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Let $f\colon X\to Y$ be a continuous function which is not null-homotopic. This doesn't necessarily mean the induced map on singular cohomology is non-zero: for example if $c\colon S^{2n+1}\to BO(2n+1)$ classifies the tangent bundle, then the induced map on cohomology is trivial (since the tangent bundle is stably trivial and $H^*(BO(2n+1))$ only contains stable characteristic classes) but this tangent bundle is typically not trivial and so $c$ would not be null-homotopic.

If $f$ is not null-homotopic is there always a cohomology theory $h_f$ such that $f^*\colon h_f^*(Y) \to h_f^*(X)$ is not zero? Is there a cohomology theory which can detect ALL maps which are not null-homotpic?

William
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    Shouldn't you ask that $f$ is not stably null-homotopic? I am not sure how a cohomology could distinguish between maps which are stably the same. –  Feb 15 '19 at 06:12
  • What is your understanding of a cohomology theory (reduced or unreduced)? Do you consider maps between CW-complexes or maps between arbitrary spaces? – Paul Frost Feb 15 '19 at 10:38
  • Mike Miller is right. For any reduced cohomology theory $h^*$ we have a natural equivalence $\sigma : h^k \to h^{k+1} \circ S$, where $S$ denotes the suspension functor. And it is well-known that reduced and unreduced theories are equivalent (at least on CW-complexes). – Paul Frost Feb 15 '19 at 10:51
  • I forgot about that, I would need $f$ to not be stably null-homotopic for it to have any hope of inducing a non-zero map on cohomology. But if $f$ is already assumed to not be stably null-homotopic, couldn't we just take the cohomology theory defined by $\Sigma^\infty Y$ (unless $\Sigma^\infty f$ is a phantom map somehow)? – William Feb 15 '19 at 15:07
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    Yes. :) That was where I was headed next. I don't think you can have something that detects every non-null map, but I don't know a proof. It remains interesting to me. –  Feb 15 '19 at 15:28
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    This doesn't answer the question, but for finite complexes $X$ and $Y$, Freyd's generating hypothesis asserts that the map $[X,Y] \to \operatorname{Hom}{\pi* \mathbb{S}}(\pi_* X, \pi_* Y)$ is a monomorphism. If true, this means that stable homotopy detects all stable null-homotopic maps. This question, which is central to stable homotopy theory, remains open after 50 years. – JHF Feb 15 '19 at 17:02

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