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Stream: learning: questions

Topic: Functors on Set preserve epis without AC

view this post on Zulip Ralph Sarkis (Jan 24 2024 at 14:04):

Do functors on Set\mathbf{Set} preserve epimorphisms (surjections) even without the axiom of choice ? (With the axiom of choice, all epimorphisms are split, and hence are preserved by all functors.)

view this post on Zulip Andrew Swan (Jan 24 2024 at 15:07):

No. You can still show that the epimorphisms in Set\mathbf{Set} are precisely the surjections. For each set II you have a functor (−)I:Set→Set(-)^I : \mathbf{Set} \to \mathbf{Set} given by exponentiation. (−)I(-)^I can only preserve surjections if II is projective, since for any surjection f:X→If : X \to I if f∘−:XI→IIf \circ - : X^I \to I^I is surjective there must exist an element ss of X→IX \to I which is mapped to 1I∈II1_I \in I^I, which is precisely a section of ff.