Show that \(X\) is homotopy-equivalent to a point \(\iff\) \(\text{id}_X \simeq g\) for some constant map \(g\).
Solution
1. Suppose \(X\) is homotopy-equivalent to a point. Then \(\operatorname{id}_X\) is homotopic to a constant map.
Let \(p:X\to\{*\}\) and \(i:\{*\}\to X\) be homotopy inverses. Since \(p\circ i=\operatorname{id}_{\{*\}}\) automatically, the other homotopy-inverse relation is \[ i\circ p\simeq\operatorname{id}_X. \] But \(i\circ p\) is the constant map with value \(i(*)\).
2. Conversely, suppose \(\operatorname{id}_X\simeq g\) for a constant map \(g(x)=x_0\). Then \(X\) is homotopy-equivalent to a point.
Let \(p:X\to\{*\}\) be the unique map and let \(i:\{*\}\to X\) send \(*\) to \(x_0\). Then \[ p\circ i=\operatorname{id}_{\{*\}}, \] while \[ i\circ p=g\simeq\operatorname{id}_X. \] Hence \(p\) and \(i\) are homotopy inverses.
3. Thus\[ \boxed{X\simeq *\iff \operatorname{id}_X\text{ is homotopic to a constant map}.} \]
Combine 1 and 2.