Strong Morita equivalence - Question about proof in Beer's "On Morita equivalence of nuclear $C^*$-algebras"

Question

I'm going over the proof of this theorem about strong Morita equivalences on page 253 of "On Morita equivalence of nuclear $C^*$-algebras" by Walter Beer (http://bit.ly/1fOZiOw), I want to make sure I understand it pretty well, I had a couple of question about a few lines in the proof, if this is trivial it's escaping me at the moment so I'll just go ahead and ask, here's theorem...

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Theorem. Let $A$ and $B$ be $C^*$-algebras with identity. Then $A$ and $B$ are strongly Morita equivalent as $C^*$-algebras if and only if they are Morita equivalent as rings.

Proofs. Assume first that $A$ and $B$ are strongly Morita equivalent. Let $_{A}X_{B}$ be an imprimitivity bimodule, and $_{B}\widetilde{X}_{A}$ its dual [30, 6.17]. Then the $A$- and $B$-valued inner products on $X$ induce bimodule homomorphisms $\phi$, $\psi$ from $_A X \otimes_B \widetilde{X}_A$ to $_A A _A$, resp. from $_B \widetilde{X} \otimes_A X_B$ to $_B B _B$. Condition 1 of [30,6.10] yields that (X, $\widetilde{X}$, $A$, $B$, $\phi$, $\psi$) is a Morita context in the sense of [5]. Since we assume that the ranges of the inner products have norm dense linear span in $A$ and $B$, it follows that $\phi(X \otimes_B \widetilde{X})$ is dense in $A$, and that $\psi(\widetilde{X} \otimes_A X)$ is dense in $B$...

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The proof is longer but I stopped right there, I wanted to know...

1 - Why do the $A$ and $B$-valued inner products on $X$ induce bimodule homomorphisms $\phi$, $\psi$ from $_A X \otimes_B \widetilde{X}_A$ to $_A A _A$, and $_B \widetilde{X} \otimes_A X_B$ to $_B B _B$ ?

which leads to...

2 - Why is $\phi(X \otimes_B \widetilde{X})$ dense in $A$, and $\psi(\widetilde{X} \otimes_A X)$ dense in $B$ ?

Thank you!

Answer 1

Let $_AX_B$ be the bimodule, then by definition, $\widetilde{X}$ is the same space with right $A$ action as $x\cdot a=a^{*}x$ and similarly for left $B$ action. The inner product on $\widetilde{X}$ is still $<x,y>_A=_A<x,y>$

Now the map from $_AX_B\otimes \widetilde{_BX_A}$ to $_AA_A$ is given on elementary tensors by $x\otimes y \rightarrow _A<x,y>$, now it's direct calculation that this map is a $A-A$ bimodule homomorphism to $_AA_A$, the assumption that $X$ is full imply the image is dense.