In a regular category, an exact sequence is a diagram which is both a coequalizer and a kernel pair: $$R\overset r{\underset s\rightrightarrows} X\to Y$$

Wiki says that in the abelian case, the above sequence is exact in the regular sense iff the following sequence is exact in the abelian sense: $$0\to R\xrightarrow{r-s}X\to Y\to 0$$

This seems strange to me. First of all, $r,s$ are jointly monic but I don't see why $r-s$ should be a mono. Second, if it is, then $P$ would have to be both the domain of kernel of $X\rightarrow Y$ and of its kernel pair, but that seems to contradict the answer to this question.

Is wiki wrong?

  • $\begingroup$ It is indeed wrong. $\endgroup$ – Zhen Lin Nov 26 '15 at 11:36

As Zhen Lin says in the comments, wiki is wrong.

The exactness of the sequence $R\overset r{\underset s\rightrightarrows} X\xrightarrow f Y$ is equivalent to exactness - in the abelian sense - of the sequence $0\to R\xrightarrow{\begin{pmatrix}r\\s\end{pmatrix}}X\oplus X\xrightarrow{(f,-f)} Y\to 0$.

You can prove $(P,r,s)$ is the kernel pair of $f$ iff $\begin{pmatrix}r\\s\end{pmatrix}$ is the kernel of $(f,-f)$. This is consistent with the linked question - the kernel pair of $f$ is bigger than the kernel of $f$.


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