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Let $\mathcal{C}$ be a category and $\{A_i,f_{ij}:A_i\rightarrow A_j\},\{B_i,g_{ij}:B_i\rightarrow B_j\}$ be two directed systems such that their direct limits exists. Furthermore, assume that there are morphisms $A_i\xrightarrow{\varphi_i} B_i$ for all $i$.

Do we then have always an obvious morphism $\varinjlim A_i\rightarrow \varinjlim B_i$?

Suppose we also have following commutative diagram

enter image description here

As direct limit of $B_i$ exists, there are maps $\eta_i:B_i\rightarrow \varinjlim B_i, \eta_j:B_j\rightarrow \varinjlim B_i$ such that $\eta_i=\eta_j\circ f_{ij}$.

We have following commutative diagram

enter image description here

This gives maps $\eta_i\circ \varphi_i:A_i\rightarrow \varinjlim B_i$ and $\eta_j\circ \varphi_j:A_j\rightarrow\varinjlim B_i$. We have $g_{ij}\circ \varphi_i=\varphi_j\circ f_{ij}$. Which omplies that $\eta_j\circ g_{ij}\circ \varphi_i=\eta_j\circ \varphi_j\circ f_{ij}$ i.e., $\eta_i\circ \varphi_i=\eta_j\circ \varphi_j\circ f_{ij}$.

Thus, by universal property of direct limits, there exists $\Phi:\varinjlim A_i\rightarrow \varinjlim B_i$ such that $\eta_i\circ \varphi_i=\Phi\circ \tau_i$ where $\tau_i:A_i\rightarrow \varinjlim A_i$ comes from definition of direct limit. So, we have a morphism on direct limits with following commutative diagram

Read above $\tau_j$ as $\eta_i$. enter image description here

Any comments on this map is welcome. Do we have similar map if I do not have that commutative diagram condition in case of $\varphi_i:A_i\rightarrow B_i$?

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  • $\begingroup$ There's just a small typo at the end of the sentence after the first diagram: you should have $\;\eta_i\circ \varphi_i=\eta_j\circ \varphi_{\color{red}j}\circ f_{ij}$. $\endgroup$ – Bernard Sep 12 '17 at 15:07
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    $\begingroup$ If you just have a random collection of morphisms between objects rather than a morphism of diagrams, IMO it would be pretty weird if they had any reasonable relation to a morphism of colimits, except in trivial cases. $\endgroup$ – Hurkyl Sep 12 '17 at 15:07
  • $\begingroup$ @Bernard You mean second diagram. I have edited that :) Thanks $\endgroup$ – user312648 Sep 12 '17 at 17:23
  • $\begingroup$ By morphism of diagrams you mean morphisms of directed sets i.e., with the commutative diagrams that i have specified.? @Hurkyl Are you saying we should assume they come with commutative diagrams? $\endgroup$ – user312648 Sep 12 '17 at 17:25
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    $\begingroup$ Yes, in order to get a canonical morphism $\varinjlim A_i\rightarrow\varinjlim B_i$, you need that the $\varphi_i:A_i\rightarrow B_i$ commutes with the $f_{ij}:A_i\rightarrow A_j$ (ie $\varphi_j\circ f_{ij}=g_{ij}\circ\varphi_i$). $\endgroup$ – Roland Sep 15 '17 at 21:18

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