The Hurewicz homomorphism is a surjective homomorphism from $\pi_n(X) \to H_n(X)$ if $\pi_{n-2}(X)=0$ according to Wikipedia.

But if it is surjective then how could the following (contradiction) I constructed be true?

$\pi_2(\mathbb{RP}^5)=\pi_2(S^5)=0$ so $\mathbb{RP}^5$ is $2$-connected. Thus $\pi_4(\mathbb{RP}^5)\to H_4(\mathbb{RP}^5)$ is a surjective homomorphism. But $H_4(\mathbb{RP}^5)=\mathbb{Z}/2$ and $\pi_4(\mathbb{RP}^5)=\pi_4(S^5)=0$. So the Hurewicz homomorphism can't be surjective.

  • $\begingroup$ Thanks for the answers. They are helpful. I am deleting the question out of embarrasment. I don't think I would have made this sort of mistake if I had not stayed up the whole night. $\endgroup$ Jan 21, 2016 at 12:18
  • $\begingroup$ Please don't do that. It is considered rude to delete a question after getting the help you asked for. I'm sure that's not your motivation, but still, it is a practice that should be avoided. $\endgroup$ Jan 21, 2016 at 12:18
  • $\begingroup$ ok i wont do it $\endgroup$ Jan 21, 2016 at 12:18
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    $\begingroup$ Frankly I prefer questions like this rather than the ton of questions "Please solve this quadratic equation for me I can't be bothered to type it into wolfram alpha". Even if the answer was easy at least you put some thought into this. $\endgroup$ Jan 21, 2016 at 12:28

2 Answers 2


A topological space $X$ is called $k$-connected if $\pi_i(X) = 0$ for $0 \leq i \leq k$. The Hurewicz theorem states that if $X$ is $(n-2)$-connected, then the Hurewicz homomorphism $h_* : \pi_n(X) \to H_n(X)$ is surjective.

In your example, while $\pi_2(\mathbb{RP}^5) = 0$, $\pi_1(\mathbb{RP}^5) = \mathbb{Z}/2\mathbb{Z} \neq 0$, so $\mathbb{RP}^5$ is not $2$-connected and therefore the Hurewicz theorem does not apply.

  • $\begingroup$ Just for a bit of extra precision on the ticked answer, I suggest replace $1 \leq i \leq k$ by $0 \leq i \leq k$, especially as homotopy groups are defined only for spaces with base point. $\endgroup$ Jan 21, 2016 at 14:19
  • $\begingroup$ You're right, I should have included that. $\endgroup$ Jan 21, 2016 at 14:27

For the hurewiczs homomorphism, ALL the homotopy groups $\pi_i, i\leq n$ have to be zero to have a surjective morphism $\pi_{n+1}(M)\rightarrow H_n(M)$. In your case, $\pi_1(RP^5)$ is not zero.


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