I was reading a note on Von Neumann Algebra, and I am not able to understand this phrase as:
Let $K$ be a closed vector subspace of a Hilbert space $H$ and let $p$ be the projection of $H$ onto $K$. If $u \in B(H)$, let $u_p=u_K$ be the compression of $u$ to $K$. Then consider the map, $\phi : pB(H)p \to B(K)$ by $\phi(u)=u_K.$ This is a $*$-isomorphism. And if $A$ is a $*$-algebra on $H$ and $p \in A'$, set $A_p =\{u_p:u \in A\}$.
In the above statements I am unable to understand the definition of the map $u_p$. I know compression of $u$ to $K$ can be viewed as the composition $~pu|_K:K \to K,$ where $u|_K$ is the restriction of $u$ to $K$. And also I am not getting the definition of the map $\phi$. Due to which I am not also understand the set $A_p$. The definition of the map $u_p$ and $\phi$ is not clear to me. Can you please help me to understand this? Thank you very much for your valuable time . Thanks again.
Given $u \in B(H)$, we can consider the element $pup \in pB(H)p$. Note that $pup(K) \subseteq p(H) = K$ and it makes sense to consider the restriction map $$\phi: pB(H)p \to B(K): u \mapsto u_K$$ so $u_K$ is nothing else than the restriction of $pup$ to the closed subspace $K$. Thus $\phi$ just restricts certain bounded operators $H\to H$ to bounded operators $K \to K$.
Let us now prove that $\phi$ is a $*$-isomorphism. Assume $x \in pB(H)p$ and $\phi(x)= 0$. Then $x\vert_K = 0$. But an element of $pB(H)p$ is also $0$ on $K^\perp$, so $x=0$. Hence, $\phi$ is injective.
To see surjectivity, let $y \in B(K)$. Define a bounded operator $u: H \to H$ by $$u(k + k^\perp) = y(k)$$ where $k \in K, k^\perp \in K^\perp.$ In other words, $u = y \oplus 0_{K^\perp}\in B(H)$. Then $u=pup \in pB(H)p$ and clearly $\phi(u) = u_K = y$ proving surjectivity.
That $\phi$ preserves the $*$-algebra structure is trivial.