Projective Modules, The Tensor of Two Functions

The Tensor of Two Functions

Let C₁ be the category of pairs of left and right R modules, where pairs of module homomorphisms act as morphisms. Let C₂ be the category of abelian groups, where group homomorphisms act as morphisms. The tensor product induces a functor from C₁ into C₂.

consider a morphism in C₁. specifically, let f₁ map A₁ into B₁, and let f₂ map A₂ into B₂. Build the tensor products S = A₁×A₂, and T = B₁×B₂. Remember that S and T are abelian groups, members of C₂. Thus tensor product is a functor that carries objects in C₁ to objects in C₂. How about the functions?

Build g, from A₁ cross A₂ into T, by applying f₁ and f₂, then the bilinear map from B₁ cross B₂ onto T. Verify that g is a bilinear map. By the universality of S, there is a unique group homomorphism h(S) into T that agrees with g. By definition h is the tensor product of f₁ cross f₂, and it is the corresponding morphism in the category C₂.

Show that the composition of morphisms in C₁ leads to the composition of morphisms in C₂. Either way we have a group homomorphism in C₂ that makes the diagram commute, and since S is universal, the morphisms must agree. Therefore our map is a functor between categories.

If R is commutative, a similar functor carries C₁ into C₃, the category of R modules and R module homomorphisms.

Direct Sum

Let A be the direct sum of left modules A₁ A₂ A₃ etc. The homomorphism that carries A into B defines, and is defined by, the homomorphisms from each A_i into B.

Let another homomorphism carry C into D. What is the induced homomorphism from AC into BD? (I'm using juxtaposition to indicate the tensor product.)

since tensor and direct sum commute, AC is the direct sum of A_iC. Thus any homomorphism from AC defines, and is defined by, component homomorphisms from A_iC into BD. Select an index i, and restrict the domain to A_iC. Let xy be a generating pair of this restricted domain; hence x comes from A_i and y comes from C. These map forward to B and D, and down to BD, and f_i(xy)must agree. This dovetails perfectly with the induced homomorphism from A_iC into BD. In other words, the induced function from AC into BD is the direct sum of the induced functions from A_iC into BD.

Tensoring with the Identity Map

Tensor A and B with a module M, and tensor A → B with the identity map from M onto M. If x in A is mapped to y in B, and w is an element of M, then xw in AM is mapped to yw in BM. This is a useful technique, as we'll see in the next couple of sections.

What happens when M = R? We know that AM is isomorphic to A, and BM is isomorphic to B. This isomorphism is realized by equating x in A with x cross 1 in AM. Since 1 maps to 1, x still maps to y in AM → BM. Tensoring with R doesn't change a thing.

What if M is the direct sum of modules M_i? Let w be an element in the direcct sum. Remember that xw leads to yw. Break w up into its components and xw_i leads to yw_i. this is what we would get if we tensored A → B with M_i. Therefore, tensoring A → B with a direct sum M is the direct sum of the individual tensor products with M_i.

As a corollary, tensor with a freee module and obtain the direct sum of multiple instances of A → B.

What if M is itself a tensor product - say M = S×T? Tensoring is associative, so AM = AST. Let's look at the homomorphisms. Start with x → y from A → B and pair this with v in S. This gives xv → yv. Tensor with T and find xvw → yvw. On the other hand, join x → y with the pair generator vw in M and find xvw → yvw. Tensoring with M is equivalent to tensoring with S, and then with T.