Let $A$ be a von Neumann algebra and let $H$ be a (separable) Hilbert space. It is known (see e.g., Section IV, Thm. 5.5 of Takesaki I) that there exists a Hilbert space $K$ such that $A \subset \mathbb{B}(K)$ such that any normal $*$-homomorphism $\varphi : A \to \mathbb{B}(H)$ can be written as $$ \varphi(a) = v^* a v,$$ where $v: H \to K$ is a partial isometry with $v^*v = \mathrm{id}_H$ and $v v^* \in A^\prime \subseteq \mathbb{B}(K)$.

We are led to consider the sets $\mathrm{Hom}(A, \mathrm{B}(K))$ with its u-topology (defined by seminorms $\varphi \mapsto \|\omega \circ \varphi\|_{A_*}$, where $\omega \in \mathrm{B}(H)_*$ is an element of the predual of $\mathrm{B}(H)$) and $$V(A, H) := \{v \in \mathrm{B}(H, K) \mid v^*v = \mathrm{id}_H, vv^* \in A^\prime\}$$ with the strong operator topology. By the result stated above, the map $V(A, H) \to \mathrm{Hom}(A, \mathrm{B}(H))$ is surjective.

Q: Is this map a Serre fibration?

If not, what are the problems here, and can we assume something on $A$ or change the topologies somehow to ensure this?

So maybe let me write something about what I tried and where I failed.

(1) Let us first prove that the map $V(A, H) \to \mathrm{Hom}(A, \mathrm{B}(H))$ is continuous. To this end, let $\omega \in \mathrm{B}(H)_*$ be positive, which can be written as $\omega(A) = \mathrm{tr}(WA)$ for some positive trace-class operator $W = \sum_{i=1}^\infty w_i e_i \otimes e_i^*$. Let $v_n \to v$ be a convergent sequence in $V(A, H)$ and let $\varphi_n$, $\varphi$ be the corresponding sequences of homomorphisms $A \to \mathrm{B}(H)$. We have to show that $\mathrm{sup}_{\|a\| \leq 1} |\omega \circ \varphi_n(a) - \omega \circ \varphi(a)|$ converges to zero. If $\|a\| \leq 1$, then $$|\omega \circ \varphi_n(a) - \omega \circ \varphi(a)| = |\mathrm{tr}(W v_n^* av_n - Wv^*av)| \\ \leq \sum_{i=1}^\infty w_i |\langle e_i, v_n^* a v_n e_i - v^*ave_i\rangle|\\ \leq \sum_{i=1}^\infty w_i \Bigl(|\langle (v_n-v)e_i, a v_n e_i\rangle| + |\langle ae_i, (v_n - v) e_i\rangle|\Bigr)\\ \leq 2 \sum_{i=1}^\infty w_i \|(v_n-v)e_i\|, $$ which converges to zero.

(2) The fibers are as follows: It is not hard to see that if two elements $v_1, v_2 \in V(A, H)$ induces the same homomorphism $\varphi : A \to \mathrm{B}(H)$, then $v_1 = wv_2$ for a partial isometry $w \in A^\prime$ with $ww^* = v_1 v_1^*$, $w^*w = v_2 v_2^*$. Conversely, given $v$ implementing $\varphi$, then replacing it by $wv$ for a partial isometry $w \in A^\prime$ with $w^*w = vv^*$ gives another element of $V(A, H)$ implementing the same $\varphi$.

(3) So what I tried was the following: Fix a projection $p \in A^\prime$, and look at the subset $V_p(A, H)$ of all $v$ such that $vv^* = p$, and let $\mathrm{Hom}_p(A, \mathrm{B}(H))$ be the set of those homomorphisms that are implemented by such a $v$. Then after fixing a basepoint $v_0 \in V_p(A, H)$, then by (2), we have identifications $$ V_p(A, H) \approx \mathrm{U}(pH)$$ given by sending $u \in \mathrm{U}(pH)$ to $uv_0$, and two elements $u_1$, $u_2$ correspond to the same element of $\mathrm{Hom}(A, \mathrm{B}(H)$ if and only if $u_2 u_1^* \in \mathrm{U}(A^\prime)$. Hence there is a continuous bijection $$\mathrm{U}(pKp)/(\mathrm{U}(pKp)\cap \mathrm{U}(A^\prime)) \to \mathrm{Hom}(A, \mathrm{B}(H)).$$ However, it is now clear to me how to show that this is a homeomorphism. Also, I am not sure how to attack this if we don't fix $p$ in advance.