Families of representations of von Neumann algebras 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?
Edit: So really, what I am interested in is the question whether the map above admits some kind local sections (this is precisely what the definition of Serre fibration is about). In other words, I would like to know if a family of representations (parametrized by some suitably nice space) can be lifted to a family of implementing partial isometries, at least locally. This would be a family version of the result of Takesaki cited above.

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.
 A: Theorem.
The map $V(A,H)\to\operatorname{Hom}(A,\mathbb{B}(H))$ is open.
We write $\omega_{\xi,\eta}$ for the linear functional $x\mapsto \langle x\xi,\eta\rangle$
and $\omega_\xi$ for $\omega_{\xi,\xi}$.
It is an elementary fact that if $\omega_\xi=\omega_{\eta}$ on a von Neumann algebra $A$,
then $a\xi\mapsto a\eta$ extends to a partial isometry $u\in A'$ such that $u\xi=\eta$.
We can trim $u^*u$ a little, if necessary, and make it satisfy $1-u^*u \sim 1-uu^*$ (Murray--von Neumann equivalence),
at the cost of $\|u\xi-\eta\|<\epsilon/2$, where $\epsilon>0$ is arbitrary small.
Then $u$ extends to a unitary element in $A'$, still denoted by $u$,
which satisfies $\|u\xi-\eta\|<\epsilon$.
The following perturbation lemma is well-known and follows from
the theory of standard form (see [Takesaki, Section IX]).
Lemma.
For any $\epsilon>0$, there is $\delta>0$ which satisfies the following.
For any von Neumann algebra $A\subset B(K)$ and any unit vectors
$\xi,\eta\in K$, if $\|(\omega_\xi-\omega_\eta)|_A\|<\delta$,
then there is a unitary element $u\in A'$ such that
$\|u\xi - \eta\|<\epsilon$.
We postpone the proof of this lemma and prove the theorem.
Let $v_0\in V(A,H)$ and an SOT neighborhood
$$G=\{ v\in V(A,H) : \forall i\ \|(v-v_0)\xi_i\|<\epsilon\}$$
be given.
Here $\xi_1,\ldots,\xi_n\in H$ are unit vectors and
$\epsilon>0$.
Take $\delta>0$ from the Lemma for $n^{-1/2}\epsilon$.
Now suppose $v\in V(A,H)$ is such that
$$\| (\omega_{\xi_i,\xi_j}\circ\operatorname{Ad}_v 
- \omega_{\xi_i,\xi_j}\circ\operatorname{Ad}_{v_0})|_A \|<\delta/n$$
for all $i,j$.
We consider the unit vector
$\xi=n^{-1/2}\left[\begin{smallmatrix} \xi_1 & \cdots & \xi_n\end{smallmatrix}\right]^T\in H^n$ and view $\mathbb{B}(H^n)=\mathbb{M}_n\otimes\mathbb{B}(H)$.
Then
$$\|(\omega_{(1\otimes v)\xi} - \omega_{(1\otimes v_0)\xi})|_{\mathbb{M}_n\otimes A}\|<\delta.$$
Thus by Lemma, one finds a unitary element
$u\in A'\cong (\mathbb{M}_n\otimes A)'\cap\mathbb{B}(H^n)$ such that
$\|(1\otimes u)(1\otimes v)\xi - (1\otimes v_0)\xi\|<n^{-1/2}\epsilon$.
This implies that $uv\in G$, which finishes the proof.
Proof of Lemma.
We may assume $K = p(L^2A \otimes \ell_2)$, where $L^2A$
is a standard representation of $A$ and
$p\in (A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$.
Fix a unit vector $\delta_0\in\ell_2$.
There are unique vectors $|\xi|$ and $|\eta|$ in the positive cone $(L^2A)_+$
such that
$\omega_\xi=\omega_{|\xi| \otimes\delta_0}$ and
$\omega_\eta=\omega_{|\eta| \otimes\delta_0}$ on $A \otimes \mathbb{C}1$
(see [Takesaki, Theorem IX.1.2.(iv)]).
There are partial isometries $v$ and $w$ in $(A\otimes \mathbb{C}1)'\cap\mathbb{B}(L^2A\otimes\ell_2)$ such that $\xi=v|\xi|$ and $\eta=w|\eta|$.
By the generalized Powers--Stormer inequality ([Takesaki, Theorem IX.1.2]),
one has $\| |\xi| - |\eta| \|^2 \le \|(\omega_\xi-\omega_\eta)|_A\|$.
Hence $t:=wv^*\in p(A\otimes \mathbb{C}1)'p = A'\cap \mathbb{B}(K)$ satisfies
$$\| t \xi - \eta \| = \| w(v^*\xi-w^*\eta)\|\approx 0.$$
Let $t=u|t|$ be the polar decomposition. Since $\|t\|\le1$ and
$\|t\xi\|\approx\|\eta\|=1$, one has $|t|\xi \approx \xi$ and $u\xi\approx\eta$.
We can further replace the partial isometry $u\in A'$ with a
unitary element without affecting $u\xi$ much.
