Equivalence of projections in smaller von Neumann algebra

Question

I came across the following assertion and I can't understand why it's true.

We are given with two finite equivalent projection $e\sim f$ in some von Neumann algebra $A$ (with a unit of course). It's known that the projection $q=e\vee f$ is also finite, so we infer the algebra $qAq$ is a finite algebra (that is, the unit element is a finite projection).

Now suppose that $q-e\sim q-f$ in $qAq$, show that $q-e\sim q-f$ also in $A$.

Thanks in advance.

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Comment: Technically it is true that $q-e\sim q-f$ in $qAq$ so we don't really need to assume that.

Answer 1

Every element in $qAq$ is also in $A$. In particular, the partial isometry that realizes the equivalence.