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When studying algebraic topology, we often use the Whitehead theorem for homotopy or homology version.

But, I don't know $\pi_n(f)=0$ implies $f\simeq \ast$?. Similarly, $H_n(f)=0$ implies $f\simeq \ast$? Where $\ast$ means a constant map.

If not, what spaces satisfy my question?

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I think you ask $\pi_n(f)=0$ for any $n$, isn't it? – Michele Triestino Mar 23 '11 at 8:14
If these answers are sufficient for your purposes, why not accept one of them? – Yemon Choi Oct 2 '11 at 1:59

The answer to both is no: A map between two Eilenberg-Mac Lane spaces of different degrees has to induce 0 on homotopy (since both space have only one homotopy group each, and they are in different degrees), but there are many nontrivial maps between them (they form the ring of cohomology operations). For a counterexample to the second statement, just take two different-dimensional spheres and a nontrivial map between them, like the Hopf map $S^3 \to S^2$. When you restrict to finite CW-complexes in the homotopy setting, the question becomes much more subtle and the answer is actually not known, even stably; it is one of the classical big open conjectures of algebraic topology, called the generating hypothesis. The relationship between $Hom(H^*(Y),H^*(X))$ and $[X,Y]$ is very well studied; in general the latter is the target of a spectral sequence called the (unstable) Adams spectral sequence whose $E^2$-term consists of $Ext^*(H^*(Y),H^*(X))$, the derived functors of $Hom$, in the category of unstable algebras over the Steenrod algebra. If you want to learn this, I recommend starting with the stable setting, which is significantly easier. Then an answer to your question for what spaces $f^*=0$ implies $f\sim *$ could be: when $H^*(X)$ is injective as an unstable algebra, or when $H^*(Y)$ is projective. This doesn't happen very often, but it does happen, for example, when $X$ is the classifying space of a $p$-group and $Y$ is $p$-complete -- you'll get into Lannes' $T$-functor theory.

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Via Dold-Kan, the $\pi_n$-question in the particular case of simplicial abelian groups can be reformulated in the category of complexes of abelian groups: Given a morphism $f : X\to Y$ of bounded below complexes of abelian groups such that $\text{H}_\ast f = 0$ for all $n$, is $f = 0$ in the derived category? This is not the case in general.

Take $A$ and $B$ two abelian groups such that $\text{Ext}^1(A,B)\ne 0$, e.g. $A = B = \mathbf{Z}/2$. Let $P$ be a projective resolution of $A$. Let $Q$ be a projective resolution of $B$. Then Hom_{K(Z)}(P,Q[1]) = Hom_{D(Z)}(P,Q[1]) = Hom_{D(Z)}(A,B[1]) = Ext^1(A,B) =/= 0 (TeX didn't work). Let $f : P\to Q[1]$ be a morphism of complexes that is not homotopic to zero. Then it is nonzero in the derived category. But $\text{H}_n f = 0$ for all $n$, since $\text{H}_n(P) = 0$ if $n\ne 0$ and $\text{H}_n(Q[1]) = 0$ if $n\ne 1$.

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