# Cohomological Whitehead theorem

Let $X$ and $Y$ be CW complexes (resp. Kan complexes) and let $f : X \to Y$ be a continuous map (resp. morphism of simplicial sets). The following seems to be a folklore result:

Theorem. The following are equivalent:

1. $f : X \to Y$ is a homotopy equivalence.
2. $f_* : \pi_0 (X) \to \pi_0 (Y)$ is a bijection; and for all $n \ge 1$ and all points $x$, $f_* : \pi_n (X, x) \to \pi_n (Y, f (y))$ is an isomorphism.
3. $f_* : \pi_{\le 1} (X) \to \pi_{\le 1} (Y)$ is an equivalence of groupoids; and for all $n \ge 0$ and all locally constant abelian sheaves (or equivalently, local systems of abelian groups) $\mathscr{A}$ on $Y$, $f^* : H^n (Y, \mathscr{A}) \to H^n (X, f^* \mathscr{A})$ is an isomorphism.

The equivalence of conditions (1) and (2) is the usual Whitehead theorem; I am interested in the equivalence of (2) and (3). Unfortunately, I have not been able to find a proof in the literature. Can someone provide a reference or a sketch proof?

• Passing to universal covers and using the Serre spectral sequence (where you get a locally constant abelian sheaf of coefficients) and applying 3 shows that you have an isomorphism on cohomology of the universal covers. Since they are simply connected, you get a homotopy equivalence, and now by comparing the covers you see that the original map was a homotopy equivalence. – Justin Young Mar 18 '14 at 18:06
• I suppose you are appealing to the cohomological Whitehead theorem for simply connected spaces? But I do not know how that is proved either. – Zhen Lin Mar 18 '14 at 18:36
• If you know the homology one, it is implied by this answer, for example: math.stackexchange.com/questions/600323/… – Justin Young Mar 18 '14 at 22:14
• No, I do not know how any of variants of the Whitehead theorem are proved; but I am willing to assume the standard version with homotopy groups. For some reason or other, I have the impression that the equivalence of (2) and (3) can be proved using some kind of Postnikov decomposition and the Hurewicz theorem... – Zhen Lin Mar 18 '14 at 22:30
• The homology one follows fairly easily from the relative Hurewicz theorem (zero homology implies zero homotopy for the relative group, and then apply the LES). – Justin Young Mar 19 '14 at 5:48

## 2 Answers

The simplicial version of the result appears as part of Proposition 4 in [Quillen, Homotopical algebra, Ch. II §3]. Curiously, the proof is essentially the same as the one sketched by Justin Young above.

One of the uses of the local coefficient or covering space version of the arguments sketched above are in conjunction with Acyclic Model arguments, and there have been occasions when authors have initially neglected to move to covering spaces to obtain a homotopy rather than a homology equivalence. However Section 10.3.1 of the book Nonabelian Algebraic Topology (EMS Tracts in Mathematics vol 15, 2011) has a version of Acyclic Models based on crossed complexes, rather than chain complexes, and this yields homotopy equivalances, by a direct use of Whitehead's theorem.

Also Section 12.4 is on $$$$Local Coefficients and Local Systems". Relations between crossed complexes and chain complexes with a groupoid of operators are spelled out over several subsections of 7.4 and 9.5. See also Section 8.4.