Exact sequence of direct sums of $\mathbb{Q}/\mathbb{Z}$

Question

Suppose $$0 \to (\mathbb{Q}/\mathbb{Z})^m \to (\mathbb{Q}/\mathbb{Z})^{m + n} \to (\mathbb{Q}/\mathbb{Z})^n$$ is a left exact sequence of abelian groups. Is it also right-exact (i.e. is $(\mathbb{Q}/\mathbb{Z})^{m + n} \to (\mathbb{Q}/\mathbb{Z})^n$ surjective?) I want to say that the cokernel is finite and therefore $0$ since $(\mathbb{Q}/\mathbb{Z})^n$ has no finite quotients. But how do you get around the fact that these groups are not finitely generated (in which case it would be easy by just looking at ranks)?

Answer 1

Let $I$ be the image of the map $(\mathbb{Q}/\mathbb{Z})^{m + n} \to (\mathbb{Q}/\mathbb{Z})^n$. Since $\mathbb{Q}/\mathbb{Z}$ is injective, the short exact sequence $$0 \to (\mathbb{Q}/\mathbb{Z})^m \to (\mathbb{Q}/\mathbb{Z})^{m + n} \to I\to 0$$ splits. It follows that for each $k>0$, there are exactly $k^n$ elements $x\in I$ such that $kx=0$ (since $(\mathbb{Q}/\mathbb{Z})^m$ has $k^m$ such elements and $(\mathbb{Q}/\mathbb{Z})^{m + n}\cong (\mathbb{Q}/\mathbb{Z})^m\oplus I$ has $k^{m+n}$ such elements). That is, $I$ contains all $k^n$ of the elements $x\in(\mathbb{Q}/\mathbb{Z})^n$ such that $kx=0$. Since $k$ is arbitrary, this means $I$ is all of $(\mathbb{Q}/\mathbb{Z})^n$.