Given spaces \(X\) and \(Y\), let \([X, Y]\) denote the set of homotopy classes of maps of \(X\) into \(Y\).
Let \(I = [0, 1]\). Show that for any \(X\), the set \([X, I]\) has a single element.
Show that if \(Y\) is path connected, the set \([I, Y]\) has a single element.
Solution
Let \(f,g:X\to I\) be arbitrary. Since \(I=[0,1]\) is convex, define \[ H(x,t)=(1-t)f(x)+tg(x). \] Then \(H:X\times I\to I\) is continuous, \(H(x,0)=f(x)\), and \(H(x,1)=g(x)\). Thus every two maps \(X\to I\) are homotopic, so \([X,I]\) has one element.
Let \(f,g:I\to Y\). Since \(Y\) is path connected, choose a path \(\alpha:I\to Y\) from \(f(0)\) to \(g(0)\). Every map from an interval is homotopic to the constant map at its initial value: for \(f\), \[ F(s,t)=f((1-t)s) \] is a homotopy from \(f\) to the constant map \(c_{f(0)}\), and similarly \(g\simeq c_{g(0)}\). The constants \(c_{f(0)}\) and \(c_{g(0)}\) are homotopic by \[ A(s,t)=\alpha(t), \] independent of \(s\). Hence \[ f\simeq c_{f(0)}\simeq c_{g(0)}\simeq g. \] Therefore \([I,Y]\) also has a single element.