Show that a simplicial map \(f: K \to L\) is a homotopy equivalence if \(f^{-1}(x)\) is contractible for all \(x \in L\).
Solution
1. For each simplex \(\sigma\) of \(L\), the preimage \(f^{-1}(\sigma)\) is contractible.
\(f^{-1}(\sigma)\) is the union of the contractible fibers \(f^{-1}(x)\) over \(x \in \sigma\), and it deformation retracts onto any single fiber (the fibers are contractible and the preimage of a simplex is a product-like structure).
2. More precisely, \(f^{-1}(\sigma)\) is homotopy equivalent to \(f^{-1}(x_0)\) for any \(x_0 \in \sigma\), hence contractible.
1 (the preimage of a simplex deformation retracts onto the preimage of a point).
3. By the Vietoris–Begle theorem (or the fact that a map with contractible fibers is a homotopy equivalence for CW complexes), \(f\) induces an isomorphism on all homotopy groups.
the Vietoris–Begle mapping theorem (a map with contractible fibers induces isomorphisms on homology, and for simply connected spaces, on homotopy).
4. Since \(K\) and \(L\) are CW complexes (simplicial complexes), and \(f\) induces isomorphisms on all homotopy groups, \(f\) is a homotopy equivalence (Whitehead’s theorem).
3 and Whitehead’s theorem.
5. Q.E.D.
4.