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I am trying to prove that $E'$ is an extension of $Q$ by $N'$

\begin{array} 00 &\longrightarrow & N & \overset{i_1} \longrightarrow & E & \overset{\pi_1} \longrightarrow& Q &\longrightarrow& 0 \\ & & \downarrow {\alpha}& &\downarrow {\beta} & & \downarrow {id}& &&\\ & & N' & \overset{i_2} \longrightarrow & E' & \overset{\pi_2} \longrightarrow & Q & & \end{array}

Where $E' = N' \oplus E \, /\ mod \, S$ where $S = \{(-\alpha(x),i_1(x)) \in N' \oplus E, \forall \, x \in N \}$

Now I think the answer is given here: Proposition 2

and can follow much of it, I am looking for a more general way using the snake lemma. What I have devised is

\begin{array} & & & 0 & & 0 & & 0 & & \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ 0 &\longrightarrow & ker (\alpha) & \overset{f} \longrightarrow & \ker (\beta) & \overset{g} \longrightarrow& ker (id) &\longrightarrow& 0 \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ 0 &\longrightarrow & N & \overset{i_1} \longrightarrow & E & \overset{\pi_1} \longrightarrow& Q &\longrightarrow& 0 \\ & & \downarrow {\alpha}& &\downarrow {\beta} & & \downarrow {id}& &&\\ &\ & N' & \overset{i_2} \longrightarrow & E' & \overset{\pi_2} \longrightarrow& Q & & \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ 0 &\longrightarrow & coker (\alpha) & \overset{j} \longrightarrow & coker (\beta) & \overset{k} \longrightarrow & coker(id) &\longrightarrow & 0 \\ & & \downarrow & &\downarrow & & \downarrow & &&\\ & & 0 & & 0 & & 0 & & \\ \end{array} Which then gives $\delta: ker (id) \rightarrow coker (\alpha)$ so \begin{array} 00 &\longrightarrow & ker (\alpha) & \overset{f} \longrightarrow & \ker (\beta) & \overset{g} \longrightarrow& ker (id) & \overset{\delta} \longrightarrow& coker (\alpha) & \overset{j} \longrightarrow & coker (\beta) & \overset{k} \longrightarrow & coker(id) &\longrightarrow & 0 \end{array}

Which because $ker(id) = coker(id) = 0$ we get that:

$ker (\alpha) \simeq ker(\beta)$ and $coker (\alpha) \simeq coker(\beta)$

But am not sure in which way to proceed..

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1 Answer 1

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You can use the fact that the cokers are isomorphic to show that every element $e'$ of E' can be expressed as $e'=i_2n'+\beta e$ for some $n' \in N$, $e \in E$ (because the class $[e'] \in coker(\beta)$ corresponds to $[n'] \in coker(\alpha)$ for some $n' \in N$. Then to show the universal property of a pushout, this allows you to "define" a map $f:E' \to M$ (where $M$ is some module with maps $p:N' \to M$ and $q:E \to M$ making the square with $N$ commute) by $$f(e')=p(n')+q(e)$$ I put define in quotes because you have to check that this $f$ is well-defined, i.e independent of the choice of $n', e$. For this, show that if $e'=i_2n'_2 + \beta e_2$, then there exists $n \in N$ such that $n_2'=n'+\alpha(n)$ and $e_2=e-\alpha(n)$. This is where you use $ker(\alpha) \simeq ker(\beta)$. But it seems a bit more involved so I didn't check all the details.

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  • $\begingroup$ Ok, I think I have most of it, i.e. showing that $E'$ is a the pushout and showing uniqueness; but I am stuck why $E'$ would HAVE to be part of an exact sequence meaning that if I am trying to show that $E'$ is an extension of $Q$ by $N'$ I can't assume that I have $0$'s coming in to $N'$ and out of $Q$ for the (Possibly) exact sequence involving $E'$. I edited the diagram to express what I mean. $\endgroup$
    – Relative0
    Commented Oct 25, 2015 at 21:21

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