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".

${\rm S}_6$ is not the automorphism group of a finite group.
See H.K. Iyer, On solving the equation Aut(X) = G, Rocky Mountain J. Math. 9 (1979), no. 4, 653--670, available online
here.

This paper proves that for any finite group $G$, there are finitely many 
finite groups $X$ with ${\rm Aut}(X) = G$, and it explicitly solves the
equation for some specific values of $G$.
In particular, Theorem 4.4 gives the complete solution for $G$
a symmetric group, and when $n = 6$ there are no such $X$.
