Even if you ask that ff induces trivial maps on all (singular) homology and cohomology groups, there are still easy manifold examples. (This actually arises as an exercise in Hatcher's AT).
For instance, let f:T3rightarrowS2f:T3rightarrowS2 be the composition T3rightarrowS3rightarrowS2T3rightarrowS3rightarrowS2, where the map from T3T3 to S3S3 is simply collapsing the 2-skeleton to a point, and the map from S3S3 to S2S2 is the Hopf map.
As others have mentioned, since T3T3 is a K(mathbbZ3,1)K(mathbbZ3,1), if follows that ff induces trivial maps on homotopy groups.
Since the Hopf map induces trivial maps on homology and cohomology, it follows that ff does as well.
Finally, to see that ff is NOT nullhomotopic, assume it is. Since the map from S3S3 to S2S2 is a fiber bundle, it has the homotopy lifting property. Hence, we can lift the homotopy of ff to a homotopy G:ItimesT3rightarrowS3G:ItimesT3rightarrowS3 where G0G0 is the above map from T3T3 to S3S3 and G1G1 is is a map from T3T3 to S1subseteqS3S1subseteqS3, the preimage of a point in S2S2 under the Hopf map.
But G0G0 has degree 1, while G1G1 has degree 0, a contradiction.
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