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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.

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    $\begingroup$ You're very close. You know that $\alpha'(a) = 0$, so what does that tell you about $a$ using the exactness of the top row? $\endgroup$ Commented Jan 25, 2022 at 20:44
  • $\begingroup$ @AlexWertheim Well, that tell me that $a \in Ker (\alpha')= Im (\alpha)$. So there is an $a' \in A' $ such that $\alpha(a')=a$. Then I obtained $ \beta(f'(a))= f(\alpha(a)) \in Im(f) =Ker(g)$ but still dont know where to go from here :/ $\endgroup$
    – Sok
    Commented Jan 25, 2022 at 21:11
  • $\begingroup$ That's good progress. You're only missing a small part, but I've written a complete answer below which shows a path. $\endgroup$ Commented Jan 25, 2022 at 21:18

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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.

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  • $\begingroup$ Thank you so much for your insightful approach to my answer Alex. I uttterly understood your answer. I guess $\gamma$ and $f$ injective is the same level of difficulty? If so, I will try it by my own with our same approach to keep practicing $\endgroup$
    – Sok
    Commented Jan 25, 2022 at 21:55
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    $\begingroup$ Glad I could help, @Sok. Yes, I think it is about the same difficulty, and is well within your capabilities. If you get stuck, feel free to comment and I can make suggestions if needed. $\endgroup$ Commented Jan 25, 2022 at 22:00

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