$f:X \to Y$ is a continuous map. Suppose it induces an isomorphism $f_*:H_*(X) \to H_*(Y)$. Does that imply $f:X \to Y$ is a homotopy equivalence? Suppose that $f:X \to Y$ is a continuous map. Suppose it induces an isomorphism $f_*:H_*(X) \to H_*(Y)$. Does that imply $f:X \to Y$ is a homotopy equivalence?
I think that this is not the case. This is because the following is not True.

Let $f_*:C_*\to D_*$ be a chain map such that $f_*$ induces an isomorphism in homology. Then $f_*$ is a chain homotopy equivalence,

Can anyone give me a simple example.
 A: In case when $Y=\{*\}$ is a singleton your question can be stated as:

Suppose $X$ has all homology groups trivial. Is it contractible?

Unfortunately this is false. The double comb space is famously known to have all homotopy, homology and cohomology groups trivial, even though it is not contractible.
A: No; for example, there are non-contractible spaces that are acyclic (have trivial reduced homology), such as the Poincare homology sphere minus a point. You can take $X$ to be any such space and $Y$ to be a point.
However, this is true if $X$ and $Y$ are simply connected CW complexes, by a combination of the relative Hurewicz theorem and Whitehead's theorem. This is "Whitehead's theorem for homology."
Edit: As for chain complexes, there's a Whitehead's theorem in this setting also: a quasi-isomorphism between two bounded-below complexes of projective modules is a chain homotopy equivalence (see, for example, Proposition 4.1 in these notes). These complexes are roughly the analogues of CW complexes.
A: A simple example in chain complexes is the following: consider the infinite complex
$$\cdots  \to \mathbb Z/4\to \mathbb Z/4 \to  \mathbb Z/4\to\cdots$$
where every map is multiplication by $2$. This is acyclic but not contractible over $\mathbb Z/4$. A similar example is afforded by $\mathbb k[t]/(t^2)$ and multiplication by $t$ over this ring.
