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I would like to ask you about a special kind of direct systems $ (A_i, f_{i}^{j} )_{ i,j \in ( I , \leq ) } $ looking like a cochaîn complex $ (A_i , f_{i}^{j} )_{ i,j \in ( \mathbb{N}^* , \leq ) } $ such that : $ f_{i+1}^{i+2} \circ f_{i}^{i+1} = 0_{A_{i}} $ for all $ i \in \mathbb{N}^* $.

How do we obtain generally and explicitly his direct limit : $ A = \displaystyle \lim_{ \to } A_i $ ? Can you give me an example showing me that ?

I would like to ask you the same question but, replacing a direct limit by an inverse limite, and a cochain complex by a chain of complex.

Thanks in advance for your help.

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    $\begingroup$ Don't you just get $A = 0$? $\endgroup$ – Michael Albanese Mar 26 '16 at 23:02
  • $\begingroup$ Thank you, but how to prove that ? $\endgroup$ – Lina45 Mar 26 '16 at 23:04
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The direct limit is always $0$. More generally, suppose you have a direct system $(A_i, f^j_i)$ such that for each $i$ there is some $j\geq i$ such that $f^j_i=0$ (in your case, you can take $j=i+2$). Then $\varinjlim A_i=0$. Indeed, given any object $B$, a map from the direct system to $B$ consists of maps $g_i:A_i\to B$ for each $i$ such that $g_i=g_jf^j_i$ whenever $i\leq j$. Choosing $j$ such that $f^j_i=0$, you get that $g_i=0$ for all $i$. It follows that every map from the direct system to any object factors uniquely through the $0$ object, so $0$ is the direct limit.

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  • $\begingroup$ Thank you very much Sir. :-) $ A $ looks like : $ \Big( \coprod A_i / \Big) \sim $ up to isomorphism, no ? What is the significance of : $ \Big( \coprod A_i \Big) / \sim \ = 0 $ in this case ? $\endgroup$ – Lina45 Mar 26 '16 at 23:12
  • $\begingroup$ The point is that the equivalence relation kills everything. The equivalence relation says that if $x\in A_i$, then $x\sim f^j_i(x)$ for any $j\geq i$. But choosing $j$ such that $f^j_i=0$, this says $x\sim 0$. $\endgroup$ – Eric Wofsey Mar 26 '16 at 23:13
  • $\begingroup$ Does it mean that : $ A = \coprod A_i $ since there is only one equivalence class $ A = \{ \overline{0} \} = \Big( \coprod A_i \Big) / \sim $, so : $ A = \coprod A_i $, no ? $\endgroup$ – Lina45 Mar 26 '16 at 23:28
  • $\begingroup$ No? How would that mean $A=\coprod A_i$? $\endgroup$ – Eric Wofsey Mar 26 '16 at 23:47
  • $\begingroup$ $ A = \{ ( i , x_i ) \ | \ i \in \mathbb{N} \ \wedge \ x_i \in A_i \ \} = ( A_i )_{ i \in \mathbb{N} } $, no ? $\endgroup$ – Lina45 Mar 26 '16 at 23:52

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