Algebraic Topology: Fundamental Group

Homotopy of paths
A homotopy of paths in a space $X$ is a family $f_t: I\to X$ , $0\leq t\leq 1$ , such that
(i) The endpoints $f_t(0)=x_0$ and $f_t(1)=x_1$ are independent of $t$ .
(ii) The associated map $F:I\times I\to X$ defined by $F(s,t)=f_t(s)$ is continuous.

When two paths $f_0$ and $f_1$ are connected in this way by a homotopy $f_t$ , they are said to be homotopic. The notation for this is $f_0\simeq f_1$ .

Example: Linear Homotopies
Any two paths $f_0$ and $f_1$ in $\mathbb{R}^n$ having the same endpoints $x_0$ and $x_1$ are homotopic via the homotopy $\displaystyle f_t(s)=(1-t)f_0(s)+tf_1(s).$

Simply-connected
A space is called simply-connected if it is path-connected and has trivial fundamental group.

A space $X$ is simply-connected iff there is a unique homotopy class of paths connecting any two parts in $X$ .
Path-connectedness is the existence of paths connecting every pair of points, so we need to be concerned only with the uniqueness of connecting paths.

( $\implies$ ) Suppose $\pi_1(X)=0$ . If $f$ and $g$ are two paths from $x_0$ to $x_1$ , then $f\simeq f\cdot \bar{g}\cdot g\simeq g$ since the loops $\bar{g}\cdot g$ and $f\cdot\bar{g}$ are each homotopic to constant loops, due to $\pi_1(X,x_0)=0$ .

( $\impliedby$ ) Conversely, if there is only one homotopy class of paths connecting a basepoint $x_0$ to itself, then all loops at $x_0$ are homotopic to the constant loop and $\pi_1(X,x_0)=0$ .

$\pi_1(X\times Y)$ is isomorphic to $\pi_1(X)\times \pi_1(Y)$ if $X$ and $Y$ are path-connected.
A basic property of the product topology is that a map $f:Z\to X\times Y$ is continuous iff the maps $g:Z\to X$ and $h:Z\to Y$ defined by $f(z)=(g(z),h(z))$ are both continuous.

Hence a loop $f$ in $X\times Y$ based at $(x_0,y_0)$ is equivalent to a pair of loops $g$ in $X$ and $h$ in $Y$ based at $x_0$ and $y_0$ respectively.

Similarly, a homotopy $f_t$ of a loop in $X\times Y$ is equivalent to a pair of homotopies $g_t$ and $h_t$ of the corresponding loops in $X$ and $Y$ .

Thus we obtain a bijection $\pi_1(X\times Y, (x_0,y_0))\approx \pi_1(X,x_0)\times \pi_1(Y,y_0)$ , $[f]\mapsto([g],[h])$ . This is clearly a group homomorphism, and hence an isomorphism.

Note: The condition that $X$ and $Y$ are path-connected implies that $\pi_1(X,x_0)=\pi_1(X)$ , $\pi_1(Y,y_0)=\pi_1(Y),\pi_1(X\times Y,(x_0,y_0))=\pi_1(X\times Y)$ .

Author: mathtuition88

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