Does the existence of cokernels imply the existence of coequalizers? Consider a pointed category with kernels and cokernels. I am wondering if such a category has coequalizers for every pair of parallel arrows. The category I am talking about is not additive. Maybe one example would be very useful for me.
 A: No, not necessarily.  For instance, consider the following subcategory of the category of pointed sets.  The objects are the sets $0=\{*\}$ and $A=\{*,a,b\}$.  The morphisms are the identity maps, the constant maps with value $*$, and all maps $f:A\to A$ such that $f^{-1}(\{*\})=\{*\}$.
This category has kernels and cokernels.  The only nontrivial case to check here is that the maps $f:A\to A$ and $g:A\to A$ such that $f(a)=f(b)=a$ and $g(a)=g(b)=b$ have cokernels.  But the map $A\to 0$ is a cokernel of both $f$ and $g$ since any map on $A$ in this category which sends either $a$ or $b$ to $*$ must send every point to $*$.
However, this category does not have a coequalizer of the identity $1_A:A\to A$ and the map $f:A\to A$ described in the previous paragraph.  Indeed, suppose a map $h$ were a coequalizer of $1_A$ and $f$.  Then $hf=h1_A$, so $h(a)=h(b)$.  Since $f^2=f1_A$, there is a unique map $i$ from the codomain of $h$ to $A$ such that $ih=f$.  This means $h(a)\neq h(*)$, so the codomain of $h$ is $A$ and $h$ can only be $f$ or $g$.  But if $h=f$, then the map $i$ is not unique, since both $i=f$ and $i=1_A$ work.  Similarly, if $h=g$, then $i$ is not unique, since $i$ could be either $f$ or the map which swaps $a$ and $b$.
