Non-commutative geometry – Dirac operator on an equivalent Morita algebra

Let $ (A, H, D) $ to be a spectral triple and let $ B $ to be an algebra that is Morita equivalent to $ A $. Then there is a project, finely generated generated $ A $ module $ E $ such as $ B = End_A (E) $. endow $ E $ with $ A $valued domestic product $ ( cdot, cdot) $ and forms a Hilbert space $ H = $ H $ with the inner product given by
$$ langle x otimes xi, y otimes eta rangle_ {H}: = langle (x, y) xi, eta rangle_H $$
We would like to impose the structure of a spectral triple at $ B $: a naive choice is to put $ D (x otimes xi) = x otimes D xi $ However, this map is not well defined because $ H $ is a tensor product more than $ A $

In order to solve this problem, use a connection $ nabla $ it is a map $ nabla: E to E otimes_A Omega_D ^ 1 $ or $ Omega ^ 1_d = { sum_ {j = 1} ^ N a_jdb_j: a_j, b_j in A, N in mathbb {N} } $ and $ db: =[D,b]$. This card must satisfy $ nabla (xa) = nabla (x) a + x $ otimes (note that because of Jacobi's idenity for switches, the space $ Omega_D ^ 1 $ is in fact $ A-A $-bimodule then $ E otimes_A Omega ^ 1_D $ is a right $ A $-module). Having fixed a connection we define
$$ (x otimes xi) = x otimes D xi + nabla (x) xi $$
(or $ nabla (x) $ xi is understood as follows: $ nabla (x) = sum_j x_j otimes omega_j $ then $ nabla (x) xi = sum_j x_j otimes omega_j ( xi) $ which is good since each $ omega_j in Omega ^ 1_D $ always acts on $ H $).
The natural action of $ B $ sure $ H $ is as follows $ b (x otimes xi) bx otimes xi $ which is fine since $ b in B = End_A (E) $.

How to show that for everyone $ b $ B $ the switch $[D’,b]$ (extends to a /) is an operator bounded on $ H $?

A simple calculation shows that this switch evaluated on $ x otimes xi $ is equal to $ nabla (bx) xi-b ( nabla x ( xi)) $ and I do not see how to get this operator to be bounded – do we have to assume something about $ nabla $ (a kind of compatibility with the $ B $-action)?