Let C^ = Cop to Set, for a natural transformation f:X-Y in C^, how to prove if f is an epimorphism, then fc is surjective for all objects c in C? Anyone can help me with that? Thank you in advance:)
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3 Answers
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It's a good question - possibly homework? - but good homework anyway.
One thing that makes it a good question is that if you change "epimorphism" to "monomorphism" and "surjective" to "injective" then it becomes much easier: you can solve it directly by testing at representables. (In other words, apply the definition of monomorphism to natural transformations from C(-, c) to X, using the Yoneda Lemma.)
But for the question posed, I think you need to apply some fairly serious theory. That's a bit surprising for something that seems so simple, but I don't know any way around it. Andreas's solution gives one very neat method. A slightly different method, which doesn't involve Kan extensions, is to use the theorem that "limits and colimits in a presheaf category are computed pointwise". This roughly speaking means that the forgetful functor from your presheaf category to $\mathbf{Set}^{\mathrm{op} \mathbf{C}}$ creates limits and colimits. And, as in Andreas's solution, you have to apply the (much simpler) fact that a map is an epimorphism if and only if a certain square associated to it is a pushout.
You can find your question on problem sheet 9 of this course. Brief hints are also given there. The relevant parts of the associated notes are 4.1.31, 5.1.6 and 5.1.8.
answered Jan 22, 2011 at 20:08
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$\begingroup$ Thank you for your answer. I think I can prove if a morphism preserves colimits then it preserves epis. but still not sure how to relate natural transformation f to fc(morphism in set). Can you explain more about that? Thanks. $\endgroup$ Commented Jan 22, 2011 at 23:43
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$\begingroup$ A piece of evidence for suggestion that this is not entirely trivial is that if you replace Set by another category K it may not be true. By Tom's argument it will be true if K is has pushouts, but there are examples of categories K and C for which the evaluation maps $ev_c:[C,K]\to K$ do not always preserve epimorphisms. $\endgroup$ Commented Jan 23, 2011 at 23:25
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Evaluation at the object c is a functor from C^ to Set that has adjoints on both sides (Kan extensions), so it preserves colimits (among many other things). Epimorphisms can be characterized in terms of colimits (when these exist, as they do in the categories under consideration): A morphism f is an epimorphism iff its pushout along itself consists of a pair of isomorphisms. (Apologies for this overkill where surely a simpler argument is possible.)
answered Jan 22, 2011 at 17:44
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What is the definition of epimorphism you are using here? Right-cancellable presheaf-morphism lead to right-cancellable functions on component "by definition", I think..
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3$\begingroup$ A morphism e in a category is called an epimorphism if it has the right cancellation property: xe = ye implies x = y. I don't think it's as simple as "by definition". $\endgroup$ Commented Jan 22, 2011 at 19:53