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Suppose $F:C\to D$ is a left adjoint. Let $U:{\mathsf{Cat}}\to {\mathsf{Ab}}$ be the left adjoint to the fully faithful functor ${\mathsf{Ab}}\to {\mathsf{Cat}}$ that views an abelian group as a category with one-object. If $U(C)$ is isomorphic to $U(D)$, is $C$ isomorphic to $D$?

This question probably has something to do with the 2-categorial structure on ${\mathsf{Cat}}$ and ${\mathsf{Ab}}$ and whether $U$ preserves this structure. However I'm not too familiar with 2-category theory.

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It seems to me that $U$ should send a lot of categories to the trivial group. Have you computed, for example, $U$ of the category of sets, or $U$ of the category of topological spaces? Specifically, are you sure that they're not a counterexample for your question? –  Andreas Blass Jun 18 '12 at 14:54
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$F$ doesn't really appear in the question, so you only require that there is some left adjoint functor $C \to D$? Or do you actually mean the following: If $U(F) : U(C) \to U(D)$ is an isomorphism, is the same true for $F$? Have you checked some examples? I doubt that anything along these lines may be true ... –  Martin Brandenburg Jun 18 '12 at 14:59
    
My previous answer is wrong, I will fix it. –  Benjamin Steinberg Jun 18 '12 at 16:00
    
This is trickier than it seems. What happens if we assume F is an equivalence and U(F) gives the iso? –  Benjamin Steinberg Jun 18 '12 at 16:10
    
I put down a new answer. Hope this one works. –  Benjamin Steinberg Jun 18 '12 at 17:14

1 Answer 1

up vote 3 down vote accepted

The universal abelian group for a finite poset is the free abelian group on the edges of the Hasse diagram. Consider the posets $P=\lbrace 0,a,b,1\rbrace$ with $0\lneq a,b\lneq 1$ and $a,b$ incomparable and $Q=\lbrace 0,1,2,3,4\rbrace$ with the usual order. Then the universal group in both cases is then free abelian of rank 4. Consider the map $F\colon P\to Q$ sending $0,a,b$ to $0$ and $1$ to $4$. This is an order-preserving function between complete lattices preserving infs and so has a left adjoint by the adjoint functor theorem which I am too lazy to write down. The posets are not isomorphic.

Added.

The left adjoint seems to be $G(0) = 0$ and $G(j) = 1$ for $j=1,2,3,4$.

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