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For $A$ and $B$ two abelian groups, their tensor product $A \otimes B$ is a new abelian group such that a group homomorphism $A \otimes B \to C$ is equivalently a bilinear map out of $A$ and $B$.
Tensor products of abelian groups were defined by Hassler Whitney in 1938.
Let Ab be the collection of abelian groups, regarded as a multicategory whose multimorphisms are the multilinear maps $A_1, \cdots, A_n \to B$.
The tensor product $A, B \mapsto A \otimes B$ in this multicategory is the tensor product of abelian groups.
Equivalently this means explicitly:
For $A, B$ two abelian groups, their tensor product of abelian groups is the abelian group $A \otimes B$ which is the quotient of the free abelian group on the product of their underlying sets $A \times B$ by the relations
$(a_1,b)+(a_2,b)\sim (a_1+a_2,b)$
$(a,b_1)+(a,b_2)\sim (a,b_1+b_2)$
for all $a, a_1, a_2 \in A$ and $b, b_1, b_2 \in B$.
In words: it is the group whose elements are presented by pairs of elements in $A$ and $B$ and such that the group operation for one argument fixed is that of the other group in the other argument.
If one generalises to abelian semigroups, this definition of the tensor product also defines the tensor product of abelian semigroups.
The 0-ary relations $(0,b)\sim 0$ and $(a,0)\sim 0$ follow automatically; one needs them explicitly only if one generalises to abelian monoids.
By definition of the free construction and the quotient there is a canonical function of the underlying sets
(where $U \colon Ab \to Set$ is the forgetful functor).
On elements this sends $(a,b)$ to the equivalence class that it represents under the above equivalence relations.
The following relates the tensor product to bilinear functions. It is a definition or a proposition dependening on whether one takes the notion of bilinear function to be defined before or after that of tensor product of abelian groups.
A function of underlying sets $f : A \times B \to C$ is a bilinear function precisely if it factors by the morphism of through a group homomorphism $\phi : A \otimes B \to C$ out of the tensor product:
Equipped with the tensor product $\otimes$ of def. and the exchange map $\sigma_{A, B}: A\otimes B \to B \otimes A$ generated by $\sigma_{A, B}(a, b) = (b, a)$, Ab becomes a symmetric monoidal category.
The unit object in $(Ab, \otimes)$ is the additive group of integers $\mathbb{Z}$.
To see that $\mathbb{Z}$ is the unit object, consider for any abelian group $A$ the map
which sends for $n \in \mathbb{N} \subset \mathbb{Z}$
Due to the quotient relation defining the tensor product, the element on the left is also equal to
This shows that $A \otimes \mathbb{Z} \to A$ is in fact an isomorphism.
Showing that $\sigma_{A, B}$ is natural in $A, B$ is trivial, so $\sigma$ is a braiding. $\sigma^2$ is identity, so it gives Ab a symmetric monoidal structure.
The tensor product of abelian groups distributes over the direct sum of abelian groups
Let $(A, \cdot)$ be a monoid in $(Ab, \otimes)$. The fact that the multiplication
is bilinear means by the above that for all $a_1, a_2, b \in A$ we have
and
This is precisely the distributivity law of the ring.
For $n \in \mathbb{N}$ positive we write $\mathbb{Z}_n$ for the cyclic group of order $n$, as usual.
For $a,b \in \mathbb{N}$ and positive, we have
where $(-,-)$ denotes the greatest common divisor.
A proof is spelled out for instance as (Conrad, theorem 4.1).
The original definition is due to Hassler Whitney:
An exposition (in the case of vector spaces) is in
and, in the further generality of the tensor product of modules, in
Last revised on May 20, 2021 at 10:18:42. See the history of this page for a list of all contributions to it.