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

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**Extra resources for A Gentle Introduction to Category Theory - the calculational approach**

**Sample text**

In Set , an isomorphism is a bijective function, a monomorphism is an injective function, and an epimorphism is a surjective function, and vice versa. ) So, in Set a morphism is an isomorphism if and only if it is both monic and epic. This does not hold in general: in the category suggested by • −→ • (containing three morphisms in total), the non-identity morphism is both epic and monic, and not an isomorphism. A morphism has at most one inverse. For suppose that f : A → B has inverses g, h: B → A .

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.

Here is a counterexample. Let A be the category determined by: A • f ✲ B • g ✲ C • and put h = f ; g . Then these are diagrams in A : A • f f ✲ ✲ B • A • ✡ f ✲ B • h g ✲ C • ✗✠ In the left diagram there are two edges (morphisms) from A to B , whereas in A there is only one. In the right diagram there are two edges (morphisms) from A to C , labelled f ; g and h respectively, whereas in A there is only one; by definition h = f ; g . Extreme cases of diagrams are diagrams with zero, one, or more nodes only, and no edges at all.