Consider a short exact sequence of directed systems of modules $$0\rightarrow (A_i)\overset{(\Phi_i)}{\rightarrow} (B_i)\overset{(\Psi_i)}{\rightarrow} (C_i)\rightarrow 0,$$ i.e. every $$0\rightarrow A_i\overset{\Phi_i}{\rightarrow} B_i\overset{\Psi_i}{\rightarrow} C_i\rightarrow 0$$ is short exact, $f_{ji}\circ\Phi_i=\Phi_j\circ f_{ji}$ and $f_{ji}\circ\Psi_i=\Psi_j\circ f_{ji}$ for all $i\le j$.
One can show that then $$0\rightarrow \varinjlim A_i\overset{\Phi}{\rightarrow}\varinjlim B_i\overset{\Psi}{\rightarrow}\varinjlim C_i\rightarrow 0$$ is short exact, where $\Phi$ and $\Psi$ are constructed and unique by the universal property.
Recall that for a direct system $(M_i,f_{ji}:M_i\to M_j)$ we have $$\varinjlim M_i=(\bigoplus_i M_i)/U,$$ where $U$ is spanned by $m_i-f_{ji}(m_i)$, $i\le j$.
In our lecture we have proven that $\Phi$ is injective but I have an issue in understanding the proof. It goes like this:
Let $\varinjlim A_i\ni a=\left[\sum_{j=1}^n\lambda_j a_j\right]$ with $\Phi (a)=0$, $a_j\in A_{i_j}$. Since we have a directed system we can find $k\ge i_1,...,i_n$, hence $a=[a_k]$ for a $a_k\in A_k$. By assumption $\Phi(a)=[\Phi_k(a_k)]=0$, hence $\exists N\ge k$ such that $f_{Nk}(\Phi_k(a_k)))=0$. Since $\Phi_N$ is injective and $0=f_{Nk}(\Phi_k(a_k))=\Phi_n(f_{Nk}(a_k))$, get $f_{Nk}(a_k)=0$, thus $a=[a_k]=[f_{Nk}(a_k)]=0$.
My questions are:
- Why do we have $a=[a_k]$ for some $a_k\in A_k$?
- Why is $\Phi(a)=[\Phi_k(a_k)]$ and why is there such an $N$?
- Why is $[a_k]=[f_{Nk}(a_k)]$?
EDIT: For example $\ker\Psi\subset\text{im}\Phi$. Assume we have $\Psi(b)=0$ for $b=[\sum b_{i_j}]$, $b_{i_j}\in B_{i_j}$. If we chose $k\ge i_1,...,i_n$, we get $b=[b_k]$ for a $b_k\in B_k$. Hence $$0=\Psi (b)=\Psi ([b_k])=\Psi (h_k (b_k))=h_k (\Psi_k (b_k))\in\varinjlim C_i,$$ where $h_k$ is the natural map $C_k\to\varinjlim C_i$. Since $h_k (\Psi_k (b_k))\in\varinjlim C_i$, we get $h_k (\Psi_k (b_k))=h_N(c_N)$ for some $N$. But $\Psi_N$ is surjective thus there is a $b_N\in B_N$ such that $h_k (\Psi_k (b_k))=h_N(\Psi_N(b_N))=\Psi (h_N(b_N))$.
Two questions: Is my chosen $N$ greater or equal to $k$? How can I conclude?
You take $a = \displaystyle\sum_{l=1}^n a_{i_l} / U$ (I'll write $/U$ to denote the class mod $U$) where $a_{i_l} \in M_{i_l}$, by definition of the direct sum. Taking $k\geq i_1,...,i_n$, which is possible because $I$ is a directed system. Then $a_{i_l} /U = f_{k i_l}(a_{i_l})/U$ by definition of $U$, so that $a = \displaystyle\sum_{l=1}^n (f_{k i_l}(a_{i_l})/U)$. But this is also $(\displaystyle\sum_{l=1}^n f_{k i_l}(a_{i_l}))/U$, which is $a_k/U$ with $a_k = \displaystyle\sum_{l=1}^n f_{k i_l}(a_{i_l}) \in M_k$. So $a=a_k/U$ for some $k$ and $a_k \in M_k$.
$\Phi(a) = \Phi(a_k /U)$. But $a_k/U$ is the image of $a_k$ in the morphism $\iota_k : M_k \to \operatorname{Lim} M_i$ (sorry I don't know how to write the arrow). So $\Phi(a) = \Phi\circ \iota_k (a_k)$. But the commutativity of the associated diagrams in the definition of $\Phi$ from the $\Phi_i$'s implies that $\Phi\circ \iota_k =j_k \circ \Phi_k$ where $j_k : B_k \to \operatorname{Lim}B_i$ is the canonical morphism.
