Let R be a ring and $F',F,F'',G',G,G''$ left R-modules. Assume we are given R-module homomorphism $i:F'\to F,p:G'\to G,p':G\to G''$ and $a:F'\to G',b:F\to G,c:F''\to G''$ such that the following conditions are satisfied:
(i) $b\circ i=i'\circ a, c\circ p=p'\circ b$.
(ii)the sequence $F'\to F\to F''\to 0$ is exact
(iii)the sequence $0\to G'\to G\to G''$ is exact.
question: (a)define an R-module homomorphism $s:Ker(c)\to Coker(a)$.
(b)show that $i,p,s,i',p'$ give rise to an exact sequence $Ker(a)\to Ker(b) \to Ker(c)\to Coker(a)\to Coker(b)\to Coker(c)$.
I still have no idea how to construct a homomorphism in this case, if I can find that homomorphism then I think maybe I can prove the second question! Here is what I am thinking, given any $x\in Ker(c)$, we can find an element $f\in F$ such that p(f)=x, then set $s(x)=i^{-1}\circ b(f)$, but I cannot prove why it is well-defined, then if it is well-defined, maybe I think it is a homomorphism! Can someone help me?
There is essentially only one thing you can do: view $x \in \ker c$ as an element of $F''$; use surjectivity of $p$ to find an element in $F$ that maps to $x$ through $p$; push this to $G$ through $b$; use the commutativity of the right square and exactness to show that this element is in $\ker p' = \operatorname{im} i'$; pick the unique preimage of this element in $G'$; push this to its coset in $\operatorname{coker} a$. In symbols: $$s(x) = (i')^{-1}(b(p^{-1}(x))) + \operatorname{im} a.$$Show that this map is well-defined and a homomorphism. (To show that the map is well-defined, consider two preimages $y,z \in F$ of $x$, i.e., $p(y) = p(z)$. Use commutativity of the left square and exactness to show that the difference $(i')^{-1}(b(y)) - (i')^{-1}(b(z))$ belongs to $\operatorname{im} a$.)
N.B. This is an extremely well-known result called the snake lemma. (There is even a movie scene in which it is shown being taught.)