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If $f,g:X_1\rightarrow X_2$ are homotopic, then they induce the same maps of homotopy groups $f_*=g_* : \pi _n(X_1)\rightarrow \pi _n(X_2) $. The opposite is not true. For instance if $X_1=\mathbb{RP}^n$, $X_2=K(\mathbb{Z}_2,2)$, then the only non trivial homotopy group of $X_2$ is $\pi_2(X_2)=\mathbb{Z}_2$ and since $\pi_2(\mathbb{RP}^n)=0$ any map $f: \mathbb{RP}^3\rightarrow K(\mathbb{Z}_2,2)$ induces trivial maps $f_*$, however since $H^2(\mathbb{RP}^3,\mathbb{Z}_2)=\mathbb{Z}_2$ is equivalent to the set of homotopy classes of maps $f:\mathbb{RP}^3\rightarrow K(\mathbb{Z}_2,2)$ there must exist a homotopically non trivial $f$.

I was wondering if there exists some additional condition on top of $f_*=g_*$ which guarantees that $f$ and $g$ are homotopic.

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    $\begingroup$ You probably mean "homotopic" not "homotopy equivalent"? The induced maps on homotopy and homology do not suffice. That said, obstruction theory gives you a relative invariant that allows you to conclude two maps are homotopic. Have you looked at the Whitehead book, or perhaps Milnor and Stasheff? The general idea appears there. Induced maps are the most basic invariant, then there are relative obstructions, secondary obstructions, etc. It's a fairly involved story, but basic. $\endgroup$
    – Ryan Budney
    May 16 at 19:20
  • $\begingroup$ Yes I meant homotopic, thank you for pointing this out, I corrected the mistake. Thank you also for the suggestions on where to look. $\endgroup$
    – Andrea Antinucci
    May 17 at 9:33
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    $\begingroup$ Well, there's Freyd generation conjecture, which states that on finite spectra functor of homotopy groups is actually faithful. This can be rephrased in unstable terms to give some sufficient conditions. Alas, this conjecture is currently far out of reach. $\endgroup$
    – Denis T
    May 17 at 10:03

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