Strong Morita Equivalence and Morphisms Between $ C^{*} $-Algebras If $ A $ and $ B $ are $ C^{*} $-algebras, then they are strongly Morita equivalent if there exist a $ (B,A) $-bimodule $ E $ and an $ (A,B) $-bimodule $ F $ such that
$$
E \otimes_{A} F \cong B \quad \text{and} \quad
F \otimes_{B} E \cong A,
$$
where the isomorphisms are between $ (B,B) $-bimodules and between $ (A,A) $-bimodules respectively. All bimodules are meant to be Hilbert bimodules.
I was wondering:

Question. If two $ C^{*} $-algebras are strongly Morita equivalent and one knows the bimodules that implement the strong Morita equivalence, then does there exist a morphism between the $ C^{*} $-algebras that can be defined using the bimodules?

I thank everyone in advance for their help.
 A: You cannot says anything about morphisms between $A$ and $B$ in general, but from the two bi-modules you can construct a third algebra $C$, such that both $A$ and $B$ embeds into $C$ with the embeddings inducing Morita equivalence between $A$ and $C$ and $B$ and $C$ compatible with the equivalence between $A$ and $B$. The algebra $C$ is constructed as an algebra of $2\times2$ matrices, with diagonals coefficient in $A$ and $B$ and non diagonal coefficient in the two bi-modules.
You will find details about this for example in Blackadar "Operator algebras" II.7, especially II.7.6.9
A: Well, here is a remarkable theorem of Brown-Rieffel-Green:
If $A,B$ are $\sigma$-unital and stable, then they are strong Morita equivalent iff  they are isomorphic. 
However,if they are not stable, there may not be a morphism, as $\mathbb{C},\mathbb{K}$ are strongly Morita equivalent via the Hilbert space $\mathbb{H}$, but there is no good map for us, nor it is reasonable to expect a map to be constructed via $\mathbb{H}$.
Something trivial can be $A,B$, $B$ stable, $\sigma$-unital, and $B$ stable, then $A$ has a map to $A\otimes \mathbb{K}$, by $a\rightarrow a\otimes p $, where $p$ is a rank one projection,$A\otimes \mathbb{K}$ is isomorphic to $B\otimes \mathbb{K} \cong B$ by the previous theorem.
