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Recall that, in general, maps of CW complexes $X\to Y$ which induce the same maps of all homotopy groups $\pi_*(X,x)\to \pi_*(Y,y)$, need not be homotopic.

Assume however, that we have a continuous self-map $f: (X,x) \rightarrow (X,x)$ of a connected CW complex $X$ which induces identity morphisms on all homotopy groups. Is it true that $f$ is actually homotopic to the identity map on $X$?

Note that in this setting Whitehead's theorem says $f$ is a homotopy equivalence.

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    This is a standard theorem whose proof you can look up in any algebraic topology book. I recommend Hatcher's book. Unless you have specific questions about understanding such a proof, I'm voting to close. – 2017-02-01
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    In Hatcher there is only a proof of Whitehead theorem " weak equivalence implies homotopy equivalence" and if you read carefully you will see that the question is not about the proof of Whitehead theorem. I understand that probably my question very easily follows from W-ed Theorem. if this is the case, can you please explain how it actually follows. As from what I see I think you read the question as " How to prove that weak equivalence implies homotopy equivalence". 2) Also if it happens that I am wrong could you please tell me in which chapter of Hatcher exactly this is proven? ( Not Wh-ed) – 2017-02-01
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    While this is not exactly Whitehead's theorem, it is proven by exactly the same method: You construct a homotopy $F: X\times I\to Y$ extending the map $F: X\times \{0, 1\}\to Y$, by induction on skeleta. If you understand the proof of theorem 4.5 in Hatcher, you should be able to do this. – 2017-02-02
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    Moishe Cohen, thank you, my doubt was if one should actually reprove the Whithead or this sort of uniqueness is a purely logical consequence of Whitehead. I looked not only in Hatcher, but in other well known algebraic topology books ( such as May) and nowhere found this sort of " refined Whitehead"- 1-1 correspondense between induced morphisms on homotopy groups and homotopy classes of maps. While the proof is analogous to usual Whithead, it seems little strange to me that nobody pays attention. – 2017-02-02
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    Note that, in general, maps of CW complexes which induce equal homomorphisms of all homotopy groups need not be homotopic, so extra care is needed in the proof, but for maps which are homotopy-equivalences everything works. – 2017-02-02

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