Example for a functor which preserves direct sums but doesn't preserve split exact sequnces.

Question

Is there a functor from the category of abelian gorups to itself such that $F0=0$ on objects and morphisms and $$ F(M\oplus N)\cong F(M)\oplus F(N) $$ for all abelian groups $M$ and $N$, but such that $F$ doesn't preserve split exact sequences, i.e. there exist abelian groups $A$ and $B$ such that for the canonical split exact sequence $$ 0\to A\overset{i}{\to} A\oplus B\overset{p}{\to} B\to 0 $$ the indecued sequence $$ 0\to F(A)\overset{Fi}{\to} F(A\oplus B)\overset{Fp}{\to} F(B)\to 0 $$ isn't exact?

I just can't wrap my head around this. I tried to come up with a counter example with something like $$ FM:= \begin{cases} 0 & \text{if }M\text{ is finite,}\\ \mathbb{Z}^{\oplus M} & \text{else,} \end{cases} $$ and my idea was to put $Ff=0$ when $f$ has finite image and if $f$ has infinite image then $Ff$ is the unique morphism such that $Ff(\delta_m)=\delta_{f(m)}$, where $\delta_m\in \mathbb{Z}^{\oplus M}$ is the function equal to $1$ at $m$ and $0$ everywhere else, but this doesn't work because if two morphisms have infinite image then their composition can nonehteless have finite image (e.g. be $0$).

Answer 1

The original sequence splits, i.e., there is $s\colon B\to A\oplus B$ with $p\circ s=\operatorname{id}_B$. Then $F(s)\colon F(B)\to F(A\oplus B)$ has the property $F(p)\circ F(s)=F(\operatorname{id}_B)=\operatorname{id}_{F(B)}$, i.e., the image sequence splits.

In other words, splitting is preserved under functors, only exactness may be "at risk".

Answer 2

I found now an example.

For $G:\operatorname{Ab}\to\operatorname{Ab}$ the functor which sends $A$ to $A/\operatorname{Tor}(A)$ (and maps to the induced maps), and $H$ the functor which sends $A$ to $\mathbb{Z}^{\oplus A}/\mathbb{Z}\delta_0$, the composed functor $F=H\circ G$ gives a counterexample.

More precisely, $H$ is the functor which sends an ableian group $A$ to $\mathbb{Z}^{\oplus A}/\mathbb{Z}\delta_0$, and which sends a morphism of abelian groups $f:A\to B$ to the unique morphism $Hf:HA\to HB$ defined by $$ Hf:HA \to HB\\ [\delta_a]\mapsto [\delta_{f(a)}], $$ where I used the $[]$-brackets to denote the class of an element of $\mathbb{Z}^{\oplus A}$ in $HA$.

The key thing to notice is that then, if we put $F=H\circ G$, then $F(A\oplus B)$ and $F(A)\oplus F(B)$ are free abelian groups of equal rank. Indeed, we have $$ \operatorname{rank}(F(A\oplus B))=|G(A\oplus B)|-1=|GA|\cdot |GB|-1 $$ where we used that $G$ is additive, and $$ \operatorname{rank}(F(A)\oplus F(B))=(|GA|-1)+(|GB|-1). $$ Now if either $A$ or $B$ is equal to its torsion subgroup, then those two ranks are clearly equal, so suppose that both $A$ and $B$ are not equal to their torsion subgroup. In this case both $GA$ and $GB$ contain an element of infinite order and thus are infinite, so the ranks are again equal by basic cardinal arithmetic. Therefore we conclude that $$ F(A\oplus B)\cong F(A)\oplus F(B) $$ for all abelian groups $A$ and $B$. However, if we pass the split exact sequence $$ 0\to\mathbb{Z}\overset{i}{\to}\mathbb{Z}^2\overset{p}{\to}\mathbb{Z}\to 0, $$ where $i:m\mapsto(m,0)$ and $p:(m,n)\mapsto n$, then applying $F$ is the same as applying just $H$ as all the groups involved are torsion free. But then, the element $[\delta_{(1,0)}-\delta_{(1,1)}]$ will be in the kernel of $Hp$, but not in the image of $Hi$.