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Consider the following commutative diagram with exact rows (of $R$-modules and $R$-linear maps):

$$ \newcommand{\ra}[1]{\kern-1.5ex\xrightarrow{\ \ #1\ \ }\phantom{}\kern-1.5ex} \newcommand{\da}[1]{\bigg\downarrow\raise.5ex\rlap{\scriptstyle#1}} \begin{array}{c} 0 & \ra{} & M' & \ra{f} & M & \ra{g} & M'' & \ra{} & 0\\ & &\da{\alpha'} &&\da{\alpha} &&\da{\alpha''}\\ 0 & \ra{} & N' & \ra{\smash{f'}} & N & \ra{\smash{g'}} & N'' & \ra{} & 0 \end{array} $$

Suppose $\alpha^{'}$ is an isomorphism. I want to show that there is an exact sequence: $$0 \longrightarrow M \xrightarrow{\ (g,\alpha)\ } M^{''} \oplus N \xrightarrow{(\alpha^{''},-g')} N^{''} \longrightarrow 0$$

Two questions: to show the last map is surjective, can we simply let $n$ be in $N^{''}$ then since $g^{'}$ is surjective we can find $x \in N$ such that $g(x)=n$. So take $(0,-x) \in M^{''} \oplus N$.

Question 2: I need to show that the image of the first (nonzero map) is equal to the kernel of $(\alpha^{''},-g')$. I'm stuck in showing that the kernel is contained in the image, can you please help with this part?

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Unfortunately, MathJax (and hence math.SE) does not support \xymatrix. See this meta thread for some possibilities of how to typeset one. Also, \begin{center} and \end{center} are not supported (I believe); you can make a displayed (and hence centered) equation by using $$ instead of $. – Arturo Magidin Apr 21 '12 at 4:30
I was almost done implementing Jack Schmidt's solution when you edited; I've re-edited. If you prefer your edited version instead, you can roll-back. – Arturo Magidin Apr 21 '12 at 4:38
@Arturo Magidin: thanks! that looks way better. – user6495 Apr 21 '12 at 4:39
I have made some edits to improve the layout of the commutative diagram, but it won't be available until peer reviewed, since I don't have editing privileges. – Davide Cervone Apr 21 '12 at 14:01
up vote 1 down vote accepted
  1. Yes; what you state would show that the map $(\alpha'',-g')$ is onto, except for the typo: you used $g$ when you meant $g'$.

  2. Suppose $(m'',n)$ lies in the kernel. That means that $\alpha''(m'')=g'(n)$. Since $g$ is onto, there exists $m\in M$ such that $g(m)=m''$, so $$g'(n) = \alpha''(m'') = \alpha''(g(m)) = g'(\alpha(m)).$$ Therefore, $n-\alpha(m)\in\mathrm{ker}(g')$; therefore, there exists $n'\in N'$ such that $f'(n') = n-\alpha(m)$. Let $m'\in M'$ correspond to $n'$. Then $$\alpha(f(m')) = f'(\alpha'(m')) = f'(n') = n-\alpha(m).$$ Therefore, $\alpha(f(m')+m) = n$. So $$(g,\alpha)(m,f(m')+m) = (g(m),\alpha(f(m')+m)) = (m'',n),$$ so $(m'',n)\in\mathrm{Im}(g,\alpha)$.

    Added. Note that we only require $\alpha'$ to be onto (not necessarily an isomorphism); this guarantees the existence of $m'\in M'$ with $\alpha'(m') = n'$, which suffices for the inclusion here.

  3. Now, if $m\in M$, then $$(\alpha'',-g')\bigl((g,\alpha)(m)\bigr) = (\alpha'',-g')(g(m),\alpha(m)) = \alpha''(g(m)) - g'(\alpha(m)) = 0$$ so $\mathrm{Im}(g,\alpha)\subseteq \mathrm{ker}(\alpha'',-g')$.

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thanks, a question: $n \in N$ right? and the domain of $f$ is $M'$ yes? so how can we evaluate $f(n)$? – user6495 Apr 22 '12 at 19:17
@user6495: There were some errors in that paragraph (a few letters in the wrong place). Fixed. – Arturo Magidin Apr 22 '12 at 20:11

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