I've posted two other questions* $[1]$ $[2]$, discussing and asking about the Tensor Product construction, in particular the "canonical construction", which is the one $[4]$:
$$ \frac{F(V \times W)}{S} := V\otimes W. \tag{1}$$
Now, due to isomorphism $L$ induced by the machinery of tensor product we can say then that the vector space $Z$ and $V\otimes W$ are isomorphic. Therefore, we can say that elements of $Z$ are tensors,i.e. $(V\otimes W, \otimes) \approx (Z, b)$:
$$\begin{array}{rcl} V \times W & \overset{\otimes}{\to} & V \otimes W \\ b & \searrow & \downarrow L \\ \, & \, & Z \end{array} Figure \hspace{1mm}1$$
My doubt is: Since $(V\otimes W, \otimes) \approx (Z, b)$, can I construct another commutative diagram like Figure 2, and say that the pair $(Z, b)$ is the tensor product for another bilinear map $\phi$?
$$\begin{array}{rcl} V \times W & \overset{b}{\to} & Z \\ \phi & \searrow & \downarrow L' \\ \, & \, & \mathbb{R} \end{array} Figure \hspace{1mm}2$$
$$ --- \circ---$$
$*$ I'm not referring my own questions aiming more answers, it's just to show self-study progression.
$[1]$ Doubt on the understanding of the role of Quotient Spaces on Tensor Product construction
$[3]$ Understanding the Details of the Construction of the Tensor Product
$[4]$ ROMAN.S; Advanced Linear Algebra
$[5]$ CONRAD.K; Tensor Products I. https://kconrad.math.uconn.edu/blurbs/linmultialg/tensorprod.pdf
Note that we cannot state that $b = \langle \cdot , \cdot \rangle$ because $Z = L(V^{} \times W^{},\mathbb{R})$. Conversely, we cannot say that $V\otimesW = L(V^{} \times W^{},\mathbb{R}) $ in the first place, because we need to construct the tensor product.
– BasicMathGuy Apr 19 '20 at 19:46