Why is $\hom(U, \hom(V, W))$ isomorphic to $\hom(U \otimes V, W)$?

The tensor product is unique up to isomorphism because if $(W, \varphi)$ and $(W', \varphi')$ are two tensor products of $U$ and $V$, there exists a unique linear space isomorphism $\tau: W \to W'$ such that the diagram commutes, meaning $W' = \tau(W)$ and $\varphi' = \tau \varphi$. This shows that any two realizations differ only by a unique isomorphism.

Conditions: $U$ and $V$ are linear spaces over the number field $\mathbb{P}$.; $(W,\varphi)$ is a tensor product of $U$ and $V$.; $(W',\varphi')$ is another tensor product of $U$ and $V$.

The notation $U \otimes V$ denotes the entire tensor product space, which is uniquely determined up to isomorphism. In contrast, $u \otimes v$ denotes a basic tensor element within that space, generated by applying the bilinear map $\varphi$ to vectors $u \in U$ and $v \in V$.

Conditions: $U$ and $V$ are linear spaces over the number field $\mathbb{P}$.; $u$ is in $U$ and $v$ is in $V$.; $\varphi$ is the bilinear map in the tensor product construction.

The universal property of the tensor product states that for linear spaces $U, V, W$ over a field $P$, if there exists a bilinear map $\varphi: U \times V \to W$ such that for any bilinear map $f: U \times V \to P$ (where $P$ is an arbitrary linear space over $P$), there exists a unique linear map $\tau: W \to P$ making the diagram commute (i.e., $f = \tau \varphi$), then the pair $(W, \varphi)$ is called a tensor product of $U$ and $V$.

Conditions: $U, V, W$ are linear spaces over the field $P$.; $\varphi$ is a bilinear map from $U \times V$ to $W$.; $f$ is a bilinear map from $U \times V$ to an arbitrary linear space $P$ over $P$.; $\tau$ is the unique linear map from $W$ to $P$.

The Kronecker product $A \otimes B$ is an $mp \times nq$ block matrix because it replaces each element $a_{ij}$ of the $m \times n$ matrix $A$ with the $p \times q$ matrix $a_{ij}B$. This results in $m$ row blocks and $n$ column blocks, each of size $p \times q$, making the total dimensions $mp \times nq$.

Conditions: $A$ is an $m \times n$ matrix.; $B$ is a $p \times q$ matrix.; $a_{ij}$ is an element of $A$.

The tensor product notation $u \otimes v$ satisfies distributive laws and scalar mobility rules because it is constructed via a bilinear map $\varphi$. Bilinearity implies linearity in each variable separately, leading to $(a_1u_1+a_2u_2)\otimes v=a_1u_1\otimes v+a_2u_2\otimes v$, $u\otimes(b_1v_1+b_2v_2)=b_1u\otimes v_1+b_2u\otimes v_2$, and $a(u\otimes v)=(au)\otimes v=u\otimes(av)$.

Conditions: $u, u_1, u_2$ belong to $U$.; $v, v_1, v_2$ belong to $V$.; $a, a_1, a_2, b_1, b_2$ are scalars in the field $\mathbb{P}$.