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I am currently trying to understand a proof of the Five lemma. For reference, the Five lemma is as follows:

In an abelian category, consider the commutative diagram

$A \longrightarrow B \longrightarrow C \longrightarrow D \longrightarrow E$

$\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\, \downarrow $

$A' \rightarrow\,\, B' \rightarrow\,\, C' \rightarrow\,\, D' \rightarrow\, E'$

with exact rows (in the category theoretic sense), the left vertical arrow epic, the right vertical arrow monic, and the second and fourth vertical arrows isomorphisms. Then the central vertical arrow is also an isomorphism.

(You'll have to forgive my diagram, and the one that follows - I couldn't figure out how to TeX a proper commutative diagram. If anyone would care to replace my horrible ad-hoc solution with a properly formatted one then please go ahead.)

Now the proof I have starts as follows:

We write out the image factorisation of all the horizontal maps.

$A \longrightarrow I_1 \longrightarrow B \longrightarrow I_2 \longrightarrow C \longrightarrow I_3 \longrightarrow D \longrightarrow I_4 \longrightarrow E$

$\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow \,\,\,\,\,\,\,\,\,\,\,\,\downarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\, \downarrow$

$A' \rightarrow\,\, I_1' \rightarrow\,\, B' \rightarrow\,\, I_2' \rightarrow\,\, C' \rightarrow\,\, I_3' \rightarrow\, D' \rightarrow\, I_4' \rightarrow\,\, E'$

Then $I_1 \to I_1'$ and $I_4$ to $I_4'$ are epic and monomorphic, therefore isomorphic. Could anyone explain why these maps are epic and monomorphic? It's probably something to do with the image factorisation, but since we're only factorising horizontally I don't see how we can say much about the vertical maps. The horozontal arrows also have some mono/epimorphism adornments (alternating mono/epi), but again I'm afraid I'm not sure how to TeX them. Your explanations would be greatly appreciated - thanks.

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Consider looking at the answers to this meta question for some hints on typing commutative diagrams. Your alignments are off. –  Arturo Magidin May 14 '12 at 0:29
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1 Answer

up vote 1 down vote accepted

You may have seen the decomposition of a long exact sequence into diagonal short exact sequences. If not, I won't try to reproduce it here, but the upshot is that every map in an abelian category decomposes into an epi to the image from the domain and a mono from the image to the codomain.

So all you need to do is remember that every square in your long diagram is commutative. Since $A\rightarrow A'$ and $A' \rightarrow I_1'$ are epic, their composition is too, and it equals the composition of $I_1 \rightarrow I'_1$ with $A \rightarrow I_1$. All you need is the fact, which holds in any category and which I bet you can prove yourself if you don't know it, that if $f\circ g$ is an epimorphism, $f$ is as well. So $I_1 \rightarrow I'_1$ is epic.

Going around the next square, we find that $(I'_1\rightarrow B')\circ(I_1\rightarrow I'_1)= (B\rightarrow B')\circ(I_1\rightarrow B)$ is monic, and apply the dual of the fact I just mentioned to see $I_1 \rightarrow I'_1$ is also monic.

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Yes, I have seen that decomposition. I was aware of what the "image factorisation" meant, but I didn't put together the fact that we were using the commutative property of each square and taking into account the vertical maps we do know about to deduce the properties of the intermediate maps. Your answer cleared that up for me, thank you. –  Spyam May 14 '12 at 2:24
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