An exercise about nuclear map Von Neumann algebra There is a quotation below:
Let $M\subset B(H)$ be a von Neumann algebra and $\{P_{i}\}_{i\in L}$ be a net of finite-rank projections which increases to the identity (in the strong operator topology). If $P_{i}$ has rank $k(i)$, then we define contractive completely positive $$\phi_{i}:M\rightarrow M_{k(i)}(\mathbb{C})\cong P_{i}B(H)P_{i}$$ by compression (i.e., $\phi_{i}(T)=P_{i}TP_{i}$) and we let $$\psi_{i}: M_{k(i)}(\mathbb{C})\rightarrow B(H)$$ 
be the natural inclusion maps. Since the predual of $B(H)$ is the trace class operators, a routine exercise shows that the maps $\psi_{i}\circ\phi_{i}$ converge to the identity (on all of $B(H)$, in fact) in the point-ultraweak topology.
I do know the how to do the "so call" routine exercise. Could someone give me some hints or show me more details? 
 A: Given $T_j\in B(H)$, we have by definition that $T_j\to0$ ultraweakly if $f(T_j)\to0$ for every $f$ in $B(H)_*=T(H)$. The duality here is given by $f(T)=\text{Tr}(Tf)$ for each trace-class operator $f$. 
So fix  $T\in B(H)$. We want to show that $P_iTP_i\to T$ ultraweakly. If $R\in B(H)$ is a finite rank operator, we can write $R=\sum_{k,j=1}^nr_{kj}E_{kj}$ for appropriate matrix units. Then
$$
\text{Tr}(R(I-P_i))=\sum_{s=1}^\infty\langle R(I-P_i)e_s,e_s\rangle=\sum_{s=1}^\infty\sum_{k,j=1}^nr_{kj}\langle(I-P_i)e_s,E_{kj}e_s\rangle
=\sum_{k,j=1}^nr_{kj}\langle(I-P_i)e_j,e_k\rangle\to0,
$$
using that $I-P_i\to0$ in the strong operator topology (and thus in the weak operator topology), and that the sum is finite. Using that the finite-rank operators are $\|\cdot\|_1$-dense in the trace-class operators, we get that $\text{Tr}(S(I-P_i))\to0$ for all trace-class operators $S$. So $P_i\to I$ ultraweakly. 
Now, writing $T-P_iTP_i=P_iT(I-P_i)+(I-P_i)TP_i+(I-P_i)T(I-P_i)$ we have, for trace-class $S$, 
$$
|\text{Tr}(S(T-P_iTP_i))|\leq|\text{Tr}(SP_iT(I-P_i))|+|\text{Tr}(S(I-P_i)TP_i)|+|\text{Tr}(S(I-P_i)T(I-P_i))|
$$
For each term we can reason like this (I'll do the first one):
If we write $S=VS_0$ the polar decomposition, with $S_0\geq0$, then $S_0$ is trace-class. So
$$
|\text{Tr}(SP_iT(I-P_i))|=|\text{Tr}(S_0^{1/2}P_iT(I-P_i)VS_0^{1/2})|\\
\leq\text{Tr}(S_0^{1/2}P_iT^*TP_iS_0^{1/2})^{1/2}\,\text{Tr}(S_0^{1/2}V^*(I-P_i)VS_0^{1/2})^{1/2}\\
\leq\|T\|\,\text{Tr}(S_0)^{1/2}\,\text{Tr}((I-P_i)VS_0V^*)^{1/2}\to0,
$$
since $VS_0V^*$ is trace-class.
After working like this for the three terms above, we get
$$
\lim_i\text{Tr}(S(T-P_iTP_i))=0.
$$
As this occurs for any trace-class $S$, we have that $P_iTP_i\to T$ ultraweakly.

Depending on the available background, here is a much easier proof. The ultraweak topology agrees with the weak operator topology on bounded sets. As $\|P_iTP_i\|\leq\|T\|$ for all $i$, we only need to prove that $P_iTP_i\to T$ weakly. So, for $h,k\in H$,
$$
|\langle P_iT(I-P_i)h,k\rangle|=|\langle T(I-P_i)h,P_ik\rangle|\leq\|T\|\,\|(I-P_i)h\|\,\|k\|\to0,
$$
since $P_ih\to h$. A similar reasoning for $(I-P_i)TP_i$ and $(I-P_i)T(I-P_i)$ shows that $P_iTP_i\to T$ weakly. 
