Silly Question about tensor products and universal property I'm struggling with the intuitive idea of tensor products.
Acctually I don't want to know the construction of tensor product, I want to know if my intuition is right.
Lets talk, without loss of generality, about bilinear maps.
Consider,then, a bilinear map $B$:
$$B: V\times W \to Z \\ (v,w) \mapsto B(v,w)$$
Consider,now, a vector space $\mathfrak{T}$ and another bilinear map, $T$:
$$T: V\times W \to \mathfrak{T} \\ (v,w) \mapsto T(v,w)$$
Such that, this bilinear map,$T$,satisfies two propreties:
1)Is just Bilinear
2)Is a universal map ,in the sense:
$$\hat{B} \circ T = B$$
Then, we can conclude (even without the diagram) that:
$$\hat{B}: \mathfrak{T} \to Z \\ T(v,w) \mapsto \hat{B}[T(v,w)]$$
So, in some (acctually quite good) sense we can say that the operation $\hat{B}$ allows you to translate multilinear maps into linear ones, via $T$.
Evenmore (and here lies my doubts), $\hat{B}$ is a linear function from a space $\mathfrak{T}$,which carries the "linearized" information from the map $T$, and then transports this information in linear fashion into and onto Z. B also do the same but in a bilinear way. 
Is it correct?
 A: I am not sure if I understood the last part completely, but I will try to help you  with your doubts.
First, let clarify what do we mean by universal property. Let us assume that the biliner map $T: V\times W \to \mathfrak{T}$ satisifies universal property. This means that for any vector space $Z$ and any bilinear map $B:V\times W\to Z$ there is a unique linear map $\hat{B}:\mathfrak{T}\to Z$ such that $\hat{B}\circ T = B.$
Let us read the above universal property in a little different way. Let us fix a vector space $Z$ and consider the linear map $\alpha_Z:L(\mathfrak{T}, Z) \to B(V\times W, Z)$ by formula
$$\alpha_Z(F) = F\circ T.$$
Important. Convince yourselt that 
$$T \text{ satisfy universal property} \iff \text{for all } Z \text{ the map } \alpha_Z \text{ is a bijection}$$
So for every $Z$ we have the inverse of $\alpha_Z$, which we may denote by $\beta_Z:B(V\times W, Z)\to L(T, Z)$. Thus, $\beta_Z(B) = \hat{B}$, where $\hat{B}$ is precisely the same $\hat{B}$ as the one in your question.
Conclusion. All informations about bilinear map $B$ are transfered to the linear one via $\beta_Z$ and this transfer of informations can be recover by applying $\alpha_Z$ (Notice that I write here about informations about maps not spaces). Since $\alpha_Z$ is a bijection, we can say more. Any bilinear map $B$ is of the form $\alpha_Z(F)$ for some linear map $F$.
