Rotman´s homological algebra regarding commutative diagram with exact rows and columns.

Question

I stumbled upon Rotman´s homological algebra, I think it is exercise 2.33, and Im stucked proving it.

Given the following commutative diagram in $R-Mod$ with exact rows and columns:

\begin{array}{c}
A' & \xrightarrow{\alpha}& A & \xrightarrow{\alpha'}& A'' &\xrightarrow{} & 0 \\
\downarrow f' & & \downarrow f & & \downarrow f'' \\
B' & \xrightarrow{\beta}& B & \xrightarrow{\beta'}& B'' & \xrightarrow{} & 0 \\
\downarrow g' & & \downarrow g & & \downarrow g'' \\
C' & \xrightarrow{\gamma}& C & \xrightarrow{\gamma'}& C'' & \xrightarrow{}& 0 \\
\downarrow & & \downarrow & & \downarrow \\
0 & & 0 & &0 &
\end{array}

Prove that if $f''$ and $\beta$ are monomorphism, then $\gamma$ is also monomorphism. Also prove that if $\gamma$ and $f$ are monomorphism, then $f''$ is also monomorphism. My attempt to prove goes like this:

Lets take an arbitray $c' \in C$ such that $\gamma(c')=0$ in order to prove that $\gamma$ is injective we need to prove that $Ker(\gamma)=0$, this is proving that $c' = 0$. Since $c' \in C'$ with $g'$ surjective there is an element $b' \in B'$ such that $g'(b')=c'$. Then, since the diagram is commuative we have that $(\gamma g')=(g \beta)$, so we have

$$g(\beta (b'))=(g \beta)(b')= (\gamma g')(b') = \gamma (g'(b'))= \gamma (c') =0.$$

This last equality means that $\beta (b') \in Ker(g) =Im(f)$. So there is an element $a \in A$ such that $f(a) = \beta (b')$. Then by the commutativity of the upper right square we have that $(\beta' f)= (f'' \alpha ')$. This implies the following equality:

$$f''(\alpha'(a))= (f'' \alpha')(a)= (\beta' f )(a)= \beta' (f(a))= \beta' (\beta(b'))=0.$$

Where the last part of the equality equal to zero because $\beta(b') \in Im(\beta) =Ker(\beta ')$. Therefore $\alpha' (a) = 0$ since $f''$ is injective. But Im run out of ideas how to proceed here in order to prove that $c' =0$. Appreciate your help with this one.

Answer 1

In general, when faced with a problem like this, you should check carefully whether you've exhausted all available information at your disposal. I'll use your notation, and pick up where you left off. I've hidden key parts of this answer with spoilers; I recommend trying to complete the problem yourself as much as you can, only consulting them as needed.

Since $\alpha(a) = 0$,

there exists $a' \in A'$ such that $\alpha(a') = a$, since $\ker(\alpha') = \operatorname{Im}(\alpha)$.

Therefore, we have

$f(\alpha(a')) = f(a) = \beta(b')$,

and so by the commutativity of the upper square, this yields

$\beta'(f'(a')) = \beta(b')$.

By using that $\beta$ is injective, we see

$f'(a') = b'$,

and since $g'(b') = c'$, we conclude that

$c' = g'(b') = f'(g'(a')) = 0$, where the last equality follows by the exactness of the left-most column.