Self map of unitary group

Question

Let $H$ be a Hilbert space and let $u_1 \in U(H)$ be a unitary operator on $H$. Consider the self-map $w: U(H) \to U(H)$ which is given by $$w(v) := v^2 u_1 v^{-1}.$$

Since $U(H)$ is connected, there is a norm-continuous path of unitaries $u_t$ such that $u_0=1_H$. Hence, $w$ is homotopic to the identity map on $U(H)$. If $H$ is finite-dimensional, this implies that $w$ is surjective. Indeed, $w$ is a self-map of degree $1$ of a closed manifold and it is well-known that such maps are surjective.

Question: Is $w$ always surjective if $H$ is infinite-dimensional?

I think this can be proved if $u_1$ is a compact perturbation of $1_H$, but I am really interested in the general case.

Answer 1

This is not really an answer but it may give somebody a starting idea. We need to solve the equation $V^2U=WV$ for given $U,W$. Rewriting it as $V=(V^{-1}WV)U^{-1}$ and noticing that for $W$ close enough to $I$ in norm, the conjugation mapping $V\mapsto V^{-1}WV=I+V^{-1}(W-I)V$ is a contraction on the unitary group with the operator norm metric, we get the result for $W$ close to $I$. Rewriting it as $V^{-1}=W^{-1}(V^2UV^{-2})$, we also get it for $U$ close to $I$. Both neighborhoods are fairly large, so there is a chance that this simple trick can be either modified or complemented by some algebra to give the general case. Alas, I do not see how to do it at the moment :( .