Can we have a continous map $f: X \longrightarrow Y$ such that $f$ induces an isomorphism on all homology groups i.e. $f_* : H_n(X) \longrightarrow H_n(Y)$ is an isomorphism of abelian groups for all $n \geq 0$ but $f$ itself is not a homotopy equivalence?
Example of a quasi-isomorphism in $\operatorname{Top}$ which is not a homotopy equivalence
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1@Neal I think that more than Whitehead's theorem is the fact that if a map $f:X\to Y$ induces an isomorphism on all homology groups, and $X$ and $Y$ are simply connected CW complexes, then $f$ is actually a homotopy equivalence. This is in Chapter Three of Hatcher somewhere. – 2012-12-10
2 Answers
Yes; in fact, it is possible for a noncontractible space (even a CW complex) to have the same homology as a point; such spaces are called acyclic. An example of such a thing can be obtained by removing a point from a homology sphere.
However, if $f$ is also required to be an isomorphism on $\pi_1$ and the spaces are connected, then $f$ is a weak homotopy equivalence. This follows from the relative Hurewicz theorem and the fact that $\pi_1(Y,X) = 0$ (assuming $f$ is an inclusion). If the spaces are CW complexes, then it follows from the Whitehead theorem that $f$ is in fact a homotopy equivalence.
edit: This last statement wasn't quite true. It's true if $X$ and $Y$ are simply connected, or if certain weaker conditions hold (see the comments below).
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0Ah, I misremembered how it went, then. Thanks. – 2012-12-11
The Warsaw circle has all its homology groups trivial, except $H_0$. There is a map from $S^0$ to the circle that induces isomorphisms on all homology groups. But, they are not homotopy equivalent.