the loop, and deforms to $g$ on the first half.  Of course, this also can be used to show that $h * f \simeq h * g$.

We can now use $*$ to help us form an operation on the homotopy classes of loops.  We will simply define ${}[f] \cdot [g]$ as ${}[f * g]$.  This operation is clearly associative, since our concatenation is associative. The identity element of this operation is the homotopy class of the constant function.  Keeping in mind that these are loops, the constant function is the function that maps everything to the basepoint.  Let $c$ denote this map.  Then we have that ${}[f] \cdot [c] = [f * c] = [f]$, where the last equality follows from the fact that with $f * c$ we traverse $f$ twice as fast as normal, then just sit at the basepoint.  This is homotopic to $f$, and so $f$ and $f * c$ live in the same homotopy class.

Finally, our operation is invertible.  To define the inverse operation we will take a loop $f$ and will let $\overline{f}$ denote the reverse loop: $\overline{f}(t) = f(1 - t)$.  This reverse loop will have the same shape as the initial loop, but traverse the loop in the opposite direction.  Now notice that $f * \overline{f}$ can be compressed into the constant map by constructing a homotopy that traverses most of $f$, then goes ahead and starts coming back through $\overline{f}$.  This homotopy slowly reels $f * \overline{f}$ back to the basepoint, which eventually turns it into the constant map.  Since $f * \overline{f} \simeq c$ ($c$ being the constant basepoint map), we have that ${}[f*\overline{f}] = [c]$, and similarly ${}[\overline{f}*f] = [c]$.

So, our operation is associtiative, has an identity, and is invertible, so we have a group of the homotopy classes of loops for a fixed basepoint.  This is known as the fundamental group and is denoted $\pi_1(X, x_0)$, where $X$ is the space and $x_0$ is the basepoint.  In the next post we’ll look at what happens when the basepoint changes.