Homology Whitehead theorem for non simply connected spaces

(One version of) the Whitehead theorem states that a homology equivalence between simply connected CW complexes is a homotopy equivalence. Does the following generalisation hold true?

Suppose $X,Y$ are two connected CW complexes and $f:X\to Y$ is a continuous map that induces isomorphisms of the fundamental groups and on homology. Then $f$ is a weak equivalence.

It would be enough to show that the map $\tilde{f}$ obtained by lifting $f$ to the universal covers (for some arbitrary choice of base points) is a homology equivalence, but I've never studied the homology spectral sequence, so I don't know the relationship between the homology of the universal cover $\widetilde{X}$ and that of $X$, and I don't know how to prove that $\tilde{f}$ is a homology equivalence.

Also, are there any good, reasonably self contained and short accounts of the homology of covering spaces, preferably online?

EDIT 1 I've seen some lecture notes (second paragraph on page 5) where this theorem is used (without further comment). They used this proposition to establish that a certain type of homotopy colimit is well defined up to weak equivalence.

EDIT 2 Theorem 6.71 from Kirk and Davis (page 179 of the document or 164 internally) is probably what I need, but it involves isomorphisms with local coefficient systems for homology for the base spaces.

EDIT 3 Thank you @studiosus for the reference. I don't quite understand yet, could you help out? If I get it (using the theorem of the source), there are finite connected two dimensional complexes $X,Y$ and a homotopy equivalence $X\vee S^2\simeq Y\vee S^2$, yet $X$ and $Y$ aren't homotopy equivalent. From the assumption it follows immediately that $$X\hookrightarrow X\vee S^2\to Y\vee S^2\twoheadrightarrow Y$$ gives an isomorphism of the fundamental groups. From $$\tilde{H}(X)\oplus\tilde{H}(S^2)\simeq\tilde{H}(X\vee S^2)\simeq\tilde{H}(Y\vee S^2)\simeq\tilde{H}(X)\oplus\tilde{H}(S^2)\;,$$ $\tilde{H}(S^2)\simeq \Bbb Z[2]$ and the fundamental theorem of finitely generated abelian groups, it follows at once that $\tilde{H}(X)\simeq\tilde{H}(Y)$ abstractly. The only problem I have is that it doesn't seem to me that we get a map $X\to Y$ that induces isomorphisms on $\pi_1$ and is a homology equivalence, actually, I don't see why there should be a homology equivalence $X\to Y$ at all.

Solution 1:

The answer is no. An easy example is provided in Allen Hatcher's Algebraic Topology, Example 4.35.

http://www.math.cornell.edu/~hatcher/AT/ATch4.pdf

There, a CW-complex $X$ is formed by attaching an appropriate $(n+1)$-cell to the wedge $S^1 \vee S^n$ of a circle and an $n$-sphere. The inclusion of the $1$-skeleton $S^1 \to X$ induces an isomorphism on integral homology and on homotopy groups $\pi_i$ for $i < n$ but not on $\pi_n$.

That example is also worked out as Problem 1 here:

http://www.home.uni-osnabrueck.de/mfrankland/Math527/Math527_HW10_sol.pdf

Solution 2:

There are two finite 2-dimensional complexes $A, B$ which are not homotopy-equivalent but are homology-equivalent, i.e., there exists a continuous map $$ f: A\to B $$ inducing isomorphism of fundamental groups and homology groups. See the last paragraph (page 522) of this paper, the actual example is due to Dunwoody.

Edit 1: The homology equivalence part follows from Dyer's paper, see the link above (read the very last paragraph of his paper). Dyer's results are by no means obvious: He proves that equality of Euler characteristics plus one extra condition ($m=1$, whatever $m$ is) imply homology equivalence of 2-d complexes with isomorphic fundamental groups. Then Dyer verifies his conditions in the case of Dunwoody's example (the only nontrivial condition is $m=1$, since equality of Euler characteristics is clear).

Edit 2: The correct version of the "homological Whitehead's theorem" is indeed requires a cohomology isomorphisms with sheaf coefficients, see the discussion and reference here.