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?