Prove exactness of $0 \to \ker(t) \to \ker(st) \to \mathrm{im}(t) \cap \ker(s) \to 0$ I am trying to prove one statement in an abelian category: consider $t \colon A \to B$ and $s \colon B \to C$, then
$$
  0
  \longrightarrow \ker(t)
  \xrightarrow{\enspace\alpha\enspace} \ker(st)
  \xrightarrow{\enspace\beta\enspace} \mathrm{im}(t) \cap \ker(s)
  \longrightarrow 0
$$
is exact. It’s not hard to prove $\ker(t)$ is also the kernel of $\beta \colon \ker(st) \to \mathrm{im}(t) \cap \ker(s)$ by using the fact that intersection is pullback.  However, I can’t find a quick answer to prove $\beta$ is an epimorphism, though I think it’s easy to see that if the abelian category is $R\text{-}\textbf{Mod}$ (we may use the embedding theorem?). Can you give me some hints? Thank you.
 A: We note that $\newcommand{\im}{\mathrm{im}} \im(t) ∩ \ker(s)$ is the kernel of the composite $\im(t) \to B \to C$.
To formally prove this, we consider the following diagram:
$$
  \require{AMScd}
  \begin{CD}
    \im(t) ∩ \ker(s) @>>> \ker(s) @>>>    0     \\
    @VVV                  @VVV            @VVV  \\
    \im(t)           @>>> B       @>>{s}> C
  \end{CD}
$$
Both squares in this diagram are pullbacks, whence the outer diagram
$$
  \require{AMScd}
  \begin{CD}
    \im(t) ∩ \ker(s) @>>> 0     \\
    @VVV                  @VVV  \\
    \im(t)           @>>> C
  \end{CD}
$$
is again a pullback.
Solution 1 (snake lemma)
We can consider the following commutative diagram:
$$
  \begin{CD}
    0  @>>> \ker(t) @>>> A      @>>> \im(t) @>>> 0  \\
    @.      @VVV         @VV{st}V    @VVV        @. \\
    0  @>>> 0       @>>> C      @>>{1}> C      @>>> 0
  \end{CD}
$$
Both rows of this diagram are short-exact, so by the Snake lemma we get an induced exact sequence
$$
  \newcommand{\longto}{\longrightarrow}
  0
  \longto \ker(t)
  \longto \ker(st)
  \longto \im(t) ∩ \ker(s)
  \longto 0
  \longto \dotsb
$$
Solution 2 (nine lemma)
Suppose that we have already costructed the sequence
$$
  0
  \longto \ker(t)
  \longto \ker(st)
  \longto \im(t) ∩ \ker(s)
  \longto 0 \,.
$$
We have seen above that $\im(t) ∩ \ker(s)$ is the kernel of $\im(t) \to C$.
We can restrict $\im(t) \to C$ to a morphism $\im(t) \to \im(st)$ without changing this kernel, because $\im(st) \to C$ is a monomorphism.
This allows us to consider the following commutative diagram:
$$
  \begin{CD}
    {}  @.    0        @.    0         @.    0                 @.    {}  \\
    @.        @VVV           @VVV            @VVV                    @.  \\
    0   @>>>  \ker(t)  @>>>  \ker(st)  @>>>  \im(t) ∩ \ker(s)  @>>>  0   \\
    @.        @VVV           @VVV            @VVV                    @.  \\
    0   @>>>  \ker(t)  @>>>  A         @>>>  \im(t)            @>>>  0   \\
    @.        @VVV           @VVV            @VVV                    @.  \\
    0   @>>>  0        @>>>  \im(st)   @>>>  \im(st)           @>>>  0   \\
    @.        @VVV           @VVV            @VVV                    @.  \\
    {}  @.    0        @.    0         @.    0                 @.    {}
  \end{CD}
$$
All three columns of this diagram are exact, and the second and third rows are exact.
It follows from the nine lemma that the first row is also exact.
