On the surjectivity of the Hurewicz homomorphism

Question

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.

Answer 1

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.

Answer 2

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.