Let $L(H)$ the bounded linear operators on a hilbert space $H$. I proved that the inclusion $$i:B(\mathbb{C}^2)\otimes B(\mathbb{C}^2)\hookrightarrow B(\mathbb{C}^4)$$ is not surjective: take $p=\begin{pmatrix} 1 & 0 & 0 &0 \\ 0 & 1 & 0 &0 \\ 0 & 0 & 1 &0 \\ 0 & 0 & 0 &0 \end{pmatrix}\in M_4(\mathbb{C})\cong B(\mathbb{C}^4)$, then it is'nt possible to write $p=a\otimes b$ with $a,b\in B(\mathbb{C}^2)$. My question is now: Is $p$ a strong operator topology limit of a sequence $(a_n\otimes b_n)_{n\in\mathbb{N}}\subseteq B(\mathbb{C}^2)\otimes B(\mathbb{C}^2)$? I.e. is there a sequence $(a_n\otimes b_n)_{n\in\mathbb{N}}\subseteq B(\mathbb{C}^2)\otimes B(\mathbb{C}^2)$ such that $\|(a_n\otimes b_n -p)x\|_{\mathbb{C}^4}\to 0,\; n\to\infty$ for all $x\in \mathbb{C}^4$?
I think the answer is yes. I tired to find matrices $a$ and $b$ such that $\begin{bmatrix} a_{1,1} & a_{1,2} \\ a_{2,1} & a_{2,2} \\ \end{bmatrix} \otimes \begin{bmatrix} b_{1,1} & b_{1,2} \\ b_{2,1} & b_{2,2} \\ \end{bmatrix} = \begin{bmatrix} a_{1,1} b_{1,1} & a_{1,1} b_{1,2} & a_{1,2} b_{1,1} & a_{1,2} b_{1,2} \\ a_{1,1} b_{2,1} & a_{1,1} b_{2,2} & a_{1,2} b_{2,1} & a_{1,2} b_{2,2} \\ a_{2,1} b_{1,1} & a_{2,1} b_{1,2} & a_{2,2} b_{1,1} & a_{2,2} b_{1,2} \\ a_{2,1} b_{2,1} & a_{2,1} b_{2,2} & a_{2,2} b_{2,1} & a_{2,2} b_{2,2} \\ \end{bmatrix}$ converges to $p$ in the strong operator topology. I don't have found suitable matrices not yet. Do you have an idea?
The inclusion $i$ is surjective, because both $B(\mathbb C^2)\otimes B(\mathbb C^2)$ and $B(\mathbb C^4)$ have the same dimension (concretely, 16).
The issue, and what you showed, is that if you restrict $i$ to the subset of elementary tensors, then $i$ is not surjective.
Your element $p$ is not a sot limit of elementary tensors. For starters, because we are in a finite-dimensional setting, the sot topology agrees with the norm topology. And since a norm-limit of elementary tensors is again an elementary tensor, elements that are not elementary tensors cannot be realized as limits of elementary tensors.