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Let $A\stackrel{f}{\to}B \stackrel{g}{\to}C$ be a complex in an abelian category, I.e. $gf=0$.

It is easy to see that there exists morphism $im(f) \to ker(g).$

Could you have a proof for it being a monomorphism?

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1 Answer 1

up vote 4 down vote accepted

It follows from the fact that

$$ im(f) \to ker(g) \to B$$

is the monomorphism

$$ im(f) \to B $$

In other words, $im(f) \to ker(g)$ isn't just a morphism of objects, but it is a morphism of subobjects of $B$.

This is, of course, rather trivial if you allow yourself to prove theorems in abelian categories by diagram chase, since $im(f) \subseteq ker(g) \subseteq B$.

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I understood. Thank you. –  Tom Oct 26 '12 at 4:59

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