Equivalent definitions of left exact functors

Question

I am new studying categories and I have the following question. Given an additive functor $F$ between abelian categories, I have tried to prove that the following left exact functor definitions are equivalent.

1. If $0 \to A \to B \to C \to 0$ is exact then $0 \to F(A) \to F(B) \to F(C)$ is exact.
2. If $0 \to A \to B \to C$ is exact then $0 \to F(A) \to F(B) \to F(C)$ is exact.

$1. \impliedby 2.$ is obvious, but I can’t prove the converse. I have found an answer but I can’t understand some details. If $0 \to A \to B \xrightarrow{f} C$ is exact then $0 \to A \to B \to \mathrm{Im}(f) \to 0$ is exact and by hypothesis $0 \to F(A) \to F(B) \to F(\mathrm{Im}(f)) \to 0$ is again exact. Now, $\mathrm{Im}(f) \to C$ is a monomorphism and $2.$ implies $F$ preserves monomorphisms then $F(\mathrm{Im}(f)) \to F(C)$ is a monomorphism, and then $0 \to F(A) \to F(B) \to F(C)$ is exact.

I don’t understand why the fact that $F(\mathrm{Im}(f)) \to F(C)$ is a monomorphism permits complete the sequence, and I don’t know why $2.$ implies $F$ preserves monomorphisms.

Answer 1

I don’t know why 2. implies $F$ preserves monomorphisms.

A sequence $0 \to K \xrightarrow{i} X \xrightarrow{f} Y$ is exact if and only if the morphism $i \colon K \to X$ is a kernel of the morphism $f$. So if the functor $F$ satisfies condition 2, then it preserves kernels, i.e., the morphism $F(i) \colon F(K) \to F(X)$ is a kernel of $F(f)$. But a morphism is an abelian category is a monomorphism if and only if its kernel is zero. And since the functor $F$ is additive, we know that $K = 0$ implies $F(K) = 0$. Therefore, altogether, $$ \text{$f$ is mono} \iff \text{$K = 0$} \implies \text{$F(K) = 0$} \iff \text{$F(f)$ is mono} \,. $$ This shows that $F$ preserves monomorphisms if $F$ satisfies condition 2.

I don’t understand why the fact that $F(\mathrm{Im}(f)) \to F(C)$ is a monomorphism permits complete the sequence, […]

In general: if $h \colon X \to Y$ is any morphism and $m \colon Y \to Z$ is a monomorphism, then $h$ and $m ∘ h$ have the same kernel. That is, a morphism $i \colon K \to X$ is kernel for $h$ if and only if it is a kernel for $m ∘ h$.

In the specific situation, the sequence $0 \to F(A) \xrightarrow{g} F(B) \xrightarrow{h} F(\mathrm{Im}(f))$ is exact and $F(\mathrm{Im}(f)) \xrightarrow{m} F(C)$ is a monomorphism, so it follows that $$ \mathrm{im}(g) = \ker(h) = \ker(m ∘ h) \,. $$ The sequence $0 \to F(A) \xrightarrow{g} F(B) \xrightarrow{m ∘ h} F(C)$ is therefore again exact. (And $m ∘ h = F(f)$ by choice of $m$ and $h$ and by the functoriality of $F$.)