REVISIT Let \(f,g : S^1 \rightarrow X\), \(P = X \cup_f B^2 \cong X \coprod B^2 / \sim\), where \(x \sim f(x)\), \(Q = X \cup_g B^2\). Show that \(f\simeq g \implies P\simeq Q\).
Solution
1. Let \(H:S^1\times I\to X\) be a homotopy from \(f\) to \(g\).
This is the hypothesis \(f\simeq g\).
2. Form\[ R=X\cup_H(S^1\times I)\cup_g B^2, \]where \(S^1\times\{0\}\) is attached to \(X\) by \(f\), \(S^1\times\{1\}\) is attached to \(X\) by \(g\), and the boundary of the disk is attached along the top circle.
This is the mapping-cylinder construction for the homotopy, followed by attachment of the \(2\)-cell at the \(g\) end.
3. Collapsing the cylinder \(S^1\times I\) toward its bottom end gives a deformation retraction\[ R\simeq X\cup_f B^2=P. \]
Under the bottomward collapse, the disk together with the cylinder is again a disk whose boundary is identified with \(X\) by \(f\). The homotopy \(H\) specifies the attaching points throughout the collapse, so it descends to the quotient.
4. Collapsing the same cylinder toward its top end gives a deformation retraction\[ R\simeq X\cup_g B^2=Q. \]
The analogous topward collapse leaves the disk attached by \(g\).
5. Therefore\[ \boxed{X\cup_f B^2\simeq X\cup_g B^2}. \]
Both spaces are homotopy-equivalent to \(R\) by <1>3–<1>4.