direct sum of two exact sequence is exact

Question

recently I have had some problems thinking about this, does the direct sum of two exact sequence is also exact? the picture is as follows: all $M_i$ be R modules, and all map is R linear map,and the diagram commute,also the left, and the right is an exact sequence, I want to prove the middle direct sum is also an exact sequence First I know that the map $\epsilon$ is an epimorphism, which defined by $\epsilon$ ($m_2'$+$m_2$)=$\sigma_1'(m_2)+\sigma_2'(m_2')$, also we also construct the map $\epsilon''$ $(m_3+m_3')=\sigma_1(m_3)+\sigma_2(m_3')$

By the hint from Daron(below), I get the answer for the im$\epsilon''$ = $Imf_1$ $\oplus$ $Imf_2$ by follows:

$\sigma_2(m'')=(w,f_2(m''))$ for some $w$ $\in$ $M_0'$, and I can show that $f_3(w)$=$0$, this shows that $Imd_1$ $\subset$ $Imf_1$ $\oplus$ $Imf_2$, and otherside is by the fact that image of $f_1$ and $f_2$ can be write as image of $d_1$.

Unfortunately, I have trouble proving the kernel case(i.e $ker $ $\epsilon$ =$ker(f_3)$ $\oplus$ $ker(f_4) $ where the left map we denote by $f_3$ ,and the right be $f_4$

I have three approaches.

1. Use snake lemma, we can get a exact sequence $0$ $\to$ $ker$ ( $M_2$ $\to$ $M_1$) $\to$ $ker$ $\epsilon$ $\to$ $ker$($M_2'$ $\to$ $M_1$) $\to$ $0$, I want to show this exact sequence is spilt, then we are done.
2. Using splitting lemma , just define the retraction map $p(m_2+m_2')$ =$m_2$ for the map g:$M_2$ $\to$ $M_2$ $\oplus$ $M_2'$

3.( the most difficult idea I want to prove) use direct calculation: here are some of my idea: First,for ker$\epsilon$ $\subset$ $ker(f_3)$ $\oplus$ $ker(f_4)$ Let $x$ $\in$ ker$\epsilon$, $x$=$(m_0',m_0'')$,and we get $\epsilon$($x$)=$\sigma_1'(m_0')+\sigma_2'(m_0'')$=$(f_3(m_0'),0)+(f_3(s),f_4(m_0''))=(0,0)$ (Note that since $f_3$ is subjective, so we also we suppose exist an $s$ $\in$ $M_2$ satisfy the condition. As a result, we get $m_0''$ $\in$ $ker f_4$, and $f_3(m_0'+s)=0$ here is my not any idea point, how can we show that $f_3(s)=0$ ?( since we thus can conclus that $m_0'$ is in kernel $f_3$.

After a hour, I think that I get the answer,and I will write it down to share with you. Note that there exist a $k$ such that $f_1(k)$=$m_0'+s$ by exactness, then s=$f_1(k)$-$m_0'$, then consider $\epsilon$ $\sigma_1(k)$=$(0,0) $ and we get $(m_0'+s,0)$ $\in$ $ker$ $\epsilon$, then compare with $(m_0',m_0'')$, we can get (s,0) is also in $ker$ $\epsilon$, then by commutative diagram, we get $f_3(s)=0$, so we are done.

Thanks to every person's concern about this question, although it still some minor detail, I think that this statement is complete.

Also, for (2) I think that this is true. as $(kerf_3,0)=(w,0)$ by the exact sequence, so we get this direct sum make sense. where $w$ $\in$ $M_0'$

Answer 1

For the sequence to be exact you need to check Im$(\epsilon'') = \ker (\epsilon)$ and likewise for the other terms in the sequence. To do this you just need to show im $(\epsilon'') = $ im$(f_1) \oplus $ im$(f_2)$ and likewise for the kernels. Then use exactness of the two starting sequences.