# Showing there is an exact sequence

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


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 $$ instead of . – Arturo Magidin Apr 21 '12 at 4:30 I was almost done implementing Jack Schmidt's solution when you edited; I've re-edited. If you prefer your edited version instead, you can roll-back. – Arturo Magidin Apr 21 '12 at 4:38 @Arturo Magidin: thanks! that looks way better. – user6495 Apr 21 '12 at 4:39 I have made some edits to improve the layout of the commutative diagram, but it won't be available until peer reviewed, since I don't have editing privileges. – Davide Cervone Apr 21 '12 at 14:01 ## 1 Answer 1. Yes; what you state would show that the map (\alpha'',-g') is onto, except for the typo: you used g when you meant g'. 2. Suppose (m'',n) lies in the kernel. That means that \alpha''(m'')=g'(n). Since g is onto, there exists m\in M such that g(m)=m'', so$$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