# A question on finite non-contractible CW complexes

The algebraic topology book I am reading recently covered the following theorem named after Whitehead and corresponding direct consequence.

THEOREM. If X is a CW complex of dimension less than n and $e: Y \rightarrow Z$ is a n-equivalence then we have a induced bijection $e_{*} : [X,Y] \rightarrow [X,Z]$.

FOLLOW UP THEOREM. If $e$ above is a n-equivalence of CW complexes of dimension less than $n$ than we have that $e$ is a homotopy equivalence.

I understand the theorems but the author makes the following statement immediately after the theorems.

"If X is a finite CW complex, in the sense that it has finitely many cells, and if dim X > 1 and X is not contractible, then it is known that X has infinitely many non-zero homotopy groups."

I was wondering if anyone could shed some light on this.

• This is not a question about the Whitehead theorem. Sep 7, 2014 at 15:41
• This is false. The torus has exactly one non-trivial homotopy group $\pi_1(T^2)\cong \mathbb{Z}^2$, $\pi_k(T^2)=0,\forall k\geq 2$. Sep 7, 2014 at 15:59
• It seems that the result is true for simply connected such spaces, and follows from a result of Serre in 1953 (according to this MO question by John Baez). Sep 7, 2014 at 16:07
• Thanks for the Serre reference. I'll take a look. Sep 7, 2014 at 16:19

Theorem [Serre]: Let $$X$$ be a path-connected, simply connected space such that:
• $$H_i(X; \mathbb{Z})$$ is a finitely generated abelian group for $$i > 0$$;
• $$H_i(X; \mathbb{Z}/2\mathbb{Z}) = 0$$ for big enough $$i$$;
• $$H_i(X; \mathbb{Z}/2\mathbb{Z}) \neq 0$$ for at least one $$i \neq 0$$.
Then there is an infinite number of integers $$i$$ such that $$\pi_i(X)$$ contains a subgroup isomorphic to $$\mathbb{Z}$$ or $$\mathbb{Z}/2\mathbb{Z}$$.