# Direct limit of a direct system looking like a cochain complex of objects.

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.

• Don't you just get $A = 0$? – Michael Albanese Mar 26 '16 at 23:02
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.
• 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 ? – Lina45 Mar 26 '16 at 23:12
• 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 $x\sim 0$. – Eric Wofsey Mar 26 '16 at 23:13
• 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 ? – Lina45 Mar 26 '16 at 23:28
• No? How would that mean $A=\coprod A_i$? – Eric Wofsey Mar 26 '16 at 23:47
• $A = \{ ( i , x_i ) \ | \ i \in \mathbb{N} \ \wedge \ x_i \in A_i \ \} = ( A_i )_{ i \in \mathbb{N} }$, no ? – Lina45 Mar 26 '16 at 23:52