This is a follow-up to this question. Since an abelian category cannot be cartesian closed, clearly the hom functor is not right adjoint to the product (by an object). However, does the product (by an object) admit a right adjoint for some objects? If so, what is it? Does it exist in general? Does it only hold for finite products?
1 Answer
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In an additive category the functor F(–) = – × A = – ⊕ A cannot have a right adjoint unless A = 0. If F had a right adjoint then it would preserve coproducts and in particular A = F(0) = F(0 ⊕ 0) = F(0) ⊕ F(0) = A ⊕ A via the fold map. This means Hom(A, K) = Hom(A, K) × Hom(A, K) for every K, but Hom(A, K) is nonempty (we have zero maps) so Hom(A, K) = • and thus A = 0 by Yoneda.
answered Dec 31, 2009 at 17:41
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3$\begingroup$ And Reid Barton scores again! $\endgroup$ Commented Dec 31, 2009 at 17:42
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$\begingroup$ There are nontrivial sets which are isomorphic to their product with themselves. What about if A and A^2 are isomorphic, e.g. if A is an infinite direct sum of isomorphic objects? $\endgroup$ Commented Dec 31, 2009 at 17:46
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2$\begingroup$ Right, that's why I said "via the fold map". The equality Hom(A, K) = Hom(A, K) x Hom(A, K) is being induced by the diagonal (I was a little sloppy in the notation). $\endgroup$ Commented Dec 31, 2009 at 17:49
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3$\begingroup$ Ah, right. One way to say this is that product with A, being a left adjoint, should preserve the colimit of the empty diagram (which is the initial object of the category). $\endgroup$ Commented Dec 31, 2009 at 17:53
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6$\begingroup$ Perhaps "Did you try taking the colimit of the empty diagram?" should go in the FAQ somewhere. $\endgroup$ Commented Dec 31, 2009 at 18:56