Prove exactness of $0 \to \ker(t) \to \ker(st) \to \mathrm{im}(t) \cap \ker(s) \to 0$

Question

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.

Answer 1

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.