Tensoring with an induced representation: proof question Let $G$ be a locally compact Hausdorff group and $H$ a closed subgroup of $G$. If $\sigma: H \to B(\mathcal{K}_\sigma)$ is a unitary representation of $G$, we can associate an "induced representation", denoted by $\operatorname{Ind}_H^G(\sigma)$.
I'll briefly recall a possible construction for this representation. All details can be found in appendix E of the book "Kazhdan's property (T)" by B. Bekka, P. de la Harpe and A. Valette or in Folland's book "A course on abstract harmonic analysis" in chapter 6. Let $p: G \to G/H$ the canonical quotient map. Start by defining the space
$$\mathcal{A}:= \{f \in C(G, K_\sigma): p(\operatorname{supp}(f)) \text{ is compact}, f(gh) = \sigma(h^{-1})f(g) \text{ for } g\in G, h \in H\}.$$
Fix a strongly quasi-invariant measure $\mu$ on $G/H$ and consider the associated continuous function $$\lambda: G\times (G/H)\to (0, \infty)$$
such that $\lambda(x,yH)= \frac{d\mu_x}{d\mu}(yH).$ The space $\mathcal{A}$ becomes an inner product space for
$$\langle f,g\rangle := \int_{G/H} \langle f(x), g(x)\rangle_\sigma d\mu(xH) $$
and we denote the Hilbert space completion of $\mathcal{A}$ by $\mathcal{H}_\mu$. Then the map
$$\pi_\mu: \mathcal{A}\to \mathcal{A}$$
defined by
$$(\pi_\mu(x)f)(y) = \sqrt{\lambda(x,yH)}f(x^{-1}y)$$
extends to a representation and this is $\operatorname{Ind}_H^G(\sigma): G \to B(\mathcal{H}_\mu)$.
Consider now the following theorem in the book "Kazhdan’s Property (T)" by B. Bekka, P. de la Harpe and A. Valette:

There are a lot of details to be filled in, but the main strategy is clear: we define
$$U: \mathcal{H}_\pi\odot \mathcal{A}\to \mathcal{L}$$ (where $\mathcal{A}$ is defined as above) and we want to show that this map extends to a unitary
$$U: \mathcal{H}_\pi\otimes \mathcal{H}\to \mathcal{L}$$
that implements the equivalence of the two representations. However, one thing is not clear to me in this proof: Why is $U$ surjective? I see that $U$ is isometric on the algebraic tensor product, so it extends uniquely to an isometry
$$U: \mathcal{H}_\pi\otimes \mathcal{H}\to \mathcal{L}$$
and it therefore has closed range. Hence, it suffices to show that range of $U$ is dense in the codomain, but I did not manage to prove this.
Lemma E.1.3 in the aforementioned book might be relevant, but I could not make it work (I also believe that the proof of this lemma contains a mistake, but it can be fixed using a partition of unity argument).
Any help/insights is highly appreciated!

EDIT: In reply to comment:

 A: I don't know if the following is a "standard" argument, but it's what I came up with.  Introduce some notation: $\newcommand{\mc}{\mathcal}\mc A_\sigma$ comes from forming $\operatorname{Ind}_H^G\sigma$ while $\mc A_1$ comes from forming $\operatorname{Ind}_H^G(\pi|_H\otimes\sigma)$.  As you note,
$$ U:\mc H_\pi\odot\mc A_\sigma\rightarrow\mc A_1; \quad
U(\theta\otimes\xi)(x) = \pi(x^{-1})\theta\otimes\xi(x) $$
forms an isometry, and so this extends to the completions.
Now pick $\xi' \in \mc A_1$ so $\xi'$ is continuous, has "compact support" in $G/H$ and $\xi'(xh) = (\pi(h^{-1})\otimes\sigma(h^{-1}))\xi'(x)$ for $x\in G, h\in H$.  Define
$$ \xi''(x) = (\pi(x)\otimes\operatorname{id})\xi'(x) \qquad (x\in G). $$
As $\pi$ is strongly continuous and unitary, a triangle-inequality argument shows that $\xi''$ is still continuous.  It has the same "support".  Also
$$ \xi''(xh) = (\pi(xh)\otimes\operatorname{id})\xi'(xh)
= (\pi(x)\otimes\sigma(h^{-1}))\xi'(x) = (\operatorname{id}\otimes\sigma(h^{-1}))\xi''(x). $$
So $\xi''$ is in the $\mc A$-space for $1\otimes\sigma$.  (We could explore this more, but I'll just quickly finish).  This construction can be reversed, so any $\xi''$ satisfying these properties comes from some $\xi'\in\mc A_1$.  Obviously this bijection $\xi' \leftrightarrow \xi''$ is linear.  For similarly $\eta',\eta''$, we see that
$$ (\xi''(x)|\eta''(x)) = (\xi'(x)|\eta'(x)) \qquad (x\in G), $$
and so $(\xi''|\eta'') = (\xi'|\eta')$ and hence our bijection is a unitary.
We now do use Lemma E.1.3. Let $\theta\in\mc H, v\in\mc K_\sigma$, let $f\in C_c(G)$, and form $\xi''_{f,\theta\otimes v}$ using $1\otimes\sigma$, that is,
$$ \xi''_{f,\theta\otimes v}(x) = \int_H f(xh) (\theta\otimes\sigma(h)v) \ dh = \theta \otimes \xi_{f,v} \in \mc H_\pi \odot \mc A_\sigma \qquad (x\in G). $$
The lemma says that such $\xi''$ are total.  However, notice that
$$ \xi' = U(\theta\otimes\xi) \ \implies \ \xi''(x) = \theta\otimes\xi(x), $$
so putting these together gives $\xi''_{f,\theta\otimes v} = U(\theta\otimes\xi_{f,v})$.  Hence $U$ has total range, as required.

Edit: There was also a question asked about the proof of Lemma E.1.3.  I think this is fine.  I freely use the notation in the proof from the book.  By the choice of $U$ and uniform continuity we know that for $x\in Ux_i$, say $x=ux_i$ for some $u\in U$,
$$ \|\eta(x)-v_i\| = \|\eta(x) - \eta(x_i)\|
= \|\eta(ux_i)-\eta(x_i)\| \leq\epsilon. $$
As $f_i$ is chosen supported on $Ux_i$, then
$$ f_i(x)>0 \ \implies\  x\in Ux_i \ \implies\  \|\eta(x)-v_i\|\leq\epsilon. $$
Thus in the sum $\sum_i f_i(x) \|\eta(x)-v_i\|$, fix $x$ and chosen each term.  If $f_i(x)=0$ then there is no contribution; if $f_i(x)>0$ then the term is $\leq f_i(x)\epsilon$.  This gives exactly
$$ \sum_i f_i(x) \|\eta(x)-v_i\| \leq \epsilon \sum_i f_i(x) $$
for each $x$, as claimed.
