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

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.