Regular categories and epis stable under pullbacks The definition of a regular epi is an arrow which coequalizes some parallel pair. Isn't this just another name for a coequalizer?
One of the (usual) axioms for a regular category says each arrow has a kernel pair. Another says regular epis are stable under pullbacks. But, some books warn this is not equivalent to stability of coequalizers under pullbacks. I don't understand this.
Generally, if some arrow is a coequalizer and has a kernel pair, it is the coequalizer of its kernel pair. Hence, in a regular category an arrow is a coequalizer iff it coequalizes its kernel pair, but what does this change?
A counterexample sometimes given in $\mathsf{Ab}$ is to pull back the coequalizer of the canonical injections $\iota _1,\iota _2:A\rightarrow A\oplus A$ along the zero morphism. But I don't see what this goes to show. Can someone explain?
 A: I think your first question "Isn't this just another name for a coequalizer?" is the source of your confusion. The answer is no. It is wrong to say that $f$ is a coequalizer. You need to say what $f$ is the coequalizer of i.e. being a coequalizer is defined with respect to a parallel pair of morphisms.
Regarding your last point. The pullback of that coequalizer diagram (i.e. the two morphism together with their coequalizer - the morphism from $A \oplus A \to A$ sending $(a,a')$ to $a+a'$) is the diagram consisting a parallel pair of zero morphisms $0 \to A$ together with the zero morphism $A\to 0$, which is clearly not a coequalizer diagram.
A: I think your confusion is ultimately a confusion about what it means for the various notions to be "stable under pullbacks".  First, "regular epis are stable under pullbacks" means that if 
$$\require{AMScd}
\begin{CD}
A @>{}>> B\\
@V{f'}VV @V{f}VV \\
C @>{}>> D
\end{CD}$$
is a pullback square and $f$ is regular epi, then $f'$ is also regular epi.  Second, "coequalizers are stable under pullbacks" means the following: Suppose you have a diagram
$$\begin{CD}
 && Z\\
& @V{a,b}VV \\
&& B\\
& @V{f}VV \\
C @>{g}>> D
\end{CD}$$
where $f$ is the coequalizer of a pair of parallel arrows $a$ and $b$.  (The arrow labelled $a,b$ should really be a double arrow, but as far as I know MathJax does not support double arrows in commutative diagrams.) Form the pullback $C\leftarrow A\to B$ of $f$ and $g$ and the pullback $C\leftarrow Y \to Z$ of $fa=fb$ and $g$.  Let $a':Y\to A$ be induced by the given map $Y\to C$ and the composition $Y\to Z\stackrel{a}\to B$, and similarly let $b':Y\to A$ be induced by $Y\to C$ and $Y\to Z\stackrel{b}\to B$.  We now have the following diagram:
$$\begin{CD}
Y @>{}>> Z\\
@V{a',b'}VV @V{a,b}VV \\
A @>{}>> B\\
@V{f'}VV @V{f}VV \\
C @>{g}>> D
\end{CD}$$
(Again, the vertical arrows on top should be double arrows.)
To say that "coequalizers are stable under pullbacks" means that in this diagram, $f'$ is the coequalizer of $a'$ and $b'$.
In particular, even if regular epis are stable under pullback, there is no particular reason to believe that $f'$ is the coequalizer of $a'$ and $b'$: all you know is that there must exist some maps $c,d:X\to A$ which $f'$ is the coequalizer of.
It is a worthwhile exercise to work out what's going on in this diagram in $\mathsf{Ab}$ when $B=Z\oplus Z$, $a,b:Z\to B$ are the two inclusions, $f:B\to D=Z$ is the coequalizer (i.e., the map $(x,y)\mapsto x+y$), and $C=0$.  You should get that $A\cong Z$ and $Y=0$, so unless $Z=0$, $f'$ is not the coequalizer of $a'$ and $b'$.
