A representative $\alpha$ of a cohomology class $[\alpha]\in H^n(X;\pi_n(X))$ is equivalent to a map from $X$ into an Eilenberg-Mac Lane space as $\alpha:X\to K(\pi_n(X),n)$. After applying a covariant functor $[S^n,\_]$, we get a homomorphism between homotopy groups: $\alpha_*:\pi_n(X)\to \pi_n(X)$.

Then, what is the condition on $\alpha$ so that $\alpha_*$ becomes an isomorphism? Rather vaguely, is there any way of describing $\alpha_*$ in terms of $\alpha$ or the other way around?

2nd edit:

Sorry for missing the motivation but appreciate the answer. I was wondering if there is any relationship between some algebraic structure in $H^n(X;\pi_n(X))$ and the property that the induced homomorphism is an isomorphism, something like "a generator induces an isomorphism". Is this true or does it make any sense to ask at all? And I also wonder if there is any general argument on what does algebraic operations in the cohomology do on induced homomorphisms. BTW, I also need this for the case when $n=1$.

3rd edit:

Thanks for the comments. I guess these previous questions were sort of ill-stated by being over simplified and generalized. The starting point was this:

First, $[BSO,K(\pi_1(O),2)]\cong H^2(BSO;\pi_1(O))$ and the cohomology is generated by the second Stiefel-Whitney class $w_2$. Then I guess that $(w_2)_*:\pi_2(BSO)\to \pi_1(O)$ should be an isomorphism but I can't be sure of that.

The same line of thought applies to $[B{\rm Spin},K(\pi_3(SO),4)]\cong H^4(B{\rm Spin};\pi_3(SO))$ and the latter cohomology is generated by the half of first Pontrjagin class $\frac{1}{2}p_1$. I also guess that this class would induce the isomorphism $(\frac{1}{2}p_1)_*:\pi_4(B{\rm Spin})\to \pi_3({\rm SO})$ but I'm not sure too.

What I was wondering was that, if $w_2$ and $\frac{1}{2}p_1$ do induce isomorphisms, then do they so because of the algebraic property of being generators (instead of using seemingly ad hoc constructions and special properties of theses spaces) so that this would generalized easily into other cases.

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    $\begingroup$ Universal coefficients? $\endgroup$ Apr 24, 2014 at 7:38
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    $\begingroup$ To the edit: any group is generated by the set of all its elements, right? So "..." seems unlikely. $\endgroup$ Apr 24, 2014 at 9:21
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    $\begingroup$ Well, if $\hom(A,A)$ is cyclic, any generator $x$ must be an isomorphism since $id_A=nx$ for some $n$. $\endgroup$ Apr 24, 2014 at 15:10
  • $\begingroup$ @FernandoMuro Ah, you're right. That was extremely trivial and I don't know why I didn't see that simple way. Many thanks! $\endgroup$
    – HBS
    Apr 24, 2014 at 15:23
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    $\begingroup$ You're welcome, this happens to everyone (at least it's happening to me right now with something I'm doing). $\endgroup$ Apr 24, 2014 at 15:31

1 Answer 1


The case $n=1$ is fairly easy: for each connected space, there is a map $X \to K(\pi_1 (X);1)$ which is an isomorphism in $\pi_1$. More generally, if $X$ is $n$-connected, then, as $H^n (X, \pi_n (X)) \cong Hom (H_n (X); \pi_n (X)) \cong End (\pi_1 (X))$, and this contains the identity map. The map $X \to K(\pi_n (X);n)$ associated with the identity induces an isomorphism on $\pi_n(X)$. For higher homotopy groups, this is in general impossible, and has to with the nontriviality of the $k$-invariants.

The map $\alpha: X \to K(\pi_n (X);n)$ induces a map $$H_n (\alpha):H_n( X) \to H_n (K(\pi_n (X);n))\cong \pi_n (X)$$ on homology. We can describe this map differently, namely as the image of $[\alpha] \in H^n (X; \pi_n (X))$ under the natural map $$H^n (X, \pi_n (X)) \to Hom (H_n (X), \pi_n (X)).$$ Now we can compose $H_n (\alpha)$ with the Hurewicz homomorphism $hur:\pi_n (X) \to H_n (X)$. The composition $H_n (\alpha) \circ hur: \pi_n (X) \to \pi_n (X)$ is the map $\alpha_{\ast}$ on homotopy induced by $\alpha$. So here is a condition on $\alpha$:

$\alpha_{\ast}$ is an isomorphism if and only if $H_n (\alpha)$ is a left-inverse for the Hurewicz homomorphism. In particular, such an $\alpha$ exists iff the Hurewicz homomorphism is split-injective.

The concrete example of $BO$ can be treated as follows. Note that $\pi_1 (BSO)=0$ and $\pi_2 (BSO)=Z/2$, hence $$[BSO,K(Z/2,2] \cong H^2 (BSO;Z/2) \cong Hom (\pi_2 (BSO);Z/2)$$ is an isomorphism. But $Hom (\pi_2 (BSO);Z/2)$ contains two elements: the zero map and the unique isomorphism. As $w_2$ is not zero, it must give an isomorphism.

Likewise, $BSpin$ is $3$-connected and so

$$H^4 (BSO;Z) = Hom (H_4(BSO);Z) \cong Hom (\pi_4 (BSpin);Z) \cong Hom (Z;Z).$$

Half on the Pontrjagin class is a generator on the left, and hence its image in $Hom (Z;Z)$ is a generator, which is why it is an isomorphism.

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    $\begingroup$ Perhaps it is worth observing here that the condition on a map into an Eilenberg-Maclane space to be an isomorphism on $\pi_{n}$ in terms of split-injectivity of the Hurewicz homomorphism can be used in practice to show that a connected commutative topological monoid $M$ is weakly equivalent to a product of Eilenberg-Maclane spaces. Namely, the map $M \rightarrow SP^{\infty}(M)$ into the infinite symmetric product induces the Hurewicz homomorphism on homotopy groups and the monoid structure gives the needed splitting $SP^{\infty}(M) \rightarrow M$. $\endgroup$ Apr 25, 2014 at 9:15

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