[Math] proving that an inclusion map from a subcomplex is a homotopy equivalence

Question

This is a pretty basic question but I have been stuck on it for a while.

Given an abstract simplicial complex X and a subcomplex A, why does * suffice to show that the map |A|->|X| induced by inclusion is a homotopy equivalence:

• Let g: (|K|,|L|) -> (|X|,|A|) be a continuous map, where K is a finite simplicial complex and L a subcomplex of K. Any such g is homotopic rel |L| to a map sending |K| into |A|.

Here |.| denotes the geometric realization.

I'm trying to understand the very first step of the proof of Proposition 2.2 of J-C. Hausmann's paper "On the Vietoris-Rips complexes and a cohomology theory for metric spaces".

Answer 1

By taking K = a simplex and L = its boundary you can show that |A| -> |X| is an isomorphism on all homotopy groups (do surjectivity and injectivity separately). Then apply Whitehead's theorem.