Download A Gentle Introduction to Category Theory - the calculational by Fokkinga, M.M.; Jeuring, J.T.; Fokkinga, Maarten M PDF

By Fokkinga, M.M.; Jeuring, J.T.; Fokkinga, Maarten M

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Example text

19 [Self] and [Uniq] ≡ [Charn] where the square brackets denote the universal quantification that was implicit in the formulations above. 17. 20 Well-formedness condition. Frequently we encounter the situation that there is a category A and another one, B say, that is built upon A . Then the well-formedness condition for the notation ([B])B (where B is a composite entity in the underlying category A ) is the condition that B an object in B ; this is not a purely syntactic condition. ) convention that in each law the free variables are quantified implicitly in such a way that the well-formedness condition, the premise of init-Type, is met.

Prove that p is a coequaliser of (h ; f, h ; g) . Exercise: let pi be a coequaliser of a pair (fi , gi ) , for i = 0, 1 . Prove that p0 + p1 is a coequaliser of (f0 + f1 , g0 + g1 ) , assuming that sums exist. 35 Interpretation in Set . Take A = Set , the default category, and let A be a set. Each parallel pair (f, g) with target A determines a binary relation Rf,g on A , and conversely, each binary relation R on A determines a parallel pair (fR , gR ) in the following way: Rf,g fR gR = = = {(f x, gx) | x ∈ src f } exl exr ⊆A×A : {(x, y) | x R y} → A : {(x, y) | x R y} → A .

The mappings + and × are bifunctors: id + id = id and f + g ; h + j = (f ; h) + (g ; j) , and similarly for × . Throughout the text we shall use several properties of product and sum. These are referred to by the hint ‘product’ or ‘sum’. Here is a list; some of these are just the laws presented before. f × g ; exl f ∆ g ; exl f × g ; exr f ∆ g ; exr f ;g∆h exl ∆ exr (h ; exl ) ∆ (h ; exr ) f ∆g ;h×j f ×g ;h×j f ∆g =h∆j = = = = = = = = = ≡ exl ; f f exr ; g g (f ; g) ∆ (f ; h) id h (f ; h) ∆ (g ; j) (f ; h) × (g ; j) f =h∧g =j inl ; f + g inl ; f ∇ g inr ; f + g inr ; f ∇ g f ∇g ;h inl ∇ inr (inl ; h) ∇ (inr ; h) f +g ;h∇j f +g ;h+j f ∇g =h∇j = = = = = = = = = ≡ f ; inl f g ; inr g (f ; h) ∇ (g ; h) id h (f ; h) ∇ (g ; j) (f ; h) + (g ; j) f =h∧g =j Exercise: identify the laws that we’ve seen already, and prove the others.

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