How do I prove if the following functor is exact?

Question

I have the following exercise:

Prove if the functor that sends an abelian group to it's $n$-torsion subgroup for $n\geq 2$ is exact.

I know that I need to take $f\colon M\to N$ and $g\colon N\to L$ of $R$-Modules in $\textrm{Ab}$ which is exact i.e. $\operatorname{im}(f)=\ker(g)$ then I need to show that for the functor $F$ also $\operatorname{im}(F(f))=\ker(F(g))$ holds. Then the functor is exact.

But somehow I first don't see how this functor works so it sends an abelian group $A$ to the set $\{g\in A:g^n=e\}$ so all the elements of order $n$.

Could someone maybe explain this a bit to me?

THanks for your help

Answer 1

With Abelian groups one usually applies additive notation.

Yes, $F(A)=\{a\in A:\,n\cdot a=0\}$.
Then, $F$ sends a homomorphism $f:A\to B$ to its restriction to $F(A)$, i e. $$F(f):=f|_{F(A)}:F(A)\to F(B)$$ as it's clear that $f(a)\in F(B)$ whenever $a\in F(A)$.

Then, if ${\rm im}(f)=\ker g$ for a $g:B\to C$, then on one hand we have $g\circ f=0$, and on the other hand, $g(b)=0\implies\exists a\in A:\,f(a)=b$.
Can you finish from here?

Answer 2

Let $f:\mathbb Z\to\mathbb Z$ be the function $n\mapsto2n$. Then the short exact sequence $$ 0\to\mathbb Z\overset{f}\to\mathbb Z\to\mathbb Z/2\to0 $$ is sent by the "torsion part" functor to $$ 0\to0\to0\to\mathbb Z/2\to0, $$ which is not exact at $\mathbb Z/2$.