Show that Corollary 4K.2 remains valid when \(X\) and \(Y\) are CW complexes and the subspaces \(U_i\) and \(V_i\) are subcomplexes rather than open sets.
Solution
Let \[ f:X\to Y \] be the map in Corollary 4K.2, and suppose the covering families \(\{U_i\}\) of \(X\) and \(\{V_i\}\) of \(Y\) consist of CW subcomplexes, with \[ f(U_i)\subset V_i, \] and with the required weak equivalences on all finite intersections.
Replace each subcomplex by a standard open regular neighborhood in a sufficiently fine barycentric subdivision. Thus for every finite set \(J\) of indices we can choose open sets \[ N(U_J)\supset U_J, \qquad N(V_J)\supset V_J, \qquad U_J=\bigcap_{j\in J}U_j,\quad V_J=\bigcap_{j\in J}V_j, \] such that \[ N(U_J)\simeq U_J, \qquad N(V_J)\simeq V_J, \] and the choices are compatible with finite intersections. After shrinking the target neighborhoods if necessary, continuity of \(f\) and the subcomplex condition give \[ f(N(U_i))\subset N(V_i). \]
For each finite \(J\), the commutative square \[ \begin{array}{ccc} U_J&\longrightarrow&V_J\\ \downarrow&&\downarrow\\ N(U_J)&\longrightarrow&N(V_J) \end{array} \] has vertical homotopy equivalences. Since the top map is a weak homotopy equivalence by hypothesis, so is the bottom map. Hence the open covers \(\{N(U_i)\}\), \(\{N(V_i)\}\) satisfy the hypotheses of Corollary 4K.2. It follows that \[ \boxed{f:X\to Y\text{ is a weak homotopy equivalence}.} \] Thus Corollary 4K.2 remains valid when the covering sets are subcomplexes rather than open sets.