Let $P,Q$ be abelian categories and $F:P\to Q$ be an additive functor. Wikipedia states two definitions on left exact functors (right dually):
- $F$ is left exact if $0\to A\to B\to C\to 0$ is exact implies $0\to F(A)\to F(B)\to F(C)$ is exact.
- $F$ is left exact if $0\to A\to B\to C$ is exact implies $0\to F(A)\to F(B)\to F(C)$ is exact.
Moreover, it states that these two are equivalent definitions. I'm quite new at this topic so I'm not sure if this is immediately clear or not. Surely, 2. $\implies$ 1., being the more general case. But I don't see how to even approach the other direction; is this merely tautological?
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$\begingroup$Assume 1. holds. First observe that $F$ preserves monomorphisms: If $i : A \to B$ is a monomorphism, then $0 \to A \xrightarrow{i} B \to \mathrm{coker}(i) \to 0$ is exact, hence also $0 \to F(A) \to F(B) \to F(\mathrm{coker}(i))$ is exact. In particular $F(i)$ is a monomorphism.
Now if $0 \to A \xrightarrow{i} B \xrightarrow{f} C$ is exact, then $0 \to A \xrightarrow{i} B \xrightarrow{f} \mathrm{im}(f) \to 0$ is exact, hence by assumption $0 \to F(A) \to F(B) \to F(\mathrm{im}(f))$ is exact. Since $F(\mathrm{im}(f)) \to F(C)$ is a monomorphism, it follows that also $0 \to F(A) \to F(B) \to F(C)$ is exact.
$\endgroup$ 3 $\begingroup$Hint: Let $0\to A\overset f\to B\overset g\to C$ be exact. It means that $f=\ker g$, now consider the canonical factorisation of $g$, letting $C':={\rm im\,} g$, and $g':B\to C'$ (in a category of modules it is just $g'(b):=g(b)$ for all $b$). Then, we have that $0\to A\overset f\to B\overset {g'}\to C'\to 0$ is exact. Apply the hypothesis on this.
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