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$\mathrm{Hom}(M,F)$ can't be determined by the underlying sets of $M,F$? where $F$ is a free module, $M$ is not a free module.

The question arises from the claim: let $G : \mathbf{Mod}_R\to\mathbf{ Set}$ be the forgetful functor which assigns to each $R$- module its underlying set. Then the functor $G$ does not have a right adjoint. About the first paragraph, I was considering if $F:\mathbf{ Set}\to\mathbf{Mod}_R$ (send a set $X$ to the free $R$-module generated by the elements of $X$) can be the right adjoint. If so, then $Hom(G(M),X)\cong Hom(M,F(X))Hom(G(M),X)\cong Hom(M,F(X))$, which induces my question on the first paragraph.

Anyway, can you show me why the right adjoint doesn't exist, thanks.

6666
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  • Why would you expect $\mathrm{Hom}(M,F)$ to be determined by the underlying sets of $M$ and $F$? The definition of module homomorphism depends on the module structure. So if $M'$ is a different module with the same underlying set as $M$, of course $\mathrm{Hom}(M,F)$ and $\mathrm{Hom}(M',F)$ will be different... – Alex Kruckman Apr 11 '17 at 17:31
  • @AlexKruckman thanks, can you give an example of what you said? – 6666 Apr 11 '17 at 17:51

1 Answers1

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"Free" things are generally left adjoints. I'm not sure what you're trying to do with your first paragraph, but if $G$ was to have a right adjoint, there's no reason that it would be a free module of any sort.

At any rate, a general and usually effective method to show that a functor is not a left adjoint is to show that it isn't cocontinuous, i.e. that it doesn't preserve colimits.

In this case it's very easy. The zero (and thus initial) object of $\mathbf{Mod}_R$ is the trivial module. $G$ should send that to $\emptyset$ if $G$ was a left adjoint, but instead it sends it to $\{0\}$ (as you would expect as $G$ is a right adjoint and thus must preserve limits).

Derek Elkins left SE
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  • About the first paragraph, I was considering if $F:\mathbf{Set}\to \mathbf{Mod}_R$ (send a set $X$ to the free $R$-module generated by the elements of $X$) can be the right adjoint. If so, then $Hom(G(M),X)\cong Hom(M,F(X))$, which induces my question on the first paragraph. – 6666 Apr 10 '17 at 16:58
  • I am confused why the underlying set of the trivial set is empty set not ${0}$? – 6666 Apr 10 '17 at 17:00
  • And I don't expect $G$ to be a right adjoint, I just want to find a right adjoint of it. – 6666 Apr 10 '17 at 17:02
  • A zero object is an object that is both an initial and a terminal object. The zero object for $\mathbf{Mod}_R$ is the trivial module consisting of $0$ and nothing else. If $G$ was a left adjoint (i.e. had a right adjoint) then it would take initial objects to initial objects and the initial object in $\mathbf{Set}$ is $\emptyset$. It doesn't so it isn't a left adjoint. It happens to be a right adjoint, in symbols $F \dashv G$, as so it should take terminal objects to terminal objects which it does as the terminal object in $\mathbf{Set}$ is a singleton set. – Derek Elkins left SE Apr 10 '17 at 18:17