To what degree can we dualize theorems regarding homotopy into theorems about cohomotopy (or is there a good source that tries to do this)?

For instance, is there some kind of Hurewicz theorem relating cohomotopy and ordinary cohomology? Is there a "cohomotopy extension property" (something that applies when relative cohomotopy groups are trivial)? If two spaces are cohomologically equivalent and have some property in cohomotopy analogous to simply-connected, are they cohomotopy equivalent?

Thanks, this is primarily a reference request, however there is the possibility that all this is impossible so no such reference exists, which would also be an acceptable answer.

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    $\begingroup$ There is a result of Hopf: if $X$ is an $n-1$-connected space (say, CW complex), then maps of $X$ into the $n$-sphere are the same as cohomology classes in $H^n(X; \mathbb{Z})$. (You can prove this quickly by noting that the relevant property is true when $S^n$ is replaced by the Eilenberg-MacLane space $K(\mathbb{Z}, n)$, and that can be obtained from $S^n$ by attaching $n+2$-cells and higher.) $\endgroup$ – Akhil Mathew Nov 9 '11 at 0:36
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    $\begingroup$ Thanks so much! This is actually an immensely useful fact, and this result solves the problem that prompted the asking of this question (at least I think it does, I will have to check all the details) math.stackexchange.com/questions/80331/…. $\endgroup$ – Jonathan Beardsley Nov 9 '11 at 0:47
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    $\begingroup$ Akhil, do you mean $n-1$ dimensional or something? Or even $n$ dimensional... and then CW approximation. For example, $K(\mathbb{Z},n)$ is $n-1$ connected, but certainly it's cohomology is not the same as it's cohomotopy... $\endgroup$ – Dylan Wilson Nov 9 '11 at 2:57
  • $\begingroup$ @Dylan: Thanks! I forgot $n$-dimensional, and you're right about the proof. $\endgroup$ – Akhil Mathew Dec 9 '11 at 14:59
  • $\begingroup$ @Dylan: And, moreover, I was being very silly -- what I meant was that maps into any $n-1$-connected space from an $n$-dimensional space. $\endgroup$ – Akhil Mathew Dec 9 '11 at 15:03

The homotopy groups can be written as covariant homotopy invariant functors $\pi_n:\mathrm{Top}_\ast\to\mathrm{Set}$. If we were to consider contravariant homotopy invariant functors $\pi^n:\mathrm{Top}_\ast^{op}\to\mathrm{Set}$, we would obtain the cohomotopy sets. How dual is it? Well, $\pi^n(S^m)=\pi_m(S^n)$. If $X$ is a CW-complex of dimension (at most) $n$, then $\pi^p(X)\to H^p(X)$ is a bijection. See the nlab. As for whether or not a "cohomotopy extension property" exists, I don't know; it seems like an interesting thing!


All homology/cohomology is with integer coefficients.

The Hurewicz map that you are looking is defined as follows I think: Given a homotopy class $f:X\rightarrow S^n$, and $[u]\in H^n(S^n)$ you get $f^*[u]\in H^n[X]$. If you take now $[u]$ to be the positive generator of $H^n(S^n)\cong \mathbb{Z}$ you get a nice map $h:\pi^n\rightarrow H^n(X)$.

If $X$ is an $n$-dimensional CW complex any map into an $n+k$-connected space (e.g. n+k dim sphere) is homotopically trivial. Combining this with the observation that $pi^n\cong H^n(X)$ for $n$-dimensional closed manifolds we see that $\pi^l(X)\cong H^l(X)$ for all $l\geq n$ for $n$ dimensional closed manifolds. Furthermore $\pi^0(X)$ measures the number of connected components of $X$ and as $S^1$ is a $K(\mathbb{Z},1)$ we have that $\pi^1(X)\cong H^1(X)$.


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