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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.

Andy
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    You actually used "bilinear map" as well, so you have to be a bit careful. You can replace "bilinear map $V\times W\to V\otimes W$" by making it a linear map from $V$ to the vector space $L(W,V\otimes W)$ of linear maps from $W$ to $V\otimes W$, though. – Arturo Magidin Oct 26 '11 at 17:06
  • This link might be useful: Category of vector spaces at wikipedia.\I am not sure, whether you meant that this should be a coproduct in Vec. Wikipedia says that: "In category theory, the coproduct, or categorical sum, is the category-theoretic construction which includes the disjoint union of sets and of topological spaces, the free product of groups, and the direct sum of modules and vector spaces." – Martin Sleziak Oct 26 '11 at 17:08
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    How do you construct a canonical linear map $V\to V\otimes W$? – hmakholm left over Monica Oct 26 '11 at 17:17
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    @Andy: It is a colimit (in fact, a coproduct) in the category of commutative rings. But it is not a colimit in the category of vector spaces or modules. Rather, it is obtained as the left adjoint of the internal hom functor. – Zhen Lin Oct 26 '11 at 17:55
  • @HenningMakholm: that's my main problem with my train of thoughts; basically, why I'm asking the question. For your first comment, I don't know, but I see a co-cone on a two-object diagram... – Andy Oct 26 '11 at 17:55
  • @ArturoMagidin: ok, but I don't see how that could help me – Andy Oct 26 '11 at 17:55
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    @Andy: I was commenting on your statement that you were using nothing except vector spaces and linear maps; in fact, you were using bilinear maps among other things, so there was something else being used. – Arturo Magidin Oct 26 '11 at 18:13
  • @Arturo: oh, ok :) – Andy Oct 26 '11 at 18:22

2 Answers2

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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.

Mariano Suárez-Álvarez
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  • This is a nice answer; I still don't have enough knowledge to go about doing something like this (I'm still at the beginning of both my classes). Hopefully I'll understand more later on. – Andy Oct 26 '11 at 18:36
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    Coming back to this question of mine with a greater understanding, I now see that this is the actual answer to my question, and thus have changed the accepted answer to this one. – Andy Apr 17 '20 at 09:08
  • In other words, $M\otimes_R N$ is the coequalizer of the two morphisms $\rho\otimes \operatorname{id}_N$ and $\operatorname{id}_M \otimes \lambda$, where $\rho:M\otimes_k R \rightarrow M$ and $\lambda:R\otimes_k N \rightarrow N$ are the module multiplication morphisms. – Max Demirdilek May 18 '23 at 13:28
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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:

  1. It is a linear map (exercise).
  2. $(f,g)\circ i_X = f$ and $(f,g)\circ i_Y = g$.
  3. 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. :-)

Agustí Roig
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