# Noncontractible space with zero homology groups?

If we define $X\subset\ell^2$ to be the unit sphere, i.e. $$X=\left\{(x_i)_{i=1}^\infty\ \Bigg|\ \sum_{i=1}^\infty x_i^2 = 1\right\}$$ then since $X$ is path connected and not compact, and since $X\setminus\{x\}$ is contractible for all $x\in X$, then a simple calculation shows that $\tilde{H}_n(X)=0$ for all $n$.

Now, I wish to show that $X$ is not contractible. Assuming we have a contraction to a point $x$, we can use a change of orthonormal basis to obtain a contraction to the point $e_1=(1,0,0,\ldots)$, i.e. a continuous map $h:X\times I\to X$ such that $h(\bullet,0)=\mathrm{id}_X$ and $h(\bullet,1)\equiv e_1$. However, I'm struggling to get a contradiction here, since I know of no stronger invariant than homology/homotopy groups, and by compactness arguments, these are all trivial.

• Unit sphere, not unit ball. And this is contractible. Bad luck.
– user98602
Commented Dec 27, 2016 at 4:13
• Huh. What's the contraction? Commented Dec 27, 2016 at 4:14
• mathoverflow.net/a/199/40804
– user98602
Commented Dec 27, 2016 at 4:14
• Ah, I see. So when can we have noncontractible spaces with zero homology groups? Commented Dec 27, 2016 at 4:17
• The Hilbert sphere is even diffeomorphic to the Hilbert space. Commented Dec 27, 2016 at 4:19