Take the 2-minute tour ×

Suppose for simplicity that we are in an abelian category. Suppose the following square is a pullback:

 P ----> A
 |       |
m|       |n
 v       v
 B ----> C

Then m and n have the same kernel, and so does p and q, I've been told (I might have something backwards). How does one prove this? All I can get is a map $ker(m) \rightarrow ker(n)$. Thanks!

share|improve this question
I never thought I'd say something like this, but "spectral sequences". Alternatively, if you assume a category of R-modules, I think injectivity and surjectivity are straightforward, given the usual characterization of P as the submodule of $B \times A$ of pairs that have the same image in $C$. (then, invoke the fact a theorem like this is true for any $R$-module category iff it's true in all abelian categories) –  Hurkyl Feb 27 '12 at 1:33

1 Answer 1

up vote 1 down vote accepted

Let $i:\ker m\to P$ the the kernel of $m$ and $j:\ker n\to A$ that of $n$. Since $npi=qmi=0$ because $mi=0$, the universal property of $j$ implies there is a map $r:\ker m\to\ker n$ such that $jr=pi$.

Now consider the maps $j:\ker n\to A$ and $t=0:\ker n\to B$. Since $nj=0=qt$, there is a unique $s:ker n\to P$ such that $ps=j$ and $ms=t=0$. This last equation implies there is a $u:\ker n\to\ker m$ such that $iu=s$.

Can you see what the compositions $ru$ and $ur$ are?

For example: since $piur=psr=jr=pi$ and $miur=0ur=0=mi$, the universal property of the cartesian square implies that $iur=i$. Since $i$ is monic, it follows that $ur=\mathrm{id}_{\ker m}$. On the other hand, $jru=piu=ps=j$, and the monicness of $j$ implies now that $ru=\mathrm{id}_{\ker n}$.

share|improve this answer

Your Answer


