TITLE: In an abelian category is it true that $\ker f \cong \ker (\operatorname{coker} (\ker f))$?
QUESTION [7 upvotes]: In an abelian category is it true that $\ker f \cong \ker (\operatorname{coker} (\ker f))$?
I am teaching myself category and was playing with the definitions of kernel and cokernel and think I established this result (and of course dually that 
$\operatorname{coker} f \cong \operatorname{coker} (\ker (\operatorname{coker} f))$ ).
The trouble is, that I haven't seen it in any of the texts and lecture notes I have been reading so perhaps I am wrong.
Here is my reasoning (which would look simpler if I could work out how to draw the commutative diagram in LaTeX):
Let $f:A\to B$.  Then there exists $i: \ker f \to A$ with $f\circ i = 0$.  Hence there exists $q: A \to \operatorname{coker} (\ker f)$ with $ q\circ i = 0$.
Hence there exists $j: \ker (\operatorname{coker}(\ker f)) \to A$ such that $q\circ j = 0$.
We thus have two maps: $0 = q\circ i : \ker f \to A \to \operatorname{coker} (\ker f)$ and $0 = q\circ j : \ker (\operatorname{coker} (\ker f)) \to A \to \operatorname{coker} (\ker f)$
and so by the universal property of kernels we must have that $\ker f \cong \ker (\operatorname{coker} (\ker f))$.

REPLY [6 votes]: In any category with zero morphisms (this is a little bit weaker than to require the existence of zero objects) one can define kernels and cokernels by the well-known universal properties. In general not every morphism will have a kernel or a cokernel, and if a (co)kernel exists, it is in general not uniquely determined. However, if a morphism $f : A \to B$ has (co)kernels, then any two  (co)kernels differ by an isomorphism between their (co)domains.
Let us recall the definition of a kernel (cokernels are defined "dually"):
A kernel of a morphism $f : A \to B$ is a morphism $k : K \to A$ such that
(1) $f \circ k = 0$
(2) For each morphism $n : N \to A$ such that $f \circ n = 0$ there exists a unique morphism $u : N \to K$ such that $k \circ u = n$.
We must not confuse the morphism $k$ with the object $K$. Let us for the moment denote $k$ as kernel morphism and $K$ as kernel object. Clearly $k$ determines $K$ (it belongs to $k$ as its domain), but $K$ alone does not determine $k$.
Thus, strictly speaking, the notation $ker(f)$ only makes sense if we mean a kernel morphism and not a kernel object. Be aware, however, that writing $ker(f)$ suggests that we can assign to $f$ a unique kernel morphism. But in general $f$ has many kernel morphisms so that we need to make a choice. In the best case we can give a universal construction assigning to each $f$ a specific kernel morphism $ker(f)$.
Despite these general findings, in many concrete applications a certain kernel object is defined as "the kernel" of $f$. This is due to the fact that $K$ comes together with an "obviuos" morphism $k : K \to A$. Consider for example the category of abelian groups. As the kernel of a homomorphism $f : A \to B$ one usually understands the subgroup $f^{-1}(0)$ of $A$ and writes $ker(f) = f^{-1}(0)$. In this  case the inclusion $i_f : f^{-1}(0) \to A$ is a kernel morphism of $f$. In fact, $i_f$ is the only reasonable choice for a homomorphism $f^{-1}(0) \to A$ which justifies the lax identification of kernel object and kernel morphism. Moreover, $f^{-1}(0)$ is uniquely determined by $f$ so that have the above-mentioned universal construction assigning to each $f$ its "standard" kernel morphism $i_f$. 
Now, with no other requirements than the existence of zero morphisms we can prove:
Theorem. Let $f : A \to B$ be a morphism which has a kernel $k : K \to A$. If $k$ has a cokernel $c : A \to C$, then $k$ is a kernel of $c$. Thus, laxly written: $ker(f) \cong ker(coker(ker(f)))$.
Your proof is essentially correct, but is focussing on (co)kernel objects. For (co)kernel morphisms we do it as follows:
Since $k$ is a kernel of $f$, we have $f \circ k = 0$. Since $c$ is a cokernel of $k$ , its universal property implies the existence of a unique $h : C \to B$ such that $h \circ c = f$.
We now show that $k$ is a kernel of $c$ by verifying the universal property.
(1) $c \circ k = 0$ because $c$ is a cokernel of $k$.
(2) Let $n : N \to A$ be any morphism such that $c \circ n = 0$. Then $f \circ n = h \circ c \circ n = 0$. Since $k$ is a kernel of $f$, there exists a unique $u : N \to K$ such that $k \circ u = n$.
The above proof shows that we can split the theorem in two parts:
(a) Let $f : A \to B$ be a morphism which has a kernel $k : K \to A$. If $f$ is the composition of two morphisms $f = h \circ c$ such that $c \circ k = 0$, then $k$ is a kernel of $c$.
(b) Let $f : A \to B$ be a morphism and $k : K \to A$ be a morphism such that $f \circ k = 0$. If $k$ has a cokernel $c : A \to C$, then $f$ can be decomposed as $f = h \circ c$.