Why is this projection of rank 1? I'm reading the following proof excerpt from a textbook that state the following:
Let $E$ be an orthonormal basis of a Hilbert space $H$.  For $e \in E$, define $p_e = e \otimes e$ for $e \otimes e: H \longrightarrow H$ given by $(e \otimes e)(z) = \langle z,e \rangle e$.  Then $p_e$ is a projection of rank $1$ with $p_eK(H)p_e = \mathbb{C}p_e$ with $K(H)$ the collection of compact operators on $H$.
Let $H'$ be another Hilbert space, and suppose that $\varphi: K(H) \longrightarrow K(H')$ is a $*$ isomorphism.  Then $q_e = \varphi(p_e)$ is a projection satisfying $q_eK(H')q_e= \mathbb{C}q_e$.  It's clear also that $q_e$ is of rank $1$ too.
Why is $q_e$ of rank one?  I know that $q_e=q_e\varphi(v)q_e$ for $v$ the rank $1$ operator that sends $x \mapsto \beta e$.  I.e, $v$ takes an $x \in H$ (which is represented in terms of the orthonormal basis $E$ on $H$), to $\beta e$, where $\beta$ is the coefficient in front of $e$ in the basis representation of $x$ (as then the range of $v$ is $\text{Span}(e)$ and $\langle v(e),e \rangle =1$, so $p_e=p_evp_e$).
However, I don't see why this arbitrary $*$ isomorphism sends the rank $1$ operator $v$ to another rank $1$ operator.
 A: The key fact is that $\varphi$ is a $*$-isomorphism $K(H)\to K(H')$. So you take $p\in B(H)$ of rank one. Now consider $\varphi(p)\in K(H')$. Because $\varphi$ is a $*$-homomorphism, $\varphi(p)$ is a projection. You know that $\varphi(p)$ is a finite-rank projection because it is compact.
Now suppose $q\leq\varphi(p)$. The $*$-isomorphism preserves order, so we get $\varphi^{-1}(q)\leq p$. As $p$ is rank-one and $q\ne0$, this forces $\varphi^{-1}(q)=p$. Thus $q=\varphi(p)$. So $\varphi(p)$ has no proper subprojections, and thus it is minimal in $K(H)$, so rank-one.
The result is not true if $\varphi$ is not surjective, nor if its range is not required to be $K(H')$. Without those restrictions, if $H'$ is infinite-dimensional one can decompose it as $H'=H\otimes H_0$ and consider $\varphi:T\to T\otimes I$. Now $\varphi$ is a $*$-monomorphism, and it maps rank-one projections to projections with rank equal to the dimension of $H_0$. If $H_0$ is chosen infinite-dimensional, then $\varphi(T)$ is not compact for all nonzero $T$.
