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**Extra resources for Categories and functors**

**Example text**

T h u s the representation functors are füll and faithful. 8 we realized that the image of a functor is not necessarily a category. T h i s , however, is the case if the functor : <3f —>- 3 is füll and faithful. FB &C) ^ M o r ^ ( 5 , C) and M o r ^ ( J ^ C , &B) ^ M o % ( C , B), there are h : B ~> C and k:C -> B with #"A = \ ^ and — \ ^ . Since tFQik) = \^ —^ \ and ^(kh) = \^ — 3F\ , we get hk = l and kh = \ . T h u s ^g&f = A B B B y B B B c B B T h e füll and faithful functors are most important, as we want to show with the following example.

N — 1) be the projections of the inner produet. L e t { / J be a family of morphisms with common domain B and ranges A . , n — 1) may be factored. F o r h and f , there is exactly one k: B -> (y^ x • • • X ^4 _i) X A with qk = h a n d = / . T h e n / > Ä =f a n d p ^ Ä = , i = 1 n — 1. T h e p ^ , . . , p -iq p are the projections. £ is uniquely determined by the given properties of the factorization. i ± n 1 i n n n n y n n n n Similarly to the proof given above, one can also break up infinite produets; speciflcally, one can split off a single factor by Y[A ^A i xJjAi j iej with /u{/}=J and j$J iej Thus, the produet is independent of the order of the factors up to an isomorphism and is associative.

L e t there be a morphism h for each i e / . Because of the property of a counterimage of § ( Q ) , there is a morphism h for all i. T h e n A exists because (J A is a union. // exists because/((J A ) is an image. T h u s we have a morp h i s m / ((J y^) — Q , fulfilling the conditions of a union. B, \ l I >B A h exists uniquely because f] is an intersection. h exists uniquely such that the diagram becomes commutative, becausef~ (C\ Bi) counterimage. T h u s t h e / ( n B ) is the intersection of t h e / ( ^ ) .