Splicing together two short exact sequences

This question is from my module theory assignment and I need help in solving the (a) only.

(a) If $0\to A\to B\xrightarrow{f} C\to 0$ and $0\to C\xrightarrow{g} D\to E\to 0$ are short exact sequences of modules, then the sequence $0\to A\to B\xrightarrow{gf} D\to E\to 0$ is exact.
(b) Show that every exact sequence my be obtained by splicing together suitable short exact sequences as in (a).

Show that isomophism of short exact sequences is an equivalence relation.

$1$st diagram means that $C=\operatorname{im}(f)$ and $2$nd diagram means that $\ker(g)=0$. But I am not able to understand what it means by the $3$rd diagram, i.e. how to interpret it? Usually in case of exact, at least $2$ functions are given simultaneously so that equality can be written in terms of image of one and kernel of other. But in this case I am not sure how to interpret as only one diagram is given.

Thanks!

From what I understand, you are only used to the concept of short exact sequences, i.e. sequences of the form

$$ 0\to A\to B\to C\to 0\,, $$

missing the general concept of an exact sequence. The latter is a sequence $(A_i,d_i)$ of modules $A_i$ and homomorphisms $d_i\colon A_i\to A_{i+1}$ (conventions vary in which direction the maps go) such that $\ker(d_{i+1})=\operatorname{im}(d_i)$ for all $i$, which ones refers to as exactness at $A_i$ as the $\ker(d_{i+1}),\operatorname{im}(d_i)\le A_i$. As I do not intend to talk about this in more generality, we may as well assume that this sequence is in fact of the form

$$ 0=A_0\xrightarrow{d_0} A_1\xrightarrow{d_1} A_2\to\cdots\to A_{n-1}\xrightarrow{d_{n-1}} A_n\xrightarrow{d_n} A_{n+1}=0\,. $$

In this case, $\ker(d_{i+1})=\operatorname{im}(d_i)$ dictates that, in particular,

$$ \ker(d_1)=\operatorname{im}(d_0)=0\quad\text{and}\quad\operatorname{im}(d_{n-1})=\ker(d_n)=A_n $$

which includes your observations as special case.

Now suppose we are given two short exact sequences

$$ 0\to A\xrightarrow{\iota} B\xrightarrow{f} C\to 0\quad\text{and}\quad0\to C\xrightarrow{g} D\xrightarrow{\pi} E\to 0 $$

and we want to show that the induced sequence

$$ 0\to A\xrightarrow{\iota} B\xrightarrow{gf} D\xrightarrow{\pi} E\to 0 $$

is then exact too. Exactness at $A$ and $E$ is still given as we use the same maps. So the relevant things to show are exactness at $B$ and at $D$.

The fastest way of doing this is to realize that

$$ \ker(gf)=\ker(f)\quad\text{and}\quad\operatorname{coker}(gf)=\operatorname{coker}(g)\,, $$

which follows from $g$ being a monomorphism and $f$ being an epimorphism (which is equivalent to being injective and surjective but requires some work), and using an alternative characterization (proposition $2.4$) of exactness in terms of cokernels. However, this might be a bit over the top. The main idea stands, nonetheless, i.e. using that $g$ and $f$ are injective and surjective (i.e. mono- and epimorphisms), resp., which then elementwise translates to

\begin{align*} ((g\circ f)\circ\iota)(x)=0\,&\iff\,(f\circ\iota)(x)=0 \\ (\pi\circ(g\circ f))(x)=0\,&\iff\,(\pi\circ g)(x)=0 \end{align*}

from where it should be clear how to apply the exactness of the original sequences.

Can you take it from here? If some terminology or ideas are unclear (mono-/epimorphisms, cokernels, etc.) please let me know. I can try to make the answer more elementary or include an explanation.