Suppose that the following diagram of $R$-modules and $R$-homomorphisms
is commutative and has exact rows. Show that, If $\alpha, \gamma$, and $g$ are onto, then so is $\beta$.
$$\begin{array}
KK & \stackrel{f}{\longrightarrow} & M & \stackrel{g}{\longrightarrow} & L\\
\downarrow{\alpha} & & \downarrow{\beta} & & \downarrow{\gamma}\\
K' & \stackrel{f'}{\longrightarrow} & M'& \stackrel{g'}{\longrightarrow} & L'
\end{array}
$$
So, I need to show that for every $m'\in M'$ there exists a $m\in M$ such that $\beta(m)=m'$. And we have the following maps which are onto.
$$\begin{array}
KK & & M & \stackrel{g}{\longrightarrow} & L\\
\downarrow{\alpha} & & & & \downarrow{\gamma}\\
K' & & M'& & L'
\end{array}
$$
Suppose $g'(m')=l'\tag1$ then using the surjectivity of $g,\gamma$ we can say that $$\exists l\in L:\gamma(l)=l'\\\exists m\in M:g(m)=l$$
And from commutativity,
\begin{align}
\gamma g(m)&=g'\beta (m)\\
\gamma(l) & =g'\beta (m)\\
l' &= g'\beta (m)
\end{align}
Does it imply $\beta(m)=m'$ using (1)? I didn't use exactness of the rows. Is my approach correct? If not, then please suggest how to mentally picture to solve this kind of problem.
Best Answer
The start of your argument is a bit confusing, due to the introduction of $\ell'$.Allow me first to rewrite slightly your arguments.
Let's start with an $m'\in M'$. We'll prove that it lies in the image of $\beta$.
We have $g'(m')\in L'$, and by surjectivity of $\gamma$, there is an $\ell\in L$ such that $g'(m')=\gamma(\ell)$.
Now you have to use commutativity of the diagram. Using surjectivity of $g$, we have $\ell=g(m)$ for some $m\in M$, so that $g'(m')=\gamma g(m)=g'\beta(m)$. So far, this is what you obtained.
Now we have $g'(m'-\beta(m))=0$, so $m'-\beta(m)\in \ker(g')=im(f')$ by exactness of the bottom row. Thus $m'-\beta(m)=f'(k')$ for some $k'\in K'$. By surjectivity of $\alpha$, we get $k\in K$ such that $m'-\beta(m)=f'\alpha(k)$, and by commutativity, $m'-\beta(m)=\beta f(k)$, that is $m'=\beta(m+f(k))$.
Conclusion. The ideas you didn't want to pursue because you didn't know where this was going were exactly the right ideas.
Sometimes, you just have to accept that the solutions of some exercises are just mechanical and that there is no deep meaning. This is typical of these "diagram chasing" exercises. The solution of such exercices is 100% of the time: apply commutativity/surjectivity/injectivity/ exactness in various orders until you get what you want...