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Omar Antolín-Camarena
Omar Antolín-Camarena
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Not neccesarilynecessarily. Consider these two functors Set → Set: F is the constant functor with value the empty set, and G is the identity functor. Then F is a subfunctor of G, G is obviously left exact, but F is not: it does not send the terminal object to the terminal object.

Not neccesarily. Consider these two functors Set → Set: F is the constant functor with value the empty set, and G is the identity functor. Then F is a subfunctor of G, G is obviously left exact, but F is not: it does not send the terminal object to the terminal object.

Not necessarily. Consider these two functors Set → Set: F is the constant functor with value the empty set, and G is the identity functor. Then F is a subfunctor of G, G is obviously left exact, but F is not: it does not send the terminal object to the terminal object.

Source Link
Omar Antolín-Camarena
Omar Antolín-Camarena
  • 5.3k
  • 1
  • 36
  • 54

Not neccesarily. Consider these two functors Set → Set: F is the constant functor with value the empty set, and G is the identity functor. Then F is a subfunctor of G, G is obviously left exact, but F is not: it does not send the terminal object to the terminal object.