- Main idea: just algebraic manipulations using the \(\pi_1\) functor and unravelling definitions.
Let \(X\) be path connected and simply connected, and let \(x,y \in X\) be two arbitrary points. Then consider two paths, \(\gamma: I \into X, \gamma(0) = x, \gamma(1) = y\) \(\alpha: I \into X, \alpha(0) = x, \alpha(1) = y\).
We would like to show \(\gamma \homotopic \alpha\). Since \(X\) is simply connected, we know that \(\pi_1(X) = 0\). This means that for any \(a,b \in \pi_1(X), a = b = e\), the identity element in this group.
So we construct two loops: one as \(\gamma \bar\alpha\), the other as \(\alpha\bar\gamma\). Apply the \(\pi_1\) functor yields \([\gamma\bar\alpha] = e = [c_x] = [\alpha\bar\gamma]\), where \([c_x]\) is the equivalence class of the constant path at \(x\), and equivalently the identity element in \(\pi_1(X)\). Lemma: If \(f\homotopic g\), then \(f\circ h \homotopic g \circ h\) for any \(h\).
But this says \(\gamma\bar\alpha \homotopic c_x\) and \(\alpha\bar\gamma \homotopic c_x\). But \(\gamma \homotopic c_x \circ \gamma \homotopic (\alpha\bar\gamma) \circ \gamma \homotopic \alpha\circ (\bar\gamma \circ\gamma) \homotopic \alpha\), which is what we desired.
Solution
Goal: Show that two paths in a simply connected space with common endpoints are homotopic rel endpoints.
1. Let \(\gamma,\alpha:I\to X\) satisfy \(\gamma(0)=\alpha(0)=x\) and \(\gamma(1)=\alpha(1)=y\).Proof: Since \(X\) is simply connected, \(\pi_1(X,x)=0\) and \(\pi_1(X,y)=0\).
2. The loops \(\gamma * \overline{\alpha}\) at \(x\) and \(\overline{\alpha} * \alpha\) at \(y\) are both null-homotopic:Proof: Triviality of the two fundamental groups gives\[ \gamma * \overline{\alpha}\sim c_x,\qquad \overline{\alpha} * \alpha\sim c_y. \]
3. Concatenate the first homotopy on the right with \(\alpha\):Proof: If \(u\sim v\) then \(u * \alpha\sim v * \alpha\), so\[ (\gamma * \overline{\alpha}) * \alpha \sim c_x * \alpha. \]Re-associating,\[ \gamma * (\overline{\alpha} * \alpha)\sim \alpha. \]
4. Replace \(\overline{\alpha} * \alpha\) by \(c_y\) and remove the unit loop:Proof: From the second homotopy,\[ \gamma \sim \gamma * c_y \sim \gamma * (\overline{\alpha} * \alpha). \]Combining with step 3 gives \(\gamma\sim\alpha\) by concatenating homotopies.Hence, with endpoints fixed, \(\gamma\) and \(\alpha\) are path-homotopic.Q.E.D.