Tensor products, existence of a unique linear map Question:
Given a bilinear map $B:  V\times W\to X $, show there exists a unique linear map          $T:V \otimes W\to X $ s.t. $B= T \circ \phi$
Background:
We define $V \otimes W $ by F[ $V\times W]/Y$, where $Y$ is chosen to make a  bilinear $\phi:$ F $[V\times W]  \to $ F[ $V\times W]/Y$, defined by $\phi((v,w))= [(v,w)]=v\otimes w$.
Lemma (haven't been proved):  If $S : V\to X$ is a linear map, $\psi: v \mapsto [v] $, and $W$ is a subspace of $ V$ s.t. $W \subset null\    S $, then $ \exists !$ linear map $Q: V/W \to X $, s.t. $ S= Q  \circ$ $  \psi $, and Q is given by  $ Q[v]= S(v).$
Outline:
Note there is a linear map $U: (v,w) \mapsto B(v,w)$, $Y \subset null \ U$, then there exists a unique linear map $T$ s.t.  $B= T \circ \phi$ $T$ is given by $T(v \otimes w) =B(v,w).$
What confuse me:
Firstly, not all elements in  $V \otimes W $ can be written in the form $v\otimes w$, namely, pure tensor. Then how is it possible that we define a linear T by  $T(v \otimes w) =B(v,w)$? I've seen some proof says that since we define on a spanning set ({$v\otimes w$} spans $V\otimes W$), then  we can extend it by linearity. But I still doubt the logic here: Shouldn't we prove $T$ is well-defined first before we prove $T$ is linear? How can we claim $T$ is linear without a well-defined $T$?
I also find it quite weird to prove the uniqueness in the lemma.  Is it sufficient to do the following:
Assume a $ Q'\neq Q$ then there exists a $[v_0]$ s.t. $Q'([v_0])\neq Q([v_0])= S(v_0)$, thus  fails $Q'$.
Finally, back to the fact that not all element in  $V \otimes W $ can be written in the form $v\otimes w$. Given that $V \otimes W $ is defined to be $[(v,w)]\in$ F[ $V\times W]/Y$, how is this true ? Is it generally true that not all elements in $V/W$ can be written as [v]?
Thanks for help!
 A: Rotman says "The safest scheme is to rely on the universal mapping problem". This universal property says that for any bilinear map $B:V \times W \to X$, there exist a unique linear map $T : V \otimes W \to A$ such that $B = T \circ \phi$, that is 
$$ B(x,y) = T \circ \phi (x,y) = T\big( \phi(x,y) \big) = T(x\otimes y) $$
That is why you "define" T on the elements $x \otimes y$.
Now suppose that $V$ has basis $\{v,w\}$, then $\{ v\otimes v, v\otimes w, w\otimes v , w \otimes w \}$ is a basis for $V \otimes_k V$. Suppose that the element $v \otimes w + w \otimes v$ has a expression of the form $x\otimes y$. Write $x$ and $y$ as linear combinations of elements of the basis of $V$, $x=av+bw$ and $y=cv+dw$. Then
$$ x\otimes y = (av+bw)\otimes (cv+dw) = (ac) v\otimes v + (ad) v \otimes w + (bc) w \otimes v + (bd) w\otimes w $$
but $x \otimes y = v \otimes w + w \otimes v$, so by comparing coefficients we have that $ac=bd=0$ and $ad=bc=1$, then $a=0$ or $c=0$, and $1=0$, a contradiction.
Rotman's book "An introduction to homological algebra" is a good book to learn about tensor products.
A: The vector space $\mathbf{F}[V\times W]$ has $V\times W$ as its basis, by definition. You can define a linear map $f\colon\mathbf{F}[V\times W]\to X$ by specifying the action on the elements of a basis, so we define
$$
f((v,w))=B(v,w)
$$
which is no problem. Now, the kernel (null space) of $f$ contains $Y$, because
$$
f((v_1+v_2,w)-(v_1,w)-(v_2,w))=
f((v_1+v_2,w))-f((v_1,w))-f((v_2,w))=
B(v_1+v_2,w)-B(v_1,w)-B(v_2,w)=0
$$
and similarly for the other elements defined to span $Y$. By the lemma, $f$ defines a unique linear map
$$
T\colon\mathbf{F}[V\times W]/Y\to X
$$
by
$$
T([x])=f(x)
$$
where $[x]=x+Y$ denotes the element in $\mathbf{F}[V\times W]/Y$ corresponding to $x$ (the canonical projection, in other words).
We define
$$v\otimes w=[(v,w)]$$
so we have 
$$
T(v\otimes w)=T([(v,w)])=f((v,w))=B(v,w)
$$
by the definition themselves.
The uniqueness of $T$ follows from the fact that $V\times W$ is a spanning set of $\mathbf{F}[V\times W]$ and so its image under the canonical projection is a spanning set of $\mathbf{F}[V\times W]/Y=V\otimes W$. Since we know what the linear map $T$ does on a spanning set, we know it's unique.
