If \(f: X \to Y\) is a map with \(X\) and \(Y\) homotopy equivalent to CW complexes, show that the pair \((M_f, X)\) is homotopy equivalent to a CW pair, where \(M_f\) is the mapping cylinder. Deduce that the mapping cone \(C_f\) has the homotopy type of a CW complex.
Solution
Choose CW complexes \(X'\) and \(Y'\) together with homotopy equivalences \[ u:X'\xrightarrow{\simeq}X, \qquad v:Y'\xrightarrow{\simeq}Y, \] and homotopy inverses \(u^{-1}\) and \(v^{-1}\). Define \[ f'=v^{-1}fu:X'\to Y'. \] By cellular approximation, replace \(f'\) by a cellular map without changing its homotopy class.
The homotopy-commutative square \[ \begin{array}{ccc} X'&\xrightarrow{f'}&Y'\\ \downarrow u&&\downarrow v\\ X&\xrightarrow{f}&Y \end{array} \] has vertical homotopy equivalences. Mapping cylinders are homotopy pushouts, so homotopy invariance of homotopy pushouts gives a homotopy equivalence of pairs \[ (M_{f'},X')\simeq(M_f,X). \] (Equivalently, one can write the maps explicitly using chosen homotopies making the square commute.)
Because \(f'\) is cellular, its mapping cylinder admits a CW structure for which \(X'\subset M_{f'}\) is a subcomplex. Hence \[ \boxed{(M_f,X)\text{ is homotopy equivalent to a CW pair}.} \]
The mapping cone is the quotient \[ C_f=M_f/X. \] Passing to the homotopy-equivalent CW pair gives \[ C_f\simeq M_{f'}/X'. \] The quotient of a CW complex by a subcomplex is again a CW complex. Therefore \[ \boxed{C_f\text{ has the homotopy type of a CW complex}.} \]