$\operatorname{Tor}^{\mathbb{Z}}_1(-,-)$ on finite abelian groups is not right exact?

Question

In his answer here Martin Brandenburg claims that the Tor functor $\operatorname{Tor}^{\mathbb{Z}}_1(-,-)$ in the category of finite abelian groups is not right exact in neither argument. Since Tor is symmetric it is enough to show this for one argument. I am struggling to continue from there. In particular I am unsure how Tor acts on morphisms. Why is the functor not right exact? Ideally someone could give an epimorphism that is not preserved under application of Tor. Is there also a higher level argument to see this quickly?

Answer 1

Let's try to see carefully what the functor $\text{Tor}_1(\mathbb{Z}/n\mathbb{Z}, -)$ does to morphisms. Given an abelian group $A$ we first tensor $A$ with the resolution $\mathbb{Z} \xrightarrow{n} \mathbb{Z}$ of $\mathbb{Z}/n\mathbb{Z}$, then take $H_1$ of the result, which is

$$\text{Tor}_1(\mathbb{Z}/n\mathbb{Z}, A) \cong \text{ker}(A \xrightarrow{n} A)$$

(whereas $H_0$, the ordinary tensor product, would be the cokernel). Now we can see, for example, that the epimorphism $\mathbb{Z}/4\mathbb{Z} \twoheadrightarrow \mathbb{Z}/2\mathbb{Z}$ is not preserved by the application of $\text{Tor}_1(\mathbb{Z}/2\mathbb{Z}, -)$ since this morphism is trivial on the kernel of multiplication by $2$.

Pushing this argument a bit further I believe it should be possible to show that if $A, B$ are two finite abelian groups then there is a natural isomorphism of functors

$$\text{Tor}_1(A, B) \cong (A^{\vee} \otimes B^{\vee})^{\vee}$$

where $(-)^{\vee}$ is the Pontryagin dual. Every finite abelian group is Pontryagin self-dual which is one way to see why $\text{Tor}_1$ happens to agree with the tensor product on isomorphism classes of objects. But this makes it clearer that one shouldn't expect Tor to be right exact, because the inner Pontryagin dual converts finite colimits in each variable to finite limits before tensoring them, and tensor product is not left exact.