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The problem is as follows:
Prove that if$\ \ \ T:{}_R\mathrm {Mod}\rightarrow \mathrm{Ab}$ is an additive left exact functor preserving direct products,then $T$ preserves inverse limits.
Suppose $\{M_i\}$ are ${}_R\mathrm {Mod}$ indexed by $\Gamma$ then we have an exact sequence$$0\rightarrow \lim_{\leftarrow} M_i\rightarrow \Pi_{i\in \Gamma}M_i\rightarrow \Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ M_{ij}$$ Since $T$ is left exact,we have $$0\rightarrow G(\lim_{\leftarrow} M_i)\rightarrow G(\Pi_{i\in \Gamma}M_i)\rightarrow G(\Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ M_{ij})$$is left exact.
Also we have $$0\rightarrow \lim_{\leftarrow} G(M_i)\rightarrow \Pi_{i\in \Gamma}G(M_i)\rightarrow \Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ G(M_{ij})$$is left exact.
How could I get $$\lim_{\leftarrow}G(M_i)\cong G(\lim_{\leftarrow} M_i) $$ via the second and the third exact sequence?Since the corresponding last two ${}_R\mathrm Mod$ are isomprphic because of preserving direct products, do I need to use short five lemma?

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You're doing too much work. It's a general fact that in a complete category all limits can be built out of products and equalizers. In a complete additive category equalizers can be computed using kernels, so all limits can be built out of products and kernels. Hence an additive functor between complete additive categories preserves all limits (not just inverse limits) iff it preserves products and kernels.

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  • $\begingroup$ Yes, all limits can be built out of products and limits. That's why I said the two objects are limits (of the corresponding objects). But are the kernel of isomorphic objects also isomorphic? That's why I wrote an answer with too much work.My teacher taught me that the diagram commute without explanation and I doubted it.When I wrote it down,I found it not that obvious.So I just post the answer to see if there are ways to help me understand its commutation. $\endgroup$ – user12580 Nov 3 '15 at 7:54
  • $\begingroup$ @Mr.Rock: taking the kernel is a functor from the arrow category $\text{Arr}(C)$ to $C$. In particular, it sends isomorphisms in the arrow category to isomorphisms in $C$. $\endgroup$ – Qiaochu Yuan Nov 3 '15 at 7:58
  • $\begingroup$ Er…I don't reach that far.I'm not familiar with the arrow category and functor category. Any hints on why the diagram commutes? Thank you. $\endgroup$ – user12580 Nov 3 '15 at 8:45
  • $\begingroup$ I've got your idea. Thank you. $\endgroup$ – user12580 Nov 3 '15 at 14:26
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We denote $M_{ij}=M_i$ for every $i,j\in \Gamma$ with $i\leqslant j$ and the morphism $\varphi^j_i:M_j\rightarrow M_i$.
Suppose $\{\Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ M_{ij},\eta_{ij}\} $ is the product of $\{M_{ij}\}$ , $\{\Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ G(M_{ij}),t_{ij}\}$ is the product of $\{G(M_{ij})\}$,$\{\Pi_{i\in \Gamma}M_i,p_i\}$ is the product of $\{M_i\}$ and $\{\Pi_{i\in \Gamma}G(M_i),q_i\}$ is the product of $\{G(M_i)\}$.

We have the following commutative diagrams: $$\begin{array} AM_{ij}\\ \uparrow{\eta_{ij}} &\nwarrow{p_i-\varphi^j_ip_j} \\ \Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ M_{ij} & \stackrel{\phi}{\leftarrow} & \Pi_{i\in \Gamma}M_i \end{array}$$ $$\begin{array} AG(M_{ij})\\ \uparrow{t_{ij}} &\nwarrow{q_i-G(\varphi^j_i)q_j} \\ \Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ G(M_{ij}) & \stackrel{\phi'}{\leftarrow} & \Pi_{i\in \Gamma}G(M_i) \end{array}$$$$\begin{array} AG(M_{i})\\ \uparrow{q_i} &\nwarrow{G(p_i)} \\\Pi_{i\in \Gamma}G(M_i) & \stackrel{l}{\leftarrow} & G(\Pi_{i\in \Gamma} M_i) \end{array}$$$$\begin{array} AG(M_{ij})\\ \uparrow{t_{ij}} &\nwarrow{G(\eta_{ij})} \\ \Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ G(M_{ij}) & \stackrel{r}{\leftarrow} & G(\Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ M_{ij}) \end{array}$$where $l$ and $r$ are isomorphisms.
Combine the four diagrams we have $$t_{ij}\phi'l=(q_i-G(\varphi^j_i)q_j)l=G(p_i)-G(\varphi^j_i)G(p_j)$$$$t_{ij}rG(\phi)=G(\eta_{ij})G(\phi)=G(\eta_{ij}\phi)=G(p_i-\varphi^j_ip_j)$$So $$t_{ij}\phi'l=t_{ij}rG(\phi)$$Thus$$\phi'l=rG(\phi)$$It suffices to say the following diagram commutes: $$ \begin{array} AG(\Pi_{i\in \Gamma}M_i) & \stackrel{G\phi}{\longrightarrow} & G(\Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ M_{ij}) \\ \downarrow{l} & & \downarrow{r} \\ \Pi_{i\in \Gamma}G(M_i) & \stackrel{\phi'}{\longrightarrow} & \Pi_{\stackrel{i,j\in \Gamma}{i\leqslant j}}\ \ G(M_{ij}) \end{array}$$ Proposition Given a commutative diagram with exact rows,$$ \begin{array} A0 &\longrightarrow & A'&\stackrel{i}{\longrightarrow} & A&\stackrel{j}{\longrightarrow}&A''\\ &&&& \downarrow{g}&& \downarrow{h} \\ 0& \stackrel{}{\longrightarrow} & B'&\stackrel{j}{\longrightarrow}&B&\stackrel{q}{\longrightarrow}&B'' \end{array}$$ there exists a unique map $f:A'\longrightarrow B'$ making the augmented diagram commute.Moreover, $f$ is an isomorphism if $g$ and $h$ are isomorphisms.
We've got all we need to prove $$\lim_{\leftarrow}G(M_i)\cong G(\lim_{\leftarrow} M_i) $$

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  • $\begingroup$ How do you know that $0 \to F(\lim_{\leftarrow} M_i) \to F(\prod_i M_i) \to F \left( \prod_{i \leq j} M_{ij} \right)$ is exact? $\endgroup$ – Blazej Sep 3 '18 at 22:00
  • $\begingroup$ I assume the functor is left exact. This answer was written when I was a beginner at category theory and I keep it to remind me of what I was week when first touching this topic. Just ignore this answer. $\endgroup$ – user12580 Sep 4 '18 at 8:13
  • $\begingroup$ Dear user12580, I am now trying to prove this theorem for myself. I understand that left exactness gives you exactness of $0 \to F(\lim_{\leftarrow} M_i ) \to F (\prod_i M_i)$, but how do you infer exactness at the "last step" of this sequence? $\endgroup$ – Blazej Sep 4 '18 at 15:21
  • $\begingroup$ See for the definition of left exact functors. What is a left exact functor in your mind? In my opinion, it is a good exercise to show several definitions are equivalent (many notions have several equivalent definitions, though some are nontrivial, most of them is just a reformulation ). $\endgroup$ – user12580 Sep 5 '18 at 2:18
  • $\begingroup$ Dear user, apparently I got confused with the definition of left exactness. $\endgroup$ – Blazej Sep 10 '18 at 16:53

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