Let $A,B$ be $C^*$-algebras. Then an imprimitivity module for $A,B$ is a bimodule $(E,\Phi)$ consisting of a right Hilbert $B$-module $E$ and a map $\Phi:A\to \mathbb{L}(E)$ such that $\langle E,E\rangle$ is dense in $B$ and $A\cong \mathbb{K}(E)$ through $\Phi$.
My goal is to construct a dual imprimitivity module $(E^\ast,\Phi^\ast)$ such that $E\otimes_B E^\ast\cong A$ and $E^\ast\otimes_A E\cong B$.
I took $E^\ast=\mathbb{K}(E,B)$ with action of $A$ by $\mathbb{K}(E)$ and of $B$ by $\mathbb{K}(B)$ by pre-, respectively postcomposition. One isomorphism is given by \begin{align} E\otimes \mathbb{K}(E,B) &\to\mathbb{K}(E)\cong A \newline e\otimes T &\mapsto eT \end{align} I already proofed tht it is an isomorphism on generators and hence an isomorphism on the whole space. The other isomorphism should be \begin{align*} \mathbb{K}(E,B)\otimes_A E &\to B \newline T\otimes E &\mapsto Te. \end{align*} I think this is the only reasonable map to $B$. However it is not clear to me why this is injective.
Surjectivity is clear because for $b\in B$ there exists an approximate unit in $B$ of the form $\langle x_i,y_i\rangle$ (because $\langle E,E\rangle$ is dense in $B$). Then the elements $b\langle x_i,-\rangle\otimes y_i$ converge to $b$.
For injectivity I need to show that $Te=T'e'$ implies $T\otimes e=T'\otimes e'$ (then in particular the map is well defined). Using the tensor product it suffices to find an appropiate map $S_{e,e'}\in\mathbb{K}(E)$ that sends $e'$ to $e$ (and the rest to zero). Then we have $T'\otimes e'=T'\otimes S_{e,e'}e=T'S_{e,e'}\otimes e$ and similarly $T\otimes e=TS_{e,e}\otimes e$ and by construction we have that $T'S_{e,e'}=TS_{e,e}$ are the unique operators in $\mathbb{K}(E)$ that send $e$ to $Te=T'e'$. BUT: I don't now if such an operator $S_{e,e'}$ exists? The usual candidat $e\langle e',-\rangle$ doesnt work since $\langle e',e'\rangle$ is just positive and not necessarily invertible in $B$. However: There always exists a map in $\mathbb{K}(E)$ that sends $x$ to itself, namely the limit of the sequence $x(\langle x,x\rangle +\frac{1}{n}1_B)^{-1}\langle x,-\rangle$. But I am not able to generalise this since I use the fact that it sends $x$ to itself (as opposed to a different element).
Can somebody help on my thoughts and correct them?
The statement that $\mathbb{K}(E,B)$ is the dual should be true, since I read it somewhere (but without proof or the isomorphisms given).