Suppose that $h:X\times Y\to \mathbb{F}$ is a bilinear map. Then prove that there exists a linear map $h_\otimes:X\otimes Y\to\mathbb{F}$ such that $h(x,y)=h_\otimes(x,y)$.
Logically, to every element $(x,y)\in X\times Y$ there is an element $x\otimes y\in X\otimes Y$, therefore we can say that there can be a one-to-one correspondence between both maps.
But how do I prove that $h_\otimes$ is a linear map ?
Let, $h:X\times Y\to\mathbb{F}$ such that $h(x,y)=f(x)g(y)$ where $f:X\to\mathbb{F}$ and $g:Y\to\mathbb{F}$
Note : I only have understanding of the linear algebra basics.
Best Answer
To understand why this is true, you need to understand the precise definition of the vector space $X \otimes Y$. I asked if you could give your own definition of $X \otimes Y$, but since you did not, you will have to be satisfied with the definition I give.
To understand the definition of $X \otimes Y$, you need to understand the idea of a free vector space and the idea of a quotient vector space. I will assume you understand what these are.
Let $\mathbf V$ be the free vector space on the set $X \times Y$. By definition, the elements of $\mathbf V$ are unique linear combinations of the form
$$\sum\limits_{(x,y) \in X \times Y} c_{(x,y)}(x,y)$$
where $c_{(x,y)} \in \mathbb F$ are scalars, almost all of which are zero.
Let $\mathbf W$ be the subspace of $\mathbf V$ spanned by the set of all elements of the form
$$(x,y) + (x,y') - (x,y+y')$$
$$(x,y)+(x',y) - (x+x',y)$$
$$\lambda (x,y) - (\lambda x,y)$$
$$(\lambda x,y) - (x,\lambda y)$$ for $x,x' \in X, y,y' \in Y$, and $\lambda \in \mathbb F$. We define the tensor product $X \otimes Y$ to be the quotient space $\mathbf V/\mathbf W$. By definition, the elements of $X \otimes Y$ are cosets
$$\sum\limits_{(x,y) \in X \times Y} c_{(x,y)}(x,y) + \mathbf W.$$
As a matter of notation, we write
$$ x \otimes y$$
instead of the coset $(x,y) + W$. It follows that every element of $\mathbf V$ can be written (nonuniquely) as a sum
$$\sum\limits_{i=1}^n x_i \otimes y_i$$
for some $x_i \in X, y_i \in Y$, and that in the vector space $X \otimes Y$ we have the relations
$$x\otimes (y+y') = x\otimes y + x \otimes y'$$
$$(x+x')\otimes y = x \otimes y + x' \otimes y$$
$$\lambda (x \otimes y) = (\lambda x) \otimes y = x \otimes(\lambda y)$$.
Now, to prove:
Proposition: for every vector space $Z$, and every bilinear map $h: X \times Y \rightarrow Z$, there is a unique linear map $h_{\otimes}: X \otimes Y \rightarrow Z$ with the property that
$$h_{\otimes}(x \otimes y) = h(x,y)$$
for every $x \in X, y \in Y$.
Proof: Uniqueness is clear, since any linear map on $X \otimes Y$ is completely determined by what it does to elements of the form $x \otimes y$. Define a function $$T: \mathbf V \rightarrow Z$$
by the formula
$$T\Bigg( \sum\limits_{(x,y) \in X \times Y} c_{(x,y)} (x,y) \Bigg) = \sum\limits_{(x,y) \in X \times Y} c_{(x,y)} h(x,y).$$
It is not difficult to check that $T$ is a linear map. Also, since $h$ is bilinear, it is not difficult to check that $T$ sends every element of $\mathbf W$ to the zero element of $Z$. Therefore, by simple facts about quotient vector spaces, $T$ descends to a linear map $h_{\otimes}$ on the quotient vector space $X \otimes Y = \mathbf V/\mathbf W$ and does what is required.