Fundamental group

Remarks

The product is well defined, and it makes $\pi_1(X, x_0)$ a group. The identity is the class of the constant loop, and the inverse of $[\gamma]$ is the class of the reversed loop $s \mapsto \gamma(1 - s)$. A pointed map $f \colon (X, x_0) \to (Y, y_0)$ induces a homomorphism $f_* \colon \pi_1(X, x_0) \to \pi_1(Y, y_0)$, $[\gamma] \mapsto [f \circ \gamma]$. If $X$ is path-connected, the group does not depend on $x_0$ up to isomorphism. Example: for a convex set $C \subseteq \R^n$ and $x_0 \in C$, every loop is path homotopic to the constant loop by the straight-line homotopy, so $\pi_1(C, x_0)$ is trivial. Non-example: loops themselves, without passing to classes, do not form a group under concatenation. Indeed $(\alpha \cdot \beta) \cdot \gamma$ and $\alpha \cdot (\beta \cdot \gamma)$ are in general different maps, which agree only up to path homotopy. A path-connected space with trivial fundamental group is simply connected.