We know that there are Whitehead theorem for homotopy and homology theory.

Is there the Whitehead theorem for cohomology theory for 1-connected CW complexes?

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    $\begingroup$ This is not too hard to work out on your own using the universal coefficient theorem. $\endgroup$ Mar 8, 2011 at 6:44

3 Answers 3


Conceptually, the following two theorems (both due to Whitehead) are Eckmann-Hilton duals.

Theorem. A weak homotopy equivalence between CW complexes is a homotopy equivalence.

Theorem. A homology isomorphism between simple spaces is a weak homotopy equivalence.

They don't look dual, but they are. See J.P. May. The dual Whitehead Theorems.
London Math. Soc. Lecture Note Series Vol. 86(1983), 46--54.

The point is that the second statement is really about cohomology, and the standard cellular proof of the first statement dualizes word-for-word to a ``cocellular'' proof of the second. Cocellular constructions are what appear in Postnikov towers, and they can be used more systematically than can be found in the literature. Yet another plug: they are central in the upcoming book "More Concise Algebraic Topology'' by Kate Ponto and myself.

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    $\begingroup$ Peter: is there a early draft version of your book that is available for consumption? $\endgroup$ Mar 12, 2011 at 6:25
  • $\begingroup$ Peter: Thanks. I will see your paper. When your new book publish? Don't your book(A consise course of algebraic topology) have the answer my question? $\endgroup$
    – Jino
    Mar 12, 2011 at 19:22

As Sean says, the key point is the Universal Coefficient Theorem, but the details are not completely obvious unless you make some finiteness assumptions.

Suppose that $f:X\to Y$ is such that $H^{\ast}(f;\mathbb{Z})$ is an isomorphism. Let $Z$ be the cofibre of $f$, so $\tilde{H}^{\ast}(Z;\mathbb{Z})=0$. If we can prove that $\tilde{H}_{\ast}(Z;\mathbb{Z})=0$ then we can appeal to the ordinary homological Whitehead theorem. MathJax is mangling my tildes: all (co)homology groups of $Z$ below should be read as reduced.

For any prime $p$, we have a universal coefficient sequence for $H^{\ast}(Z;\mathbb{Z}/p)$ in terms of $H^{\ast}(Z;\mathbb{Z})$, so $H^{\ast}(Z;\mathbb{Z}/p)=0$. As ${\mathbb{Z}/p}$ is a field we also know that $H_{\ast}(Z;\mathbb{Z}/p)$ is a free module with $H^{\ast}(Z;\mathbb{Z}/p)$ as its dual, so we must have $H_{\ast}(Z;\mathbb{Z}/p)=0$. Using the universal coefficient theorem for homology we deduce that $H_{\ast}(Z;\mathbb{Z})/p$ and $\text{ann}(p,H_{\ast}(Z;\mathbb{Z}))$ are zero, so multiplication by $p$ is an isomorphism on $H_{\ast}(Z;\mathbb{Z})$. As this holds for all $p$, we see that $H_{\ast}(Z;\mathbb{Z})$ is a rational vector space. Thus, if it is nontrivial it will contain a copy of $\mathbb{Q}$ so (via universal coefficients again) $H^{\ast}(Z;\mathbb{Z})$ will contain a copy of $\text{Ext}(\mathbb{Q},\mathbb{Z})$. This group is nonzero (in fact, enormous) by a standard calculation, so this contradicts the initial assumption.

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    $\begingroup$ Your argument proves two other versions that I see frequently in action: 1. If f induces a cohomology isomorphism with coefficients in all fields, then f is a homology isomorphism. 2. If f induces a cohomology isomorphism for all fields of positive characteristic and if the integral homologies of X and Y are finitely generated in each degree, then f is a homology isomorphism. $\endgroup$ Mar 8, 2011 at 11:18
  • $\begingroup$ Neil : How prove that $\tilde{H}^\ast (Z:\mathbb{Z})=0$? $\endgroup$
    – Jino
    Mar 12, 2011 at 19:18


The basic point is that for simply-connected spaces, you can determine the connectivity of a map by looking at the connectivity of the cofiber instead of the connectivity of the fiber.

In homology, you determine the connectivity of the cofiber by looking at $H_*(C;\mathbb{Z})$, because of the Hurewicz Theorem.

In cohomology, you appeal to: if $X$ is simply-connected, then $X$ is $n$-connected if and only if $[ X, K(G,m)] = *$ for all $m \leq n$ and all abelian groups $G$. This is because (by basic obstruction theory) if $X$ is $(n-1)$-connected, then $[X, K(G,n)] \cong \mathrm{Hom}(\pi_n(X), G )$.

This results in a long list of groups to check, admittedly; but it can be whittled down by Universal Coefficients theorems if you like.


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