Is $U^A\otimes_R N\to(U\otimes_R N)^A$ always monic?

Question

Let $R$ be a ring with identity, $U$ a right $R$-module, $N$ a left $R$-module and $A$ an arbitrary index set. My question is:

Is the canonical homomorphism $\tau:U^A\otimes_R N\to(U\otimes_R N)^A$ always monic?

Answer 1

Let $R=\mathbb{Z}$.

Let $U=\mathbb{Q}/\mathbb{Z}$, or any other torsion abelian group with no upper bound on the orders of elements.

Let $N=\mathbb{Q}$.

Let $A$ be any infinite set.

Since $U$ is a torsion group, $U\otimes_RN=0$, and so $(U\otimes_RN)^A=0$.

Since $U^A$ is not a torsion group, $U^A\otimes_RN\neq0$.

So the canonical homomorphism $U^A\otimes_RN\to(U\otimes_RN)^A$ can't be injective.

Edit to address comments:

If $N$ is finitely presented, then it is true that $\tau$ is injective; in fact, it is an isomorphism.

To simplify notation, let $FN=U^A\otimes_RN$ and $GN=(U\otimes_RN)^A$, so $\tau=\tau_N$ is a natural transformation $F\to G$ evaluated at $N$.

Let $\mathcal{C}$ be the class of modules $N$ such that $\tau_N$ is an isomorphism.

Clearly $R$ is in $\mathcal{C}$, and $\mathcal{C}$ is closed under finite direct sums, so finitely generated free modules are in $\mathcal{C}$.

Since both $F$ and $G$ are right exact functors, it is easy to see that if $\alpha: C_1\to C_2$ is a homomorphism between objects of $\mathcal{C}$, then the cokernel of $\alpha$ is in $\mathcal{C}$. Thus all finitely presented modules are in $\mathcal{C}$.

However, in general $\tau_N$ may not be injective if $N$ is merely finitely generated. Of course, if $R$ is left Noetherian this is not an issue, since then every finitely generated left $R$-module is finitely presented.

Let $k$ be a field, and $R=k\oplus V$, where $V$ is an infinite dimensional square-zero ideal, and let $U=R$, so that $GN=N^A$ and $G$ is an exact functor. Let $A$ be any infinite set.

Then $k=R/V$ is a finitely generated, but not finitely presented, $R$-module, and I claim that $\tau_k$ is not injective.

Consider the commutative diagram $$ \begin{array}{ccccccccc} &&FV&\to&FR&\to&Fk&\to&0\\ &&\downarrow&&\downarrow&&\downarrow&&\\ 0&\to&GV&\to&GR&\to&Gk&\to&0 \end{array} $$ with exact rows derived from the short exact sequence $0\to V\to R\to k\to0$. The middle vertical map $\tau_R$ is an isomorphism, and a simple diagram chase shows that is $\tau_k$ is injective, then $\tau_V$ must be surjective.

But the image of $\tau_V$ only contains elements of $GV=V^A$ where the components are all contained in some finite dimensional subspace of $V$.