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).

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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