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](http://www.ams.org/mathscinet-getitem?mr=1368659)".



proving that an inclusion map from a subcomplex is a homotopy equivalence
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.


