Abelian categories and bimodules, chapter 6: tensor products

Proof.

Form the free abelian group \(F=\mathbb Z^{(B\times M)}\) from Theorem 5.8, with generator \([b,m]\) for each pair. Let \(L\subseteq F\) be the subgroup generated by all elements

The generated subgroup consists of finite integer linear combinations of these relations: such combinations form a subgroup containing the relations and lie in every subgroup containing them. Set \(B\otimes_R M=F/L\), and write \(b\otimes m\) for the class of \([b,m]\). The quotient laws in Chapter 5 make this an abelian group. The imposed relations give biadditivity and balancing. They also imply \(0\otimes m=b\otimes0=0\), \((-b)\otimes m=-(b\otimes m)\), and \(b\otimes(-m)=-(b\otimes m)\). Therefore the integer coefficients of a finite sum of generator classes can be absorbed into their first factors, proving the finite-sum assertion.

A function \(\beta:B\times M\to A\) extends uniquely to a homomorphism \(F\to A\), by the free abelian group property. It kills \(L\) precisely when it satisfies the three displayed types of relation. The quotient property then gives a unique homomorphism \(B\otimes_R M\to A\). This proves the bijection. Given any other object with this property, applying the property in both directions produces maps preserving the defining pairs. Their composites preserve every pure tensor, and hence every finite sum, so both composites are identities.

Proof.

For fixed \(s\in S\), the map \((b,m)\mapsto(sb)\otimes m\) is biadditive. It is balanced since \(s(br)=(sb)r\), and \(((sb)r)\otimes m=(sb)\otimes rm\). The universal property gives an additive operator on the tensor product. Checking on pure tensors shows \(1x=x\), \((ss')x=s(s'x)\), and \((s+s')x=sx+s'x\). Each identity holds on finite sums by additivity, so these are all left-module laws.

The formula involving \(u,f\) is biadditive and balanced because \(u(br)=u(b)r\) and \(f(rm)=rf(m)\). It therefore gives an additive map, and \(u(sb)=su(b)\) makes it \(S\)-linear. Identity maps and composites have the required formulas on pure tensors, so they agree everywhere. The same calculation gives \((u+v)\otimes f=(u\otimes f)+(v\otimes f)\) and \(u\otimes(f+g)=(u\otimes f)+(u\otimes g)\), proving separate additivity. Finally, if \(M\) is also an \(R\)-\(T\) bimodule, \( (b\otimes m)t=b\otimes(mt)\) is a well-defined right \(T\)-action by the same balanced-map check; it commutes with the left \(S\)-action on pure tensors and hence everywhere.

Proof.

The two formulas are biadditive and balanced by the module associativity laws, so define additive maps. They preserve their stated actions by associativity and the bimodule compatibility law. Their inverses are \(m\mapsto1\otimes m\) and \(b\mapsto b\otimes1\). The first is \(R\)-linear because \(1\otimes rm=r\otimes m=r(1\otimes m)\); the second preserves both actions because \(sb\otimes1=s(b\otimes1)\) and \(br\otimes1=b\otimes r=(b\otimes1)r\). The composites are identities: \(1m=m\), \(b1=b\), \(1\otimes rm=r\otimes m\), and \(br\otimes1=b\otimes r\). These checks on pure tensors extend to finite sums. For maps of either factor the formulas commute directly with those maps, proving naturality.

Proof.

The bimodule actions follow from Proposition 6.3. To justify the displayed map, fix \(m\). The function \((c,b)\mapsto c\otimes(b\otimes m)\) is biadditive and \(S\)-balanced, since

It extends to an additive function of \(x\in C\otimes_S B\). Its dependence on \(m\) is additive: both sides of the additivity identity agree on every \(x=c\otimes b\), hence on every \(x\). It is \(R\)-balanced in \(x,m\), since \((br)\otimes m=b\otimes rm\), again first on generators and then on sums. A second application of the universal property gives the desired map.

For its inverse, fix \(c\) and use \((b,m)\mapsto(c\otimes b)\otimes m\). This is biadditive and \(R\)-balanced because \((c\otimes br)\otimes m=(c\otimes b)\otimes rm\). It extends to \(B\otimes_R M\), additively in \(c\), and is \(S\)-balanced because \(cs\otimes b=c\otimes sb\). The universal property supplies the inverse map. Both maps preserve the \(T\)-action by their formulas. Every element of either iterated tensor product is a finite sum of the displayed triple tensors, so the formulas prove their composites are identities. They commute with maps of \(C,B,M\) on those same generators, giving naturality.

Proof.

The structural maps \(1_B\otimes\iota_i\) combine to the displayed map by the coproduct property. Define its inverse on pairs by \((b,(m_i))\mapsto\sum_i j_i(b\otimes m_i)\), where \(j_i\) includes the \(i\)-th tensor factor into the direct sum. The finite support of \((m_i)\) makes this sum finite. The function is biadditive and \(R\)-balanced coordinatewise, so extends to the tensor product; it is \(S\)-linear by its formula. One composite fixes each \(j_i(b\otimes m_i)\); the other fixes \(b\otimes(m_i)\), since \((m_i)=\sum_i\iota_i(m_i)\). These elements generate the respective modules, proving both identity equations. The construction also covers the empty family, giving \(B\otimes_R0=0\).

Proof.

By Proposition 5.5, we may identify \(Q\) with \(N/\operatorname{im}f\). Put \(D=(B\otimes_R N)/\operatorname{im}(1\otimes f)\), a left \(S\)-module. The map

is well defined: replacing \(n\) by \(n+f(m)\) changes its value by the class of \(b\otimes f(m)\), which is zero in \(D\). It is biadditive and \(R\)-balanced by the tensor relations, so gives an \(S\)-linear map \(B\otimes_R Q\to D\). Conversely, \(1\otimes q\) kills \(\operatorname{im}(1\otimes f)\), hence descends to an \(S\)-linear map \(D\to B\otimes_R Q\). Their formulas are inverse on the generating tensors, so they are inverse everywhere. Under this isomorphism \(1\otimes q\) is the quotient projection onto \(D\). Its kernel is therefore \(\operatorname{im}(1\otimes f)\) and it is surjective, proving exactness and the cokernel property. Any categorical cokernel is uniquely isomorphic to this quotient, so the conclusion applies without choosing a particular quotient model.

Proof.

The subgroup \(2M\) is closed under subtraction since \(2m-2n=2(m-n)\). In the quotient action, changing \(a\) by an even integer changes \(am\) by an element of \(2M\), and changing \(m\) by \(2n\) changes \(am\) by \(2an\). Thus the action is well defined, and its module laws descend from the integer action.

The displayed function on pairs is independent of the integer representative of \(\bar a\), is biadditive, and is balanced over \(\mathbb Z\) because \((ar)m=a(rm)\). It induces an \(\mathbb F_2\)-linear map from the tensor product. Define an inverse by \(m+2M\mapsto\bar1\otimes m\). This is well defined since \(\bar1\otimes2n=(\bar1\cdot2)\otimes n=0\); it is additive and \(\mathbb F_2\)-linear by balancing. One composite fixes every class \(m+2M\); the other takes \(\bar a\otimes m\) to \(\bar1\otimes am=\bar a\otimes m\). Pure tensors generate, so these are inverse maps. Every homomorphism \(f:M\to N\) satisfies \(f(2M)\subseteq2N\), and both maps commute with its induced quotient and tensor maps by their formulas, proving naturality.