31
$\begingroup$

I've heard that taking direct limits is an exact functor in the category of modules, and I'm trying to figure out why, as I couldn't find a proof.

Suppose you have homomorphisms $\varphi_i: K_i\to N_i$ and $\psi_i: N_i\to M_i$ for $(K_i,h^i_j)$, $(N_i,g^i_j)$, and $(M_i,f^i_j)$ directed systems of modules such that $0\to K_i\to N_i\to M_i\to 0$ is exact for every $i$.

Why is $0\to\varinjlim K_i\to\varinjlim N_i\to\varinjlim M_i\to 0$ also exact?


So I let $\varphi:\varinjlim K_i\to\varinjlim N_i$ and $\psi:\varinjlim N_i\to\varinjlim M_i$ be the natural homomorphisms. Take $x\in\ker\psi$. Then $x=g^i(x_i)$ for some $x_i\in N_i$. Then $0=\psi(g^i(x_i))=f^i(\psi_i(x_i))$. I know there exists some $j\geq i$ such that $f^i_j(\psi_i(x_i))=0$ in $M_j$. But $f^i_j\circ\psi_i=\psi_j\circ g^i_j$, so $g^i_j(x_i)\in\ker\psi_j=\text{im}(\varphi_j)$. Then $g^i_j(x_i)=\varphi_j(y_j)$ for some $y_j\in K_j$, so $$ x=g^i(x_i)=g^j(g^i_j(x_i))=g^j(\varphi_j(y_j))=\varphi(h^j(y_j)) $$ and so $\ker\psi\subseteq\text{im}\varphi$.

Conversely, suppose $x\in\text{im}\varphi$. Then $x=\varphi(y)$ for some $y=h^i(y_i)$ and $y_i\in K_i$. So $x=\varphi(h^i(y_i))=g^i(\varphi_i(y_i))$. Thus $$ \psi(x)=\psi(g^i(\varphi_i(y_i)))=f^i(\psi_i(\varphi_i(y_i)))=0 $$ since $\psi_i\circ\varphi_i=0$. Then $\ker\psi=\text{im}\varphi$. (Please let me know if I've written nonsense, too many maps can cause me to get lost!)

What is bugging me is, is $\varphi$ injective and $\psi$ surjective to see that the short exact sequence is in fact exact? Is there some obvious fact I'm missing? If possible, is there an explanation in the same vein as the above (i.e. using the maps and manipulating the elements without relying on more general facts from category theory? I'm not too knowledgeable about the latter.) Thanks.

$\endgroup$
3
  • 2
    $\begingroup$ All colimits (and so direct limits in particular) commute with colimits (and so cokernels in particular), thus $\psi$ is automatically an epimorphism. You do need to prove that $\phi$ is a monomorphism though, but this is straightforward enough. $\endgroup$ – Zhen Lin Mar 16 '12 at 23:43
  • $\begingroup$ Thanks @ZhenLin. If $x\in\ker\varphi$, then $x=h^i(x_i)$ for some $x_i\in K_i$. Then $$\varphi(x)=\varphi(h^i(x_i))=g^i(\varphi_i(x_i))=0.$$ This last equality implies $g^i_j(\varphi_i(x_i))=0$ for some $j>i$, so $\varphi_i(x_i)\in\ker g^i_j$. Is there a way to conclude $x_i=0$ or something, or have I gone off track? $\endgroup$ – Jacqueline Pauwels Mar 17 '12 at 0:03
  • $\begingroup$ You know $g^i_j(\varphi_i(x_i)) = 0$, so $\varphi_j(h^i_j(x_i)) = 0$, and you know $\varphi_j$ is injective. $\endgroup$ – Zhen Lin Mar 17 '12 at 9:59
24
$\begingroup$

It suffices to show that if $K_i\to M_i\to N_i$ is exact at $M_i$ for each $i$ (and the appropriate squares commute), then $\varinjlim K_i\to \varinjlim M_i\to \varinjlim N_i$ is exact at $\varinjlim M_i$.

(To get that $\varinjlim$ sends short exact sequences to short exact sequences from this, simply apply the argument to $0\to K_i\to M_i$, exact at $K_i$; to $K_i\to M_i\to N_i$; and then to $M_i\to N_i\to 0$).

Call the first map $f_i$ (with induced map of limits $f$), the second $g_i$ (with induced map $g$); use $\kappa$, $\mu$, and $\nu$ for the structure maps.

Given $[(k,i)]$, we have $g(f([(k,i)])) = g([f_i(k),i]) = [g_i(f_i(k)),i] = [0,i]$, so the composition is trivial. That is, $\mathrm{Im}(f)\subseteq \mathrm{Ker}(g)$.

Now assume that $g([(m,i)]) = [(0,j)]$. Then there exists $t\geq i$ such that $\nu_{it}(g_i(m)) = 0$; hence $g_t(\mu_{it}(m)) = 0$, so by exactness of the original diagram we know that there exists $k\in K_t$ such that $f_t(k) = \mu_{it}(m)$. Therefore, $$f([k,t]) = [(f(k),t)] = [(\mu_{it}(m),t)] = [(m,i)],$$ so $[(m,i)]$ lies in the image of $f$. Thus, $\mathrm{Im}(f)\supseteq \mathrm{Ker}(g)$, proving equality.

Now, this proves that $\varinjlim$ is exact; as to "why" it is exact (is there some deep why it is exact)? I don't know if I can answer.

$\endgroup$
5
  • 2
    $\begingroup$ I feel the left exactness is really "mysterious" as it does not follow from category. The dual property, the right exactness of inverse limit, does not hold. $\endgroup$ – Yifeng Huang Jan 23 '17 at 19:58
  • $\begingroup$ @Y.H. : would you have an example showing that the right exactness of inverse limit (of modules) does not hold? $\endgroup$ – Watson Oct 2 '17 at 19:56
  • $\begingroup$ @Watson: You should ask that as a question (with perhaps a pointer to this comment); you are replying to a comment that is 9 months old, and the query is probably of general interest. $\endgroup$ – Arturo Magidin Oct 3 '17 at 6:15
  • $\begingroup$ @ArturoMagidin : actually I've just noticed that it was already asked. Sorry for the inconvenience. $\endgroup$ – Watson Oct 3 '17 at 21:01
  • $\begingroup$ @a-lawliet: The equivalence class of the element $(k,i)$ in the construction of the direct limit. $\endgroup$ – Arturo Magidin Oct 27 '19 at 17:12

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.