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Given that a Quiver (also known as directed multigraphs) is so important in category theory, being that every Quiver produces a free Category, and every Category has an underlying Quiver, I was wondering what a limit of a Quiver is. (See also the nice overview of Quivers and Diagrams in the answer on finding the limits of a diagram)

So if I take a category with two objects E and V and two functions s,t: E → V which given the start and end edges of vertices, then I should be able to have a functor from that to sets, which determines a quiver. Then all I need is to add a constant functor to some object of set say $L$ - the putative limit - and then find the real one. So for each vertex of the diagram, $v$, we have an arrow $p_v:L\to D(v)$. In this case it would be have the equations $p_V = s \circ p_E$, $p_V = t \circ p_E$. Which leads to $s \circ p_E = t \circ p_E$.

What would the co-version be? I suppose we turn the arrows around $p_v: D(v) \to L$ giving us $p_E = p_V \circ s$ and $p_E = p_V \circ t$, and so $p_V \circ s = p_V \circ t$

(I'd like to also ask what Monad one could build with the Free and Underlying Functors, but I guess that should be a different question).

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  • $\begingroup$ To be clear, you're just asking for the (co)limit of a functor from your two-object category to $\mathtt{Set}$? $\endgroup$ – Eric Wofsey Aug 26 '18 at 15:42
  • $\begingroup$ Two object and two arrows. Two objects alone would give us (co)products. $\endgroup$ – Henry Story Aug 26 '18 at 15:44
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(Co)limits of this form are called (co)equalizers. In sets, an equalizer of a functor $D$ from your category is just given by the set $L=\{x\in D(E): D(s)(x)=D(t)(x)\}$ with its inclusion map to $D(E)$. (So, in terms of quivers, it is just the set of loops.) Coequalizers are a bit more complicated, but a coequalizer can be constructed as the quotient of $D(V)$ by the smallest equivalence relation $\sim$ such that $D(s)(x)\sim D(t)(x)$ for all $x\in D(E)$. (In terms of quivers, this quotient is the set of connected components of the quiver.)

Note that colloquially, one typically just speaks of the (co)equalizer of a pair of morphisms $f,g:A\to B$ in some category, without talking about a functor explicitly. This refers to the (co)limit of the functor from your category which sends $E$ to $A$, $V$ to $B$, and $s$ and $t$ to $f$ and $g$.

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  • $\begingroup$ Yes, that is quite neat. Having written down my question as above I was thinking this looked very much like (co)equalisers :-) Thinking about them in terms of quivers will I think make them more intuitive for me. $\endgroup$ – Henry Story Aug 26 '18 at 16:25
  • $\begingroup$ Interestingly this means that what corresponds to products here are looping arrows which are I suppose the most independent form of arrows, a bit like products bring two things together but in the most independent way (orthogonally). And co-products that are disjoint unions here correspond to groups of nodes that are somehow linked up, each group constituting such a disjoint set. $\endgroup$ – Henry Story Aug 27 '18 at 11:30

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