Prove that if $\alpha_{3}$ is an isomorphism, then the sequence is a short exact sequence. I am having trouble making some connections in proving this result and I'd like some help sorting this out.
So here is the commutative diagram of $R$-modules and $R$-module homomorphisms.

Each row is exact and $\alpha_3$ is an isomorphism. I am to show that the sequence
$$
0 \to A_1 \overset\varphi\to A_2\oplus B_1 \overset\psi\to B_2 \to 0
$$
is a short exact sequence, where
$$
\varphi(a_1)=(f_1(a_1),\alpha_1(a_1))
$$
for each $a_1 \in A_1$ and
$$
\psi(a_2,b_1) = \alpha_2(a_2)-g_1(b_1)
$$
for $(a_2,b_1)\in A_2\oplus B_1$.
I have already worked out that $\varphi$ is an injective homomorphism (i.e. a monomorphism), so I don't believe I need help with that. I still need to show that $\operatorname{im}(\varphi)=\ker(\psi)$ and that $\psi$ is a surjective (= an epimorphism). Here is what I've reasoned so far, and maybe I am overcomplicating this:
So far I have worked out that $f_1$ and $g_1$ are monomorphisms and $f_2$ and $g_2$ are epimorphisms since they are contained in exact rows. I believe that means $\operatorname{im}(f_1) = \ker(f_2)$, which means any $f_1(a)\in \ker(f_2) \to f_1(a) = 0_{A_2}$. Is that correct? If so I have already worked out that $\varphi(A_1) = (f_1(A_1),\alpha_1(a_1))=(0_{A_2}, \alpha_1(A_1))$ since $f_1(A_1)\subseteq \ker(f_2)$. I am not sure what to do with $\alpha_1(A_1)$, however.
At this point I could use some suggestions/pointers. Or maybe I am misunderstanding this entire process.
 A: First off, the information encapsulated in any short exact sequence
$$
0 \to A \overset\varphi\to B \overset\psi\to C\to 0
$$
is three-fold:

*

*$\varphi$ is injective (=a monomorphism)

*$\psi$ is surjective (=an epimorphism)

*$\ker\psi=\operatorname{im}\varphi$
Indeed, the kernel of $\varphi$ is the image of the trivial map, hence trivial implying that $\varphi$ is injective. Also, the image of $\psi$ is the kernel of the trivial map, hence all of $C$ implying that $\psi$ is surjective. Exactness at $B$ cannot really be further simplified.

Let's consider the given sequence
$$
0 \to A_1 \overset\varphi\to A_2\oplus B_1\overset\psi\to B_2 \to 0
$$
regarding exactness. As the first component of $\varphi$ is the injective map $f_1$ we conclude that $\varphi$ is injective as well. For this note that if $\varphi(a_1)=(f_1(a_1),\alpha_1(a_1))=(0,0)$, then $f_1(a_1)=0$ which is only possible for $a_1=0$. The other parts require some more work.
We continue with exactness at $A_2\oplus B_1$ which is the equality $\ker\psi=\operatorname{im}\varphi$. For the inclusion $\operatorname{im}\varphi\subseteq\ker\psi$ note that the composition $\psi\varphi$ evaluates at $a_1\in A_1$ to
$$
(\psi\varphi)(a_1)=\psi(f_1(a_1),\alpha_1(a_1))=(\alpha_2f_1)(a_1)-(g_1\alpha_1)(a_1)
$$
which is zero since the first square in the diagram of exact rows commutes. For the reverse inclusion start with $(a_2,b_1)\in\ker\psi$. Now we are only a short diagram chase away from the result. Note that
$$
\psi(a_2,b_1)=0 \iff \alpha_2(a_2)=g_1(b_1)
$$
implying that
$$
\alpha_3(f_2(a_2))=g_2(\alpha_2(a_2))=(g_2g_1)(b_1)=0
$$
as the lower row is exact. Since $\alpha_3$ is an isomorphism (an assumption which should be included in the body not only in the title), we conclude that $f_2(a_2)=0$. By exactness we find $a_1\in A_1$ such that $f_1(a_1)=a_2$. Noting that then
$$
g_1(\alpha_1(a_1))=\alpha_2(f_1(a_1))=\alpha_2(a_2)=g_1(b_1)
$$
gives us that $\alpha_1(a_1)=b_1$. Hence $\varphi(a_1)=(a_2,b_1)$ as desired.
It remains to show that $\psi$ is surjective. For this we start with $b_2\in B_2$ and chase it around the right commuting square. Write $g_2(b_2)=(\alpha_3f_2)(a_2)$ using that $\alpha_3$ and $f_2$ are both surjective. It follows that
$$
(\alpha_3f_2)(a_2)-g_2(b_2)=g_2(\alpha_2(a_2)-b_2)=0
$$
and hence $\alpha_2(a_2)-b_2\in\operatorname{im}(g_1)$ from exactness at $B_2$. But this nothing else than saying that we have
$$
\alpha_2(a_2)-b_2=g_1(b_1)\iff b_2=\alpha_2(a_2)-g_1(b_1)=\psi(a_2,b_1)
$$
for some $b_1\in B_1$.

Such arguments by diagram chase become clearer the more you do them. The crucial point is that "there is (more or less) only one way to go" at any given time. This is, of course, less clear the first few times but usually you can only go in one direction reversing only when assumptions such as injectivity or surjectivity of certain maps are given.
