Kuiper's theorem says that the unitary group $U(H)$ of a separable infinite dimensional Hilbert space $H$ is contractible, if it is equipped with the norm topology.

Let's suppose, I do not know this theorem, but I do know that $U^{st*}(H)$ is contractible, where the latter group is $U(H)$ equipped with the strong$^*$ topology generated by the semi-norms $p_v(a) = \lVert av \rVert$ and $q_v(a) = \lVert a^*v\rVert$ for all $v \in H$. I have a continuous map $U(H) \to U^{st*}(H)$.

Is there any way to see that this is a weak equivalence by an approximation argument, thereby proving Kuiper's theorem?

side remark:

The motivation for this question comes from a setup which looks completely different, but is from a certain point of view surprisingly similar: Let $\mathcal{O}_{\infty}$ be the Cuntz algebra on countably infinite generators. $Aut(\mathcal{O}_{\infty})$ carries two topologies: One from the norm via the inclusion into bounded maps on Banach spaces $\mathcal{O}_{\infty} \to \mathcal{O}_{\infty}$, the other is the so-called point-norm topology generated by the semi-norms $p_a(\alpha) = \lVert \alpha(a) \rVert$. It is known that $Aut(\mathcal{O}_{\infty})$ is weakly contractible in the latter topology, but I would like to know it for the first.

`$U(H) \to U^{st*}(H)$`

induces an isomorphism on all homotopy groups. All groups above are considered to have the identity as their basepoint. $\endgroup$ – Ulrich Pennig May 15 '12 at 18:37