Tensor product as a colimit I've been dealing with category theory for three weeks now and we just started covering limits and colimits, meanwhile in my geometry class we defined the tensor product of vector spaces.
Then I figured that the tensor product must be a colimit in some category: we have a map $V \times W \to V \otimes W$ which is bilinear, then we have two linear maps $V \to V \otimes W$ and $W \to V \otimes W$, plus the "universal" arrow.
Since I didn't use anything but vector spaces and linear maps, I have that the tensor product is a colimit in $\mathbf{Vect}$. 
Did I get anything wrong in my reasoning? 
EDIT To clarify: I have a problem with my constructing the two linear maps from $V$ and $W$ to the tensor product (and as per the comments, I now know it's not going to work), but I would like to know how I can see, just by looking at the tensor product, what category it is a colimit of.
 A: As Zhen Lin told you, the tensor product is NOT a colimit [coproduct] in the category of vector spaces. In this category, the colimit [coproduct] of two vector spaces coincides with its cartesian product: $X\times Y$.
Proof. First, we have morphisms
$$
i_X : X \longrightarrow X\times Y \longleftarrow Y : i_Y
$$
defined by
$$
i_X (x) = (x, 0), \qquad i_Y(y) = (0,y) \ .
$$
As for the universal property of the coproduct, given two linear maps
$$
f : X \longrightarrow Z \qquad \text{and} \qquad g: Y \longrightarrow Z
$$
define the map
$$
(f,g) : X \times Y \longrightarrow Z
$$
as 
$$
(f,g)(x,y) = f(x) + g(y) \ .
$$
This map verifies:


*

*It is a linear map (exercise).

*$(f,g)\circ i_X = f$ and $(f,g)\circ i_Y = g$.

*Given any other linear map $h: X\times Y \longrightarrow Z$ such that $h\circ i_X = f$ and $h\circ i_Y = g$, we necessarily have $h = (f,g)$. Indeed, $h(x,y) = h(x,0) + h(0,y) = f(x) + g(y) = (f,g)(x,y)$.


Instead, as Zhen Lin told you, the tensor product is the colimit [coproduct] in the category of commutative algebras with unit.
Given two commutative algebras with unit $A, B$, you have maps
$$
i_A : A \longrightarrow  A \otimes B \longleftarrow B : i_B \ ,
$$
defined by
$$
i_A (a) = a\otimes 1 , \qquad i_B(b) = 1 \otimes b \ .
$$
And I let you as an exercise to verify what is needed here.  :-)
A: One thing you can do is the following:
Suppose $R$ is a commutative $k$-algebra, with $k$ a field, and that you "know" about tensor products over $k$. If $M$ and $N$ are $R$-modules, there is a short exact sequence $$M\otimes_kR\otimes_k N\xrightarrow{f} M\otimes_k N\to M\otimes_RN\to 0$$ in the category of $k$-vector spaces, with the map $f$ given by $$f(m\otimes r\otimes n)=mr\otimes n-m\otimes rn.$$ In other words, $M\otimes_RN$ is the cokernel of the map $f:M\otimes_kR\otimes_kN\to M\otimes_kN$, and therefore a colimit in this category.
N.B. This idea is useful, in that one can dualize it, and it is this way that one defines the cotensor product of two comodules over a coalgebra.
