2
$\begingroup$

Fix an arbitrary abelian category $\mathscr{A}$, and let $$0\to A\xrightarrow{f}B\xrightarrow{g}C\to 0$$ be a short exact sequence in the category of chains $\mathscr{A}_\bullet$, where $A$, $B$, and $C$ have chain maps $\varphi^A_n:A_n\to A_{n-1}$, $\varphi^B_n:B_n\to B_{n-1}$, $\varphi^C_n:C_n\to C_{n-1}$ respectively, and let $A$ and $B$ be exact. I claim that $C$ is exact.

My current approach is to try to show that $$\ker\varphi^C_{n-1} = \mathrm{coker\,}(\ker\varphi^C_n\hookrightarrow C_n) = \mathrm{coim\,}\varphi_n^C = \mathrm{im\,}\varphi_n^C.$$

So, for an arbitrary object $M\in\mathscr{A}$ and morphism $\psi:C_n\to M$ such that $$\left(\ker\varphi^C_n\hookrightarrow C_n\xrightarrow{\psi}M\right) = 0$$ I wish to show that there exists a unique $\ker\varphi^C_{n-1}\to M$ such that $$\left(C_n\twoheadrightarrow\mathrm{im\,}\varphi^C_n\hookrightarrow\ker\varphi^C_{n-1}\to M\right) = \left(C_n\xrightarrow{\psi}M\right).$$ However, despite playing around a lot with commutative diagrams, kernels, and cokernels, I haven't found a good way of doing this. What have I missed?

$\endgroup$
0
2
$\begingroup$

The long exact homology sequence is the result to prove that if two out of three complexes in a SEC are exact, so is the third. Prove this, and you're done. To do so, if suffices you prove the well known snake lemma for an arbitrary abelian category.

$\endgroup$
0
$\begingroup$

Are you able to use tools such as the long exact homology sequence, or the snake lemma, as Pedro has suggested? If not, this can be done directly. Write out what this short exact of sequences actually looks like:

$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}} % \begin{array}{llllllllllll} & & 0 & & 0 & & 0 & & \\ && \da{} & & \da{} & & \da{} & & \\ \cdots & \ra{} & A_{n+1} & \ra{} & A_n & \ra{} & A_{n-1} & \ra{} & \cdots & \\ && \da{} & & \da{} & & \da{} & & \\ \cdots & \ra{} & B_{n+1} & \ra{} & B_n & \ra{} & B_{n-1} & \ra{} & \cdots \\ && \da{} & & \da{} & & \da{} & & \\ \cdots & \ra{} & C_{n+1} & \ra{} & C_n & \ra{} & C_{n-1} & \ra{} & \cdots \\ && \da{} & & \da{} & & \da{} & & \\ & & 0 & & 0 & & 0 & & \\ \end{array} $$

Then choose an element $c_n\in C_n$ which is in the kernel, and do some diagram chasing to show there's a $c_{n+1}$ such that $\varphi_n^C(c_{n+1})= c_n$.

For instance, to start off, $c_n$ lifts to some $b_n\in B_n$; then by commutativity, $g\varphi_n^B(b_n)=\varphi_n^C g(b_n)=0$, so by exactness we can write $\varphi_n^B(b_n)=f(a_{n-1})$ for some $a_{n-1}\in A_{n-1}$. Do something similar a few more times, using the exactness of $A$ and $B$, and you'll find your $c_{n+1}$.

$\endgroup$
2
  • $\begingroup$ I figured the proof in elements would be doable, but in an Abelian category, it's not a given that we have elements (i.e. that there exists a forgetful functor that preserves isomorphisms). Consider the category of sheaves on a topological space $X$. $\endgroup$ – Monstrous Moonshine Jan 7 '17 at 17:07
  • $\begingroup$ Well, there exists.. ncatlab.org/nlab/show/Freyd-Mitchell+embedding+theorem $\endgroup$ – Berci Jan 7 '17 at 21:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.