When can a partial isometry $u$ in $\mathcal B(H \otimes K)$ be extended to a unitary in $1 \otimes \mathcal B(K)$? Let $H$ and $K$ be Hilbert spaces, and let $u$ be a partial isometry in $\mathcal{B}(H \otimes K)$ between projections $p_0 = u^\ast u$ and $p_1 = u u^\ast$ such that $p_0, p_1 \leq 1 \otimes (1-q)$ for some projection $q \in \mathcal B(K)$ equivalent to $1 \in \mathcal B (K)$. Does $p_1 u (a \otimes 1)p_0 = p_1 (a \otimes 1)u p_0$ for all $a \otimes 1 \in \mathcal{B}(H) \otimes 1$ imply that $u$ can be extended to a unitary operator in the commutant $(\mathcal B(H)\otimes 1)' = 1 \otimes \mathcal B(K)$?
I've verified the implication for several simple examples, but I'm having trouble proving it in full generality.
Update: I've reworded the question for clarity thanks to Yemon's comments. By a unitary operator extending $u$, I mean a unitary operator that agrees with $u$ on $p_0 (\mathcal H \otimes \mathcal K)$.
Update: Here's one example.
Let $K$ be infinite dimensional with basis $e_1,e_2,\ldots$, and $H$ be $n$-dimensional with bases $\xi_1, \ldots, \xi_n$ and $\zeta_1, \ldots, \zeta_n$. Let $\xi = \frac{1}{\sqrt n}\sum_i \xi_i \otimes e_i$ and $\zeta = \frac{1}{\sqrt n}\sum_I \zeta_i \otimes e_i$. Define $u\colon\eta \mapsto \langle \eta, \xi \rangle \zeta$. Note that the assumptions in the first sentence are satisfied with $q$ the projection on the span of $\{e_{n+1}, e_{n+2}, \ldots \}$. The condition that $p_1 u (a \otimes 1)p_0 = p_1 (a \otimes 1)u p_0$ for all $a \otimes 1 \in \mathcal{B}(H) \otimes 1$ is here equivalent to $\langle u (a \otimes 1) \xi, \zeta\rangle = \langle (a \otimes 1) u \xi , \zeta \rangle $ for all $a \in \mathcal B(H)$, which is in turn equivalent to $\langle(a \otimes 1)\xi, \xi \rangle = \langle (a \otimes 1) \zeta, \zeta \rangle$ for all $a \in \mathcal B(H)$. The latter equation holds because both sides are equal to the trace of $a$. Thus, all conditions are satisfied.
To see that the conclusion holds, let $K_n$ be the span of $e_1,\ldots,e_n$, and let $c_{ij}$ be the unitary $n \times n$ matrix such that $\zeta_i = \sum_j c_{ij} \xi_j$. Then $$u \xi = \zeta = \frac{1}{\sqrt n}\sum_i \zeta_i \otimes e_i = \frac{1}{\sqrt n} \sum_{ij} c_{ij}\xi_j \otimes e_i = \frac{1}{\sqrt n}\sum_j \xi_j \otimes we_j = (1 \otimes w) \xi,$$ where $w\colon K_n \rightarrow K_n$ is the unitary defined by $we_j = \sum_i c_{ij}e_i$. We extend $w$ to all of $K$ by setting $\tilde{w} = w \oplus 1$. Then $1 \otimes \tilde w$ is a unitary operator in $1 \otimes B(K)$ that extends $u$.
I realized that the other examples I was thinking about are trivial.
 A: If I get it right this time, the following Zorn argument proves the implication.
Denote by $q_0=\bigvee_{v\in \mathcal{U}(H)} (v\otimes 1)p_0(v^\ast\otimes 1)$ and similarly $q_1=\bigvee_{v\in \mathcal{U}(H)} (v\otimes 1)p_1(v^\ast\otimes 1)$. Apply Zorn's lemma to the set $I=\left\{w\in \mathcal{B}(H\otimes K)\left\vert \begin{array}{l}\,\,w\text{ is a partial isometry, }\\w^\ast w\leq q_0,\\ww^\ast\leq q_1,\\w\text{ extends u, and }\\\!\!\!ww^\ast(v\otimes 1)w=w(v\otimes 1)ww^\ast\text{ for all }v\in\mathcal{U}(H)\end{array}\right\}\right.$
ordered by extension: $w_1\leq w_2$ iff $w_2 w_1^\ast=w_1 w_1^\ast$ (in other words: $w_2$ has larger initial support and agrees with $w_1$ on the initial support of $w_1$)
A maximal element $w\in I$ must have $w^\ast w=q_0$, for otherwise there would have been a unitary $v\in\mathcal{B}(H)$ such that $f=(v^\ast\otimes 1)w^\ast w(v\otimes 1)\wedge (q_0-w^\ast w)\not=0$. Then $(v^\ast\otimes 1)w(v\otimes 1)f + w$ is a partial isometry in $I$ that extends $w$. Since $1-q_0$ and $1-q_1$ are equivalent projections in $1\otimes \mathcal{B}(K)$, we can extend $w$ to a unitary.
