I am working on a problem which asks to show that for an additive, covariant functor $F:R-\sf{Mod}\rightarrow S-\sf{Mod}$ the following characterizations are equivalent:
- $F$ reflects zero objects. (If $F(X)$ is a zero object, then $X$ is.)
- $F$ reflects zero morphisms. (If $Ff$ is a zero morphism, then $f$ is.)
- $F$ is reflects exact sequences. ($FA\rightarrow FB\rightarrow FC$ is exact implies $A\rightarrow B\rightarrow C$ is exact.)
I am working on the implication $(2)\implies (3)$ and am trying to prove that reflecting exact sequences is equivalent to reflecting short, exact sequences. To do so, start with a sequence $A\xrightarrow{f} B\xrightarrow{g} C$ such that $FA\xrightarrow{Ff} FB\xrightarrow{Fg} FC$ is exact and want to show that $\text{im}f = \text{ker}g$. Then, exactness of the second sequence implies that $F(gf) = 0$ and because $F$ reflects zero morphisms, $gf = 0$ which shows that $\text{im}f \subseteq \ker g$...and so on.
My question: Is there a way to show that reflecting exact sequences is equivalent to reflecting short, exact sequences without using any of the three conditions listed? The trouble I am having is that without the ability to reflect zero morphisms, I cannot show that $\text{im}f\subseteq\ker g$.