Consider the $\ell^2$ complex Hilbert space which is defined us $$\ell^2=\left\{x=(x_n)_{n\in \mathbb{N}^*}\subset \mathbb{C};\;\;\|x\|_2^2:=\sum_{i=1}^{\infty}|x_i|^2<\infty\;\right\}.$$
I want to prove that the unit sphere of $l^2$ and $S^1\times l^2$ are not homeomorphic, where $$S^1:=\{y\in \mathbb{C}; \|y\|=1\}.$$ Is it possible to prove that the unit sphere of $l^2$ is simply connected however $S^1\times l^2$ is not?
I assume that the topology on $\ell^2$ is the norm.
As noted, the fact that there is no homeomorphism follows from the fact that the two spaces have different fundamental groups: $\pi_1(S^1\times \ell^2)\cong \pi_1(S^1)\cong \mathbb Z$, because $\ell^2$ is contractible. However, the unit sphere $S_{\ell^2}$ does not have fundamental group $\mathbb Z$. It is also contractible (this proof comes from Blackadar's Operator Algebras):
Identify $\ell^2$ with $L^2[0,1]$. Define the operator $V_t\in \mathcal{B}(H)$ by $[V_t(f)](s)=t^{-1/2}f(s/t)$, setting $f(s/t)$ to be $0$ when $s$ is larger than $1$. Now define our homotopy by $$H(f,t)=t^{1/2}V_tf+(1-t)^{1/2}\chi_{[t,1]}$$ with the missing steps of the proof being: checking the norm continuity of everything, showing that $H(f,t)\in S_{\ell^2}$ for every $t\in [0,1]$, $f\in S_{\ell^2}$, and that $H(f,1)=f$, but $H(f,0)\equiv 1$.