Is the following assertion and the proof below correct,

or am I missing something very important?

Moreover, would the corollaries be correct then?

Besides, I would also appreciate a lot any comment, advice, suggestion about the mathematical writing like structure, titles, phrasing, layout, arguments, etc. :)

# Problem

Assertion:Given C*-algebras $A$ and $D$ and let $I\trianglelefteq A$ be a closed two-sided ideal in $A$.

Consider their algebraic tensor products $A\odot D$ and $I\odot D$, respectively.

Let $B(H)$ denote the bounded operators over a Hilbert space $H$.

Then every *-morphism $\varphi:I\odot D\to B(H)$ extends to $$ \begin{matrix} A\odot D & \xrightarrow{\Phi} & B(H) \\ \uparrow & \nearrow\varphi & \\ I\odot D \end{matrix}$$ for a unique *-morphism $\Phi:A\odot D\to B(H)$.

# Application/Motivation

As an interesting application of the assertion we would obtain the following

Corollary:Given C*-algebras $A$ and $B$.

Denote their multiplier algebras by $\mathcal{M}A$ and $\mathcal{M}B$, respectively.

Consider their algebraic tensor products $A\odot B$ and $\mathcal{M}A\odot\mathcal{M}B$.

Let $B(H)$ denote the bounded operators over a Hilbert space $H$.

Then every *-morphism $\varphi:A\odot B\to B(H)$ extends to $$ \begin{matrix} \mathcal{M}A\odot \mathcal{M}B & \xrightarrow{\Phi} & B(H) \\ \uparrow & \nearrow\varphi & \\ A\odot B \end{matrix}$$ for a unique *-morphism $\Phi:A\odot D\to B(H)$.

as well as a nice proof for the restriction theorem:

Corollary:Given C*-algebras $A$ and $B$.

Consider their algebraic tensor product $A\odot B$.

Let $B(H)$ denote the bounded operators over a Hilbert space $H$.

Then every *-morphism $\varphi:A\odot B\to B(H)$ restricts to $$ \begin{matrix} A\odot B & \xrightarrow{\varphi} & B(H) \\ \uparrow & \nearrow\varphi_A\cdot\varphi_B & \\ A\times B \end{matrix}$$ for unique *-morphisms $\varphi_A:A\to B(H)$ and $\varphi_B:B\to B(H)$.

Moreover, $\varphi_A$ and $\varphi_B$ commute.

# Proof

*Proof:* a) Existence:

Though the idea of the construction is pretty simple, the proof can get pretty lengthy in detail.

Thus let us split the proof into several parts:

i.) Reduction to non-degenerate *-morphism:

Denote the kernel of the image by $$\mathcal{N}(I\odot D):=\{v\in H:\varphi(I\odot D)v=0\}=\bigcap_{y\in I\odot D}\mathcal{N}y$$ while the range of the image by $$\mathcal{R}(I\odot D):=\varphi(I\odot D)H=\bigcup_{y\in I\odot D}\mathcal{R}y$$ and denote its closed linear span by $H_\varphi:=\overline{\operatorname{span}}\mathcal{R}(I\odot D)$.

Then we obtain the orthogonal decomposition $$H=H_\varphi\oplus H_\varphi^\bot=\overline{\operatorname{span}}\mathcal{R}(I\odot D)\oplus\mathcal{N}(I\odot D)$$ and whence the matrix representation $$\varphi=\begin{pmatrix}\pi&0\\0&0\end{pmatrix}$$ where $\pi:I\odot D\to B(H_\varphi)$ is a non-degenerate *-morphism.

For shorthand notation redefine $H:=H_\varphi$.

Thus the problem reduces to extending a non-degenerate *-morphism $\pi:I\odot D\to B(H)$.

ii.) Construction:

Fix an element $d_0\in D$ and define the map $\Pi(\_\odot d_0):A\to B(H)$ by $$\Pi(a\odot d_0)\pi(i\odot d)v:=\pi(ai\odot d_0d)v$$ for $a\in A$, $i\in I$, $d\in D$ and $v\in H$ and extend linearly.

iii.) Well-Definedness:

This is indeed well-defined by the following standard argument,

which is actually the heart of the proof:

As a first instance, note that for $d_0\geq0$ the linear map $$\pi(\_\odot d_0):I\to B(H):i\mapsto\pi(i\odot d_0)$$ is positive between C*-algebras and as such bounded.

But a C*-algebra is spanned by its positive elements,

and thus $\pi(\_\odot d_0)$ is bounded also for $d_0\in D$ not necessarily positive.

With this observation in hand we obtain that $$\pi(eai\odot d_0)\xrightarrow{e\to1}\pi(ai\odot d_0)$$ for $a\in A$, $i\in I$ and any approximate unit $e\in I$ for $I$.

Let now $a\in A$ and $\sum_ki_k\odot d_kv_k\in\operatorname{span}\pi(I\odot D)H$.

Moreover let $e\in I$ be a, now bounded, approximate unit for $I$. Then
\begin{align*}
\|\Pi(a\odot d_0)\sum_k\pi(i_k\odot d_k)v_k\|=&\|\sum_k\pi(ai_k\odot d_0d_k)v_k\|\\
=&\|\lim_e\sum_k\pi(eai_k\odot d_0d_k)v_k\|\\
=&\lim_e\|\pi(ea\odot d_0)\sum_k\pi(i_k\odot d_k)v_k\|\\
\leq&\sup_e\|\pi(\_\odot d_0)\|\cdot\|ea\|\cdot\|\sum_k\pi(i_k\odot d_k)v_k\|\\
\leq&\|\pi(\_\odot d_0)\|\cdot\sup_e\|e\|\cdot\|a\|\cdot\|\sum_k\pi(i_k\odot d_k)v_k\|,
\end{align*}
whence the operator $\Pi(a\odot d_0)$ is well-defined and bounded on $\operatorname{span}\pi(I\odot D)H$,

and thus, since $\pi$ is non-degenerate, extends to a bounded operator on $H$.

Hence, the map $\Pi(\_\odot d_0)$ is well-defined.

iv.) *-Morphism:

We may now vary the previously fixed element $d_0\in D$ as well to obtain a bilinear map $$\Pi:A\times D\to B(H):(a,d)\mapsto \Pi(a\odot d)$$ and so to obtain a linear map, also denoted by $$\Pi:A\odot D\to B(H):a\odot d\mapsto\Pi(a\odot d).$$ The resulting map is easily seen to be a *-morphism:

For $a_1,a_2\in A$, $d_1,d_2\in D$ as well as $\pi(i\odot d)v\in\operatorname{span}\pi(I\odot D)H$ we have \begin{align*} \Pi(a_1\odot d_1)\Pi(a_2\odot d_2)\pi(i\odot d)v=&\pi(a_1a_2i\odot d_1d_2d)v\\ =&\Pi(a_1a_2\odot d_1d_2)\pi(i\odot d)v, \end{align*} and thus, again since $\pi$ is non-degenerate, $$\Pi(a_1\odot d_1)\Pi(a_2\odot d_2)=\Pi(a_1a_2\odot d_1d_2).$$

Involutivity follows a similar vein.

b) Uniqueness: Here, I'm not sure about the degenerate case!