Thus denoting $V$ the same thing as $U$ but for the $B_i$'s, you get indeed $\Phi(a)= \Phi_k(a_k) /V$.
This is a general property of limits of directed systems : if $m_i / U = 0$, then there exists $N \geq i$ with $f_{Ni}(m_i) = 0$. If you look at the characterization of the limit I gave in the comments, this is quite clear.
I hope I've answered your questions, if you have others, don't hesitate to ask them.
EDIT : I have shown that $\Phi$ was injective. Now to show that the sequence $0\to A\to B \to C\to 0$ is exact, it suffices to show two things : $\operatorname{Ker}\Psi = \operatorname{Im}\Psi$ and $\Psi$ is surjective.
Remember that one has, for a given $N$ a commutative diagram $j_N\circ \Phi_N = \Phi\circ \iota_N$.
So let $b\in \operatorname{Im}\Phi$. $b = \Phi(a)$. But also, $a=\iota_N(a_N)$ for some $N, a_N$ so that $b= j_N\circ\Phi_N(a_N)$. So $\Psi(b) =\Psi\circ j_N\circ \Phi_N(a_N)$.
But looking at the definition of $\Psi$, we get $\Psi\circ j_N = g_N\circ \Psi_N$ where $g_N : C_N\to C$. Therefore, $\Psi(b) = g_N\circ \Psi_N\circ\Phi_N(a_N)$. Since $0\to A_N\to B_N\to C_N\to 0$ is exact, we get $\Psi(b) = 0$, which shows $\operatorname{Im}\Phi\subset \operatorname{Ker}\Psi$.
As for the converse, assume $\Psi(b) = 0$. Write $b= b_k/V$, so that $\Psi\circ j_k(b_k) = 0$ . Again, looking at the definition of $\Psi$ and the commutative squares that go with it, we get $g_k\circ \Psi_k(b_k) = 0$.
Denoting $w_{Nk} : C_k\to C_N$ when $N\geq k$, this last equality implies the existence of $N\geq k$ with $w_{Nk}\circ \Psi_k(b_k) = 0$ (with the characterization of the limit I gave in the comments).
The commutative diagrams that are given in the hypothesis give $\Psi_N\circ g_{Nk}(b_k)=0$ where $g_{Nk} : B_k\to B_N$ when $N\geq k$.
Since the sequence at index $N$ is exact, we get $g_{Nk}(b_k) \in \operatorname{Im}\Phi_N$. Write it as $g_{Nk}(b_k) = \Phi_N(a_N)$. Applying $j_N$ leads to $j_N\circ g_{Nk}(b_k) = j_N\circ \Phi_N(a_N) = \Phi\circ \iota_N(a_N)$ (this last equality holding because of the commutatuve diagrams in the definition of $\Phi$).
So $j_N\circ g_{Nk}(b_k) \in \operatorname{Im}\Phi$. But this is $g_{Nk}(b_k)/V = b_k/V = b$.
Therefore $\operatorname{Ker}\Psi\subset \operatorname{Im}\Phi$, which, in conclusion, give us $\operatorname{Ker}\Psi = \operatorname{Im}\Phi$.
The same sort of abstract nonsense leads to the surjectivity of $\Psi$ (I'm a bit tired so if I made any typo or serious mistake, or left some nontrivial details out in the above proof, please let me know)
2nd EDIT : I'm adding this image for a bit more (hopefully) clarity. The notations are a bit unfortunate but those are the ones I went with so might as well go all the way.
Hopefully this diagram will help you see the situation a bit more clearly : the "circle arrows" indicate that the cell commutes. The outer cells commute by definition of a direct limit, the inner lower cell commutes by definition of a directed system, and the inner upper cell commutes by definition of $\Psi$ (it was defined to be the unique arrow making it commute). What matters for the surjectivity of $\Psi$ is actually the commutativity of the upper inner cell, but drawing the whole diagram seemed a good way of summing up the situation.
What I'm saying in the comments is that if $c\in \varinjlim C_i$, then $c = g_j(c_j)$ for some $j$: but $\Psi_j$ is surjective, so $c = g_j\circ \Psi_j (b_j)$ for some $b_j$. According to the diagram I've drawn, $g_j\circ\Psi_j = \Psi\circ j_j$, so thqt $c\in \operatorname{Im}\Psi$