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I have two horizontal exact sequences of abelian groups.

$\require{AMScd}$ \begin{CD} 0 @>{}>> A_1 @>{}>> A_2 @>{}>> A_3 @>{}>> A_4 @>{}>> 0\\ @| @VeVV @VfVV @VgVV @VhVV @|\\ 0 @>{}>> B_1 @>{}>> B_2 @>{}>> B_3 @>{}>> B_4 @>{}>> 0 \end{CD}

where $f,g$ And $h$ are isomorphisms. By diagram chasing I can figure out that $e$ is also an isomorphism. But is this also a consequence of the five or four lemma?

I forgot to mention that the squares commute

Con
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roi_saumon
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1 Answers1

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You can extend your diagram on the left side by two zeroes to get the typical five lemma situation where the morphism $e$ is in the middle and will be an isomorphism as $f$ and $g$ are isomorphisms (and the morphisms between the zeroes as well): $\require{AMScd}$ \begin{CD} 0 @>>> 0 @>>> A_1 @>>> A_2 @>>> A_3 @>>> A_4 @>>> 0 \\ @V0VV @V0VV @VeVV @VfVV @VgVV @VhVV @V0VV \\ 0 @>>> 0 @>>> B_1 @>>> B_2 @>>> B_3 @>>> B_4 @>>> 0 \\ \end{CD}

Con
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  • This sounds reasonable, but I don't see what mistake I'm making. – Ruben Jul 03 '19 at 11:14
  • Oh, I had to think out of the box then! Thanks! – roi_saumon Jul 03 '19 at 11:14
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    @ThorWittich I see, I was presuming the existence of a map as an extra condition. So my conclusion should have been that either the statement was false OR that map does not exist. It is clear from your answer that the latter is true. – Ruben Jul 03 '19 at 11:19
  • @roi_saumon Actually you had to create another "box" for your diagram (as you basically add one more square to the diagram). – Con Jul 03 '19 at 15:01