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What's the number N(n) of directed circle graphs with n verties (a,b,c,d,...) and edges (f,g,h,...)?
For the first two, it's simple to check the number is two:
N(2) = 2
1) Either, say
f : a -> b
g : b -> a
2) or, say
f : a -> b
g : a -> b
N(3)=2
1) The graph where all arrows go in one direction, say
h : c -> a
f : a -> b
g : b -> c
2) The graph where one of the arrows point in another direction, say
h : c -> a
f : a -> b
g : c -> b
so that both f and g point at b.
A short thinking makes clear that a graph where no two arrows point in the same direction isn't possible.
(Bonus: What I tried to find out, in algebra, is if there is a limit that has both pullback and equalizer as special case. I found pic related, which equals the equalizer if h is an isomorphism, but as far as I see there's no condition on h which makes this into the pullback)
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Bump
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>>8089392
Isn't an equalizer already the special case of a pullback when the two arrows have the same source?
On that note, I have a somewhat stupid question (probably too obscure for /sqt/ though):
According to
https://qchu.wordpress.com/2012/09/29/monomorphisms-and-epimorphisms/
the natural map (m: d --> e in OP's example) is stated to always be a monomorphism, meaning that the pullback is a subobject of the product. Intuitively I can see why, but I can't seem to formally prove it from the definition of monomorphism.
In fact by using uniqeness of the universal map I get that mx = my for any two x,y. What am I doing wrong?
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>>8091282
d is not the product here
this is what I know:
when 1 exists,
-the product of 2 objects is the pullback of their unique arrow towards 1
-given two arrows f, g, with same domain, codomain, the equalizer [f, 1_f] and [g, 1_f=1_g] is the pullback of f by g
-if the pullback of two arrows f, g exists, then if f is monic, then g*(f) is monic
this is false when you replace monic by epic
the kernel pair of an arrow f is the pullback of f by itself
-if the kernel pair of f exists, then the projection arrows are epic
f is monic
iff
-its kernel pair exists and it is (dom f, 1_f, 1_f)
iff
its kernel pair exists and the projections are equal
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Def
A locally-small category is finitely-complete — or cartesian or left exact or lex — when it is complete for any functor having it as codomain and a finite category as domain.
theorem
A locally-small category is complete if and only if every product of any collection of objects and every equalizer of any parallel arrows exist.
theorem
For any locally-small category, the following propositions are equivalent
1 it is finitely-complete
2 it has a terminal object, all equalizers and all biproducts
3 it has a terminal object and all pullbacks
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to prove that the fibred product FP is a subobject of the product P, you must prove that FP is an equivalence class of monic arrow with codomain P,
-you know that all pullbacks are equivalent
-now you must find a monic arrow FP >->P.
in sets, this is done in forgetting the equation f.f* = g.g*, so you just project (L, R) and this it.
in category theory, you just use the universality of the product, to get an arrow FP ->P, then I guess that you show it is monic by universality of the FP.
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>>8091417
and to show that it is monic, it is enough to show that it is a section
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>>8091282
>Isn't an equalizer already the special case of a pullback when the two arrows have the same source?
No. Here a counterexample (not due to me):
The equalizer of
! : 2 -> 1
is again 2, while the pullback is 2x2=4
In more detail:
Write 1={0} and 2={a,b} and consider ! as
f: {a,b} -> {0}
g: {a,b} -> {0}
The pullback, on the other hand is the set
{(a,a),(a,b),(b,a),(b,b)}
and the arrows out of it are left and right projection.
Only when you draw the diagram with h=id, you force one of the projections be replaced with the other, and then (a,a) <=> (a,b) and (b,a) <=> (b,b) each melt to one object.
ANYWAY, what about the combinatorical question? - it's mere graph theory.
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>>8091831
but what is your goal with this question?
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