1
$\begingroup$

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 $$0\rightarrow \varinjlim A_i\rightarrow\varinjlim B_i\rightarrow\varinjlim C_i\rightarrow 0$$ is short exact.

I wonder what the maps $\varinjlim A_i\rightarrow\varinjlim B_i$ look like. They must arise out of the universal property, mustn't they? We have maps $h_i: B_i\rightarrow\varinjlim B_i$ and thus we get maps $g_i=h_i\circ\Phi_i:A_i\rightarrow\varinjlim B_i$. But in order to use the universal property to get a unique map $\varinjlim A_i\rightarrow\varinjlim B_i$ one must show that $g_i=g_j\circ f_{ji}$ for all $i\le j$. Why does this hold?

$\endgroup$
2
$\begingroup$

Essentially your question is : having maps $\Phi_i : A_i \to B_i$, how do we get a map $\Phi: \varinjlim A_i \to \varinjlim B_i$ ?

This amounts to showing that ($I$ being a small category) when $C$ is a category with $I$-colimits, then the mapping $C^I \to C$, $(A_i)_{i\in I} \to \varinjlim A_i$ can be extended to a functor.

This result holds and we can prove it in this general setting.

Well assume given $(A_i)_{i\in I}, (B_i)_{i\in I}$ with maps $\Phi_i : A_i \to B_i$ such that the associated diagrams commmute. Then as you said we get maps $g_i : A_i \to B$. But also if $i\to j$ is an arrow in $I$, $f_{ij} : A_i \to A_j$, $k_{ij} : B_i\to B_j$ with $\Phi_j \circ f_{ij} = k_{ij}\circ \Phi_i$.

Thus denoting $m_i : B_i \to B$ ($B$ being $\varinjlim B_i$) and $g_i = m_i \circ \Phi_i$, one has $g_j\circ f_{ij} = m_j\circ \Phi_j\circ f_{ij} = m_j\circ k_{ij}\circ \Phi_i$. But as $B$ is the limit of $(B_i)_{i\in I}$, $m_j\circ k_{ij} = m_i$, so that $g_j\circ f_{ij} = m_i\circ\Phi_i =g_i$, which is what we wanted.

Therefore (by the universal property of the limit) there is a unique arrow $A\to B$ making the associated diagrams commute (where $A:=\varinjlim A_i$) (the unicity of the arrow making the diagrams commute suffices to say that this does indeed give a functor)

You could do a way more visual proof, easier to follow with the diagrams drawn in and it would basically amount to diagram chasing but I am not yet good enough in TeX to be able to draw these efficiently. If anyone wants to edit my answer with the righy diagrams, they'd be more than welcome to do so.

$\endgroup$
0

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.