If $\pi_0(X)=\pi_1(X)=0$ and $H_2(X)=0$ then is it true that $\pi_2(X)=0$?

Question

I'm reading about Hurewicz theorem and just wondered if it works this way. Stupid question, but I was thinking whether n-connectedness in the theorem means just that $\pi_i(X)=0$ for all $i=0,\dots,n$ or that in addition $\pi_{n+1}(X)$ must be non-trivial.

Thank you.

Answer 1

I don't think it's a stupid question. It's just a matter of getting the definitions straight. I looked it up and $n$-connected means the first $n$ homotopy groups are trivial. There is nothing about $\pi_{n+1}$.

So for example an $n$-sphere is $n-1$ connected. Of course, we know the unstable homotopy groups ($\pi_{n+k}\,,k\lt n+2$) can be nontrivial. Consider for instance the Hopf fibration. But not that they always are.

I guess to zero in on your question better, what about $\Bbb R^n$. All its homotopy groups are trivial, but it's pretty clear it's $n$-connected.

For your title question, $1$-connected implies $H_*:\pi_2 X\to H_2 X$ is an isomorphism. I see this got answered in the comments.

Answer 2

When we say "$X$ is an $n$-connected space" we mean $\pi_i(X) = 0$ for all $i \leq n$; in particular "$0$-connected" means path-connected and "$1$-connected" means simply-connected.

It's important to not mix this up with the definition of "$f\colon X \to Y$ is an $n$-connected function". A pointed function is $n$-connected if $\pi_i(f)$ is an isomorphism for $i < n$ and $\pi_{n}(f)$ is surjective; this is equivalent to requiring that the homotopy fiber $hofib(f)$ over the basepoint is an $(n-1)$-connected space as above, and you can see these are equivalent from the long exact sequence for a fibration. (Be cautious because there is a different convention where "$n$-connected function" means that the homotopy fiber is $n$-connected, I'm using the definition from nLab.)

It follows from the Hurewicz Theorem that if $X$ is simply-connected and $H_i(X) = 0$ for $i \leq n$ then in fact $X$ is $n$-connected: if there were a $1 < i < n$ such that $\pi_i(X) \neq 0$ then if $i_0$ is the least such $i$ then $H_{i_0}(X) \cong \pi_{i_0}(X) \neq 0$ by Hurewicz. In your particular case, you are correctly deducing that $\pi_2(X) = 0$.