In the book of Kadison-Ringrose, vol 1 the authors have defined the Hilbert Tensor Product of two (complex) Hilbert spaces $\mathcal{H}_1$ and $\mathcal{H}_2$ to be a Hilbert space $\mathcal{H}_1 \otimes\mathcal{H}_2$ together with a weak Hilbert-Schmidt mapping $p:\mathcal{H}_1 \times\mathcal{H}_2\to\mathcal{H}_1 \otimes\mathcal{H}_2$ satisfying the universal property: any weak Hilbert-Schmidt mapping $L:\mathcal{H}_1 \times\mathcal{H}_2 \to \mathcal{K}$ must factor through a unique bounded linear operator from $T:\mathcal{H}_1 \otimes\mathcal{H}_2\to \mathcal{K}$, i.e., $L=Tp$. (Existence and uniqueness of $(\mathcal{H}_1 \otimes\mathcal{H}_2, p)$ is the content of Theorem 2.6.4)
After this, in Proposition 2.6.6, they established the relationship between this Hilbert tensor product, $\mathcal{H}_1 \otimes\mathcal{H}_2$, and the algebraic tensor product, $\mathcal{H}_1 \odot\mathcal{H}_2$. The latter is defined as the tensor product of $\Bbb{C}$-modules. More precisely they concluded, from Proposition 2.6.6, the following:
(Remark 2.6.7). The first part of Proposition 2.6.6 asserts, in effect, that the only finite families of simple tensors that have sum zero are those that are "forced" to have zero sum by the bilinearity of the mapping $p:(x,y)\to x\otimes y$. From this the subspace, $\mathcal{H}_0:=\{\sum_{j=1}^n x_j\otimes y_j:x_j \in \mathcal{H}_1, y_j\in \mathcal{H}_2\}$ of $\mathcal{H}_1\otimes\mathcal{H}_2$, can be identified with $\mathcal{H}_1\odot\mathcal{H}_2$...
I cannot understand what did they mean by the first line of this remark. I mean what are those simple tensors which have zero sum by the bilinearity of $p$?
Also, how does that imply that $\mathcal{H}_0$ can be identified with $\mathcal{H}_1\odot\mathcal{H}_2$?
For the second question I tried to proceed as follows: First let $i:\mathcal{H}_1\times\mathcal{H}_2\to \mathcal{H}_1\odot\mathcal{H}_2$ be the canonical map $i(a, b):=a\odot b$. Then since $p:\mathcal{H}_1\times\mathcal{H}_2\to\mathcal{H}_1\otimes\mathcal{H}_2$ is weak H-S map so in particular it is bilinear (see Definition 2.6.3) map between the $\Bbb{C}$-modules $p:\mathcal{H}_1\times\mathcal{H}_2\to\mathcal{H}_0$; by universal property of $\mathcal{H}_1\odot\mathcal{H}_2$ we get a linear map $T:\mathcal{H}_1\odot\mathcal{H}_2\to \mathcal{H}_0$ satisfying $p = Ti$, i.e., $x\otimes y = T(x\odot y)$ for $x \in \mathcal{H}_1$ and $y \in \mathcal{H}_2$. Now it is enough to prove the injectivity of $T$ , which follows from Proposition 2.6.6, for $x\otimes y=0$ then there is a $c\in \Bbb{C}$ such that $cx=0;cy=y$ which impies $c=1; x=0$ but then $x\odot y=0$. More generally, if $T(\sum_{j =1}^nx_j\odot y_j)=0$ i.e., $\sum_{j =1}^nx_j\otimes y_j=0$ which implies, by Proposition 2.6.6, $ \forall k, \sum_jc_{jk}x_j=0$ and $\forall j, \sum_k c_{jk}y_k=y_j$ from which it is easy to see that $\sum_{j =1}^nx_j\odot y_j=0$.
Is my argument right? Thanks