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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):

  1. $F$ 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.
  2. $F$ 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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2 Answers 2

up vote 6 down vote accepted

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.

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Ah, that certainly makes sense. It was much less trivial than I imagined it to be, so at least I can save myself from some borderline humiliation. :) –  Dustin Tran Feb 1 '13 at 11:18
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@nil: Use books, not Wikipedia. (@)Martin: Thanks, I've corrected it. –  Martin Brandenburg Feb 1 '13 at 11:40
    
Yes, certainly a useful tip in general. This is just one of those cases where late night browsing leads to some interesting article, and you resolve to glean something out of it. But of course I digress. –  Dustin Tran Feb 1 '13 at 11:46

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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The formula $g'(b):=g(b)$ makes sense in the setting of modules, but here we are given an abelian category. Anyway, it essentially makes no difference ... –  Martin Brandenburg Feb 1 '13 at 11:14
    
Ah, indeed. Thanks. –  Berci Feb 1 '13 at 11:20

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