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Let $F\colon \mathcal{A} \rightarrow \mathcal{B}$ be a functor of abelian categories. We will define some properties of $F$ before we state a question.

  • Let $X \rightarrow Y \rightarrow Z$ be a sequence in $\mathcal{A}$. We say $F$ reflects exact sequences if $X \rightarrow Y \rightarrow Z$ is exact whenever $F(X) \rightarrow F(Y) \rightarrow F(Z)$ is exact.

  • We say $F$ is faithfully exact if it is exact and reflects exact sequences.

  • Let $X$ be an object of $\mathcal{A}$. We say $F$ reflects zero objects if $X = 0$ whenever $F(X) = 0$.

  • Let $f\colon X \rightarrow Y$ be a morphism of $\mathcal{A}$. We say $F$ reflects zero morphisms if $f = 0$ whenever $F(f) = 0$.

Question. Let $F\colon \mathcal{A} \rightarrow \mathcal{B}$ be an exact functor of abelian categories. Are the following conditions equivalent?

(1) $F$ reflects exact sequences, i.e., $F$ is faithfully exact.

(2) Let $f\colon X \rightarrow Y$ and $g\colon Y \rightarrow Z$ be morphisms of $\mathcal{A}$ such that $gf = 0$. Suppose $F(X) \rightarrow F(Y) \rightarrow F(Z)$ is exact. Then $X \rightarrow Y \rightarrow Z$ is exact.

(3) $F$ reflects zero objects.

(4) $F$ reflects zero morphisms.

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Yes, they are equivalent.

Obviously, (1) implies (2). (2) implies (3) because $Y = 0$ if and only if $$0 \longrightarrow Y \longrightarrow 0$$ is exact. (3) implies (4): suppose $f = g \circ h$ where $g$ is monic and $h$ is epic; since $F$ is assumed exact, $F$ preserves epi–mono factorisations; but if $F f = 0$, then $\operatorname{dom} F g = 0$, so $\operatorname{dom} g = 0$, so $f = 0$ as well.

It remains to be shown that (4) implies (1). It is not hard to see that (4) implies (3): after all $X = 0$ if and only if $\textrm{id}_X = 0$. To show that $F$ reflects exactness, it is enough to check that $F$ reflects short exact sequences, and that is the same thing as checking that $F$ reflects kernels and cokernels. First, we will show that $F$ reflects monomorphisms and epimorphisms. Indeed, suppose $$0 \longrightarrow X \longrightarrow Y \longrightarrow Z$$ is exact in $\mathcal{A}$; then $$0 \longrightarrow F X \longrightarrow F Y \longrightarrow F Z$$ is exact in $\mathcal{B}$ because $F$ is exact; but if $F Y \to F Z$ is monic, then $F X = 0$, so by (3) $X = 0$ as well, and so $Y \to Z$ is monic. The dual argument shows that $F$ reflects epimorphisms. Now, in an abelian category, $f$ is an isomorphism if and only if $f$ is both monic and epic, so this implies $F$ reflects isomorphisms. Now suppose $X \to Y \to Z$ is given and $$0 \longrightarrow F X \longrightarrow F Y \longrightarrow F Z$$ is exact in $\mathcal{B}$; then $F X \to F Z$ is zero, so $X \to Z$ is zero, and therefore there is a comparison morphism $X \to \operatorname{Ker} (Y \to Z)$; but $F$ is exact, so $F X \to F \operatorname{Ker} (Y \to Z)$ is an isomorphism in $\mathcal{B}$, so $X \to \operatorname{Ker} (Y \to Z)$ is an isomorphism in $\mathcal{A}$. Thus $$0 \longrightarrow X \longrightarrow Y \longrightarrow Z$$ is exact in $\mathcal{A}$ as well. The dual argument shows that $F$ reflects cokernels. This completes the proof.

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