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I am slightly confused about the notion of exactness in a general abelian category (I want to stay clear of anything related to the Mitchell embedding theorem). Here are two definitions that I have seen:

A sequence $A \xrightarrow{f} B \xrightarrow{g} C$ is exact at $B$ if

  1. $\operatorname{Im} f = \operatorname{Ker} g$, in the sense that they are isomorphic as subobjects;

    or

  2. $gf = 0$ and the canonical morphism $\operatorname{Im} f \rightarrow \operatorname{Ker} g$ is an isomorphism.

Could someone please shed some light on the equivalence between these two notions? A priori, one seems more general/stronger than the other.

Thanks for your time.

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Where have you seen the first definition? I have only ever seen the second. I mean, that's the right definition if you abstract from a common category, say $R\text{-}\mathbf{Mod}$, right? –  Alex Youcis Nov 4 '13 at 21:39
    
@AlexYoucis The first definition can be found, for instance, in Freyd's Abelian Categories, or in Mitchell's Theory of Categories. –  user105429 Nov 4 '13 at 21:42
    
Oh, really? I don't remember Freyd defining it this way. I'm pretty sure that Weibel defines it as the latter. For 1. do you mean they are isomorphic, or they are equivalent as subobjects (i.e. there is an isomorphism commuting with the inclusion map)? –  Alex Youcis Nov 4 '13 at 21:44
    
@AlexYoucis I read "isomorphic as subobjects" (otherwise it would indeed be a trivial task to find counterexamples). –  Hagen von Eitzen Nov 4 '13 at 21:46
    
@HagenvonEitzen I didn't know that "isomorphic as subobjects" was a phrase--I've always heard equivalent. Thanks for clarifying! –  Alex Youcis Nov 4 '13 at 21:48

1 Answer 1

up vote 1 down vote accepted

$\operatorname{im}f$ is the kernel of the cokernel, hence is naturally equipped with a morphism $\to B$ and this is a monomorphism (as is always the case for kernels).

If we assume (1), then by the isomorphism of subobjects, $\operatorname{im}f$ factors over $\ker g$ (and vice versa), hence $\operatorname{im}f\to C$ is the zero morphism and finally $g\circ f=0$. To put it differently, the given isomorphism $\operatorname{im}f\to\ker g$ is the canonical morphism obtained from the fact the $\operatorname{im}f\to C$ is zero and so this is an isomorphism.

If we assume (2), then the canonical $\operatorname{im}f\to\ker g$ obtained from $g\circ f=0$ by definition is a factorization of $\operatorname{im} g\to B$ over $\ker g\to B$ and similar for its inverse, so we have $(1)$

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Thank you, Hagen! –  user105429 Nov 4 '13 at 22:24

